BUSINESS MATHEMATICS AND STATISTICS

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1 GOVERNMENT OF TAMIL NADU BUSINESS MATHEMATICS AND STATISTICS HIGHER SECONDARY FIRST YEAR VOLUME I Untouchbility is Inhumn nd Crime A publiction under Free Tetbook Progrmme of Government of Tmil Ndu Deprtment of School Eduction

2 Government of Tmil Ndu First Edition 08 NOT FOR SALE Content Cretion The wise possess ll Stte Council of Eductionl Reserch nd Trining SCERT 08 Printing & Publishing Tmil NduTetbook nd Eductionl Services Corportion (ii) II

3 HOW TO USE THE BOOK Creer Options List of Further Studies & Professions. Lerning Objectives: Lerning objectives re brief sttements tht describe wht students will be epected to lern by the end of school yer, course, unit, lesson or clss period. Additionl informtion bout the concept. Amzing fcts, Rhetoricl questions to led students to Mthemticl inquiry Eercise Assess students criticl thinking nd their understnding ICT To enhnce digitl skills mong students Web links List of digitl resources To motivte the students to further eplore the content digitlly nd tke them in to virtul world Miscellneous Problems Additionl problems for the students Glossry Tmil trnsltion of Mthemticl terms References List of relted books for further studies of the topic Lets use the QR code in the tet books! How? Downlod the QR code scnner from the Google PlyStore/ Apple App Store into your smrtphone Open the QR code scnner ppliction Once the scnner button in the ppliction is clicked, cmer opens nd then bring it closer to the QR code in the tet book. Once the cmer detects the QR code, url ppers in the screen.click the url nd goto the content pge. (iii)

4 CAREER OPTIONS IN BUSINESS MATHEMATICS Higher Secondry students who hve tken commerce with Business mthemtics cn tke up creers in BCA, B.Com., nd B.Sc. Sttistics. Students who hve tken up commerce strem, hve good future in bnking nd finncil institutions. A lot of students choose to do B.Com with speciliztion in computers. Higher Secondry Commerce students plnning for further studies cn tke up creers in professionl fields such s Compny Secretry, Chrtered Accountnt (CA), ICAI nd so on. Others cn tke up bchelor s degree in commerce (B.Com), followed by M.Com, Ph.D nd M.Phil. There re wide rnge of creer opportunities for B.Com grdutes. After grdution in commerce, one cn choose MBA, MA Economics, MA Opertionl nd Reserch Sttistics t Postgrdute level. Aprt from these, there re severl diplom, certificte nd voctionl courses which provide entry level jobs in the field of commerce. Creer chrt for Higher Secondry students who hve tken commerce with Business Mthemtics. Courses B.Com., B.B.A., B.B.M., B.C.A., B.Com (Computer), B.A. Institutions Government Arts & Science Colleges, Aided Colleges, Self finncing Colleges. Shri Rm College of Commerce (SRCC), Delhi Symbiosis Society s College of Arts & Commerce, Pune. St. Joseph s College, Bnglore Scope for further studies C.A., I.C.W.A, C.S. B.Sc Sttistics Presidency College, Chepuk, Chenni. Mdrs Christin College, Tmbrm Loyol College, Chenni. D.R.B.C.C Hindu College, Pttbirm, Chenni. M.Sc., Sttistics B.B.A., LLB, B.A., LLB, B.Com., LL.B. (Five yers integrted Course) M.A. Economics (Integrted Five Yer course) Admission bsed on All Indi Entrnce Emintion Government Lw College. School of ecellence, Affilited to Dr.Ambethkr Lw University Mdrs School of Economics, Kotturpurm, Chenni. M.L. Ph.D., B.S.W. School of Socil studies, Egmore, Chenni M.S.W (iv)

5 CONTENTS Mtrices nd Determinnts 9. Determinnts. Inverse of Mtri. Input Output Anlysis 7 Algebr 08. Prtil frctions 0. Permuttions 7. Combintions 58. Mthemticl induction 6.5 Binomil Theorem 70 Anlyticl Geometry 8. Locus 8. System of Stright Lines 86. Pir of Stright Lines 9. Circles 99.5 Conics 08 Trigonometry 56. Trigonometric rtios 6. Trigonometric rtios of Compound ngles. Trnsformtion formule 7. Inverse Trigonometric functions 5 5 Differentil Clculus Functions nd their grphs Limits nd Derivtives 7 5. Differentition techniques 89 Answers 08 Glossry 90 Books for Reference E Book Assessment DIGI Links (v)

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7 SYLLABUS. MATRICES AND DETERMINANTS (5 periods) Determinnts: Recll Types of mtrices Determinnts Minors Cofctors Properties of determinnts. Inverse of mtri: Singulr mtri Non singulr mtri Adjoint of mtri Inverse of mtri of system of liner eqution. Input Output nlysis: Hwkins Simon conditions.. ALGEBRA (0 periods) Prtil frctions: Denomintor contins non repeted Liner fctors Denomintor contins Liner fctors, repeted for n times Denomintor contins qudrtic fctor, which cnnot be fctorized into liner fctors. Permuttions: Fctoril Fundmentl principle of counting Permuttion Circulr permuttion. Combintions Mthemticl Induction Binomil Theorem.. ANALYTICAL GEOMETRY (0 periods) Locus: Eqution of locus. System of stright lines: Recll Vrious forms of stright lines Angle between two stright lines Distnce of point from line Concurrence of three lines. Pir of stright lines: Combined eqution of pir of stright lines Pir of stright lines pssing through the origin Angle between pir of stright lines pssing through the origin The condition for generl second degree eqution to represent the pir of stright lines. Circles: The eqution of circle when the centre nd rdius re given Eqution of circle when the end points of dimeter re given Generl eqution of circle Prmetric form of circle Tngents Length of tngent to the circle. CONICS: Prbol.. TRIGONOMETRY (5 periods) Trigonometric rtios: Qudrnts Signs of the trigonometric rtios of n ngle θ s θ vries from 0º to 60º Trigonometric rtios of llied ngles. Trigonometric rtios of compound ngles: Compound ngles Sum nd difference formule of sine, cosine nd tngent Trigonometric rtios of multiple ngles. Trnsformtion formule: Trnsformtion of the products into sum or difference Trnsformtion of sum or difference into product. Inverse Trigonometric Functions: Properties of Inverse Trigonometric Functions. 5. DIFFERENTIAL CALCULUS (5 periods) Functions nd their grphs: Some bsic concepts Function Vrious types of functions Grph of function. Limits nd derivtives: Eistence of limit Indeterminte forms nd evlution of limits Continuous function Differentibility t point Differentition from first principle. Differentition techniques: Some stndrd results Generl rules for differentition Successive differentition. (vi)

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9 Chpter MATRICES AND DETERMINANTS Lerning Objectives After studying this chpter, the students will be ble to understnd the definition of mtrices nd determinnts the properties of determinnts the concept of inverse mtri the concept of djoint mtri the solving simultneous liner equtions the inputoutput nlysis. Determinnts Introduction The ide of determinnt ws believed to be originted from Jpnese Mthemticin Seki Kow (68) while systemtizing the old Chinese method of solving simultneous equtions whose coefficients were represented by clculting bmboos or sticks. Lter the Germn Mthemticin Gottfried Wilhelm Von Leibnitz formlly developed determinnts. The present verticl nottion ws given in 8 by Arthur Cyley. Determinnt ws invented independently by Crmmer whose well known rule for solving simultneous equtions ws published in 750. S ek i K ow In clss X, we hve studied mtrices nd lgebr of mtrices. We hve lso lernt tht system of lgebric equtions cn be epressed in the form of mtrices. We know tht the re of tringle with vertices (, y ) (, y ) nd (, y ) is 6 ( y y) + ( y y) + ( y M t ri ces nd D et erm i n nt s

10 To minimize the difficulty in remembering this type of epression, Mthemticins developed the ide of representing the epression in determinnt form nd the bove epression cn be represented in the form y y y Thus determinnt is prticulr type of epression written in specil concise form. Note tht the quntities re rrnged in the form of squre between two verticl lines. This rrngement is clled determinnt. In this chpter, we study determinnts up to order three only with rel entries. Also we shll study vrious properties of determinnt (without proof), minors, cofctors, djoint, inverse of squre mtri nd business pplictions of determinnts. We hve studied mtrices in the previous clss. Let us recll the bsic concepts nd opertions on mtrices... Recll Mtri Definition. A mtri is rectngulr rrngement of numbers in horizontl lines (rows) nd verticl lines (columns). Numbers re enclosed in squre brckets or open brckets or pir of double brs. It is denoted by A, B, C, For emple, A G Order of mtri If mtri A hs m rows nd n columns, then A is clled mtri of order m n. For emple, If, A G then order of A is R V Generl form of Mtri ggi ggnw WWWW ggi ggn Mtri of order m n is represented s ggg ggg ggg ggg ggg ggg Sm m m ggmi ggmnw T X It is shortly written s 6 ij@ mn i,,... m; j,,... n Here ij is the element in the i th row nd j th column of the mtri. th Std. Business Mthemtics

11 Types of mtrices Row mtri A mtri hving only one row is clled row mtri. For emple A 6... i... n@ n ; B 7 A X Column mtri A mtri hving only one column is clled column mtri R V SW SW S W. For emples, A S W, B G S. W 5 S. W Sm W m T X Zero mtri (or) Null mtri If ll the elements of mtri re zero, then it is clled zero mtri. It is represented by n English lphbet O R V 0 0 For emples, O 6@ 0, O 0 0 G, O S0 0 0W T X re ll zero mtrices. NOTE Zero mtri cn be of ny order. Squre mtri If the number of rows nd number of columns of mtri re equl then it is clled squre mtri. 8 For emples, A G is squre mtri of order R V S 0 W B S 0 5 W is squre mtri of order. S S 5 W W T X Tringulr mtri A squre mtri, whose elements bove or below the min digonl (leding digonl) re ll zero is clled tringulr mtri. 0 0 For emples, A > H, B > H NOTE The sum of the digonl elements of squre mtri is clled the trce of mtri. M t ri ces nd D et erm i n nt s

12 Digonl mtri A squre mtri in which ll the elements other thn the min digonl elements re zero is clled the digonl mtri. 5 For emple, A > H is digonl mtri of order 0 0 Sclr mtri A digonl mtri with ll digonl elements re equl to K ( sclr) is clled sclr mtri. 5 For emple, A > H is sclr mtri of order Unit mtri (or) Identity mtri A sclr mtri hving ech digonl element equl to one (unity) is clled Unit mtri. For emples, I 0 0 G is unit mtri of order. R V S 0 0W I S0 0W is unit mtri of order. S0 0 W T X Multipliction of mtri by sclr If A 6 is mtri of ny order nd if k is sclr, then the sclr multipliction of A by the sclr k is defined s ka 6 kij@ for ll i, j. In other words to multiply mtri A by sclr k, multiply every element of A by k. For emple, if A 8 G then, A 8 6 G 6 Negtive of mtri The negtive of mtri A 6 ij@ m # n is defined s A 6 ij@ m# n for ll i, j nd is obtined by chnging the sign of every element For emple, if A G then A G Equlity of mtrices Two mtrices A nd B re sid to be equl if (i) they hve the sme order nd (ii) their corresponding elements re equl. th Std. Business Mthemtics

13 Addition nd subtrction of mtrices Two mtrices A nd B cn be dded, provided both the mtrices re of the sme order. Their sum A+B is obtined by dding the corresponding entries of both the mtrices A nd B. Symboliclly if A [ ij ] m n nd B [b ij ] m n, then A + B [ ij + b ij ] m n Similrly A B A + ( B) [ ij ] m n + [ b ij ] m n [ ij b ij ] m n Multipliction of mtrices Multipliction of two mtrices is possible only when the number of columns of the first mtri is equl to the number of rows of the second mtri. The product of mtrices A nd B is obtined by multiplying every row of mtri A with the corresponding elements of every column of mtri B elementwise nd dd the results. Let A 6 be n m # p mtri nd B 6 b be p # n mtri, then the product AB is mtri C 6 c of order m # n. Trnspose of mtri Let A 6 be mtri of order m # n. The trnspose of A, denoted by A T of order n # m is obtined by interchnging either rows into columns or columns into rows of A. For emple, if NOTE A 5 G then 6 A T > 5 6 It is believed tht the students might be fmilir with the bove concepts nd our present syllbus continues from the following. H Definition. To every squre mtri A of order n with entries s rel or comple numbers, we cn ssocite number clled determinnt of mtri A nd it is denoted by A or det (A) or Δ. Thus determinnt cn be formed by the elements of the mtri A. If A G then its A M t ri ces nd D et erm i n nt s 5

14 Emple. Evlute: ( )( ) ( ) 8+.. Minors: To evlute the determinnt of order or more, we define minors nd cofctors. Let A [ ij ] be determinnt of order n. The minor of n rbitrry element ij is the determinnt obtined by deleting the i th row nd j th column in which the element ij stnds. The minor of ij is denoted by M ij... Cofctors: The cofctor is signed minor. The cofctor of i j is denoted by A ij nd is defined s A ij ( ) i+ j M ij The minors nd cofctors of,, in third order determinnt re s follows (i) Minor of is M Cofctor of is A ( ) + M (ii) Minor of is M + Cofctor of is A ( ) M ( ) (iii) Minor of is M + Cofctor of is A ( ) M Emple. Find the minor nd cofctor of ll the elements in the determinnt 6 th Std. Business Mthemtics

15 Emple. Minor of M Minor of M Minor of M Minor of M + Cofctor of A ( ) M + Cofctor of A ( ) M + Cofctor of A ( ) M + Cofctor of A ( ) M Find the minor nd cofctor of ech element of the determinnt. Minor of is M 5 0 Minor of is M 5 0 Minor of is M Minor of is M 0 Minor of is M 0 Minor of 5 is M Minor of is M Minor of is M Minor of 0 is M Cofctor of is A ( ) M M 5 + Cofctor of is A ( ) M M M t ri ces nd D et erm i n nt s 7

16 + Cofctor of is A ( ) M M 6 + Cofctor of is A ( ) M M + Cofctor of is A ( ) M M 8 + Cofctor of 5 is A ( ) M M + Cofctor of is A ( ) M M Cofctor of is A ( ) + M M + Cofctor of 0 is A ( ) M M For emple If D, then NOTE Vlue of determinnt cn be obtined by using ny row or column Δ A + A + A (or) M M + M (epnding long R ) Δ A + A + A (or) M M + M (epnding long C ) Emple. Evlute: (Minor of ) (Minor of ) + (Minor of ) Properties of determinnts (without proof). The vlue of determinnt is unltered when its rows nd columns re interchnged. If ny two rows (columns) of determinnt re interchnged then the vlue of the determinnt chnges only in sign.. If the determinnt hs two identicl rows (columns) then the vlue of the determinnt is zero. 8 th Std. Business Mthemtics

17 If ll the elements in row (column) of determinnt re multiplied by constnt k, then the vlue of the determinnt is multiplied by k. 5. If ny two rows (columns) of determinnt re proportionl then the vlue of the determinnt is zero. 6. If ech element in row (column) of determinnt is epressed s the sum of two or more terms, then the determinnt cn be epressed s the sum of two or more determinnts of the sme order. 7. The vlue of the determinnt is unltered when constnt multiple of the elements of ny row (column) is dded to the corresponding elements of different row (column) in determinnt. Emple.5 Show tht + y y+ b z z+ c 0 b c y z y z y z + y+ b z+ c y z + b c b c b c b c Emple.6 Emple Evlute + Solve 0 + ( )(+) ( ) e o+ e o e o ! M t ri ces nd D et erm i n nt s 8 NOTE The vlue of the determinnt of tringulr mtri is equl to the product of the min digonl elements. 9

18 Emple & ( )( )( ) 0,, Evlute (since R / R ) Emple.9 Evlute b c b c ( b) (b c) (c ) b c b c 0 0 b b c c b b c c R " R R R " R R 0 0 b b c c ^ bh^+ bh ^b ch^b+ ch c ^ bh^b ch 0 0 c + b b+ c c ^ bh^b ch " b+ c ^+ ^ bh^b ch^c h. Eercise.. Find the minors nd cofctors of ll the elements of the following determinnts. 5 0 (i) (ii) 0 5. Evlute 0 0 th Std. Business Mthemtics

19 Solve: Find AB if A G nd B 0 G 5. Solve: Evlute: 7. Prove tht b c b c b c bc c b bc c b b+ c c+ + b 0 8. Prove tht b c b b bc c bc c bc. Inverse of mtri.. Singulr mtri : A Squre mtri A is sid to be singulr, if A 0.. Non singulr mtri: A squre mtri A is sid to be non singulr, if A! 0 Emple.0 Show tht G is singulr mtri. Let A G A 0 ` A is singulr mtri. M t ri ces nd D et erm i n nt s

20 Emple. Show tht 8 G is non singulr. Let A 8 G A 8 8 6! 0 ` A is nonsingulr mtri.. Adjoint of Mtri If A nd B re non singulr mtrices of the sme order then AB nd BA re lso non singulr mtrices of the sme order. The djoint of squre mtri A is defined s the trnspose of cofctor mtri. Adjoint of the mtri A is denoted by dj A i.e., dj A where is the cofctor mtri of A. ij T ij Emple. Find dj A for A G A G T dj A 6 A G Emple. RS SSSS Find djoint of A 0 S T A ij ( ) i+j M ij V 0 0W X NOTE (i) dja A n n is the order of the mtri A (ii) ka n k A n is the order of the mtri A (iii) n dj( ka) k dja n is the order of the mtri A (iv) A( dj A) ( dj AA ) AlI (v) AdjI II, is the unit mtri. (vi) dj (AB) (dj B)(dj A) (vii) AB A B A ( ) + M th Std. Business Mthemtics 0 0 0

21 A ( ) + 0 M A ( ) + M 0 0 ( ) A ( ) + M 0 ^ 0 ] gh A ( ) + M 0 0 A ( ) + M ] 8 g 7 A ( ) + M 0 ( ) 0 A ( ) + M A ( ) + M 0 0 [A ij ] > H T Adj A 6 A Inverse of mtri 7 0 > H L e t A be ny nonsingulr mtri of order n. If there eists squre mtri B of order n such tht A B B A I then, B is clled the inverse of A nd is denoted by A We hve A (dj A) (dj A)A ( A c A dj A m c A dj A ma I A I ( AB BA I where B A A dja A dja M t ri ces nd D et erm i n nt s

22 NOTE (i) If B is the inverse of A then, A is the inverse of B i.e., B A & A B (ii) AA A A I (iii) The inverse of mtri, if it eists, is unique. (iv) Order of inverse of A will be the sme s tht of order of A. (v) I (vi) ( AB) I, where I is the identity mtri. B A (vii) A A Let A be nonsingulr mtri, then A I + A A Emple. If A G then, find A. A G A 6 0 Since A is nonsingulr mtri, A eists Now dj A G A A dja G 6 Emple.5 6 If A G then, find A 9 th Std. Business Mthemtics

23 A Since A is singulr mtri, A does not eist. Emple.6 If A 5 5 > H then find A A ( 0) ( 6 8) + 0 ( 0) ( 5) ( ) + 0 ( ) ! 0 Since A is nonsingulr mtri, A eists. A 5; A 8; A A ; A ; A 0 A 0; A ; A 6 A > H dj A > H A A dja R 5 S 0 T V 0 W X > H M t ri ces nd D et erm i n nt s 5

24 Emple.7 If A G stisfies the eqution A ka + I O then, find k nd lso A A G A A.A. G 7 7 G G Given A ka+ I O & G k G+ G G k + k + 0 & G 0 0 G k k k k 0 & G k 8 k 0 0 G 0 & k 0 & k. Also A! 0 dj A A G A dja G G 6 th Std. Business Mthemtics

25 Emple.8 If A G, B 0 G then, show tht ( AB) B A. ` A eists. A! 0 B 0 0 +! 0 B lso eists. AB 0 G G G AB! 0 ( AB) eists. dj(ab) G ( AB) dj( AB) AB G G dj A G () A A dja G G M t ri ces nd D et erm i n nt s 7

26 dj B G 0 B B djb G 0 B A G 0 G 0 G G () From () nd (), ( AB) B A. Hence proved. Emple.9 7 R Show tht the mtricesa > H nd 5 other. B S R V 5 S 5 R V S W T S 7W S W 6 AB S 5 W S W S W S W S 6 0 W T X S W T X R V R V S 7W S 5W S W 5 S 6 5W S W S 6 0W T X T X > 0 H V W W X re inverses of ech > H > H I 8 th Std. Business Mthemtics

27 R V S 5 W S W 6 5 BA S W S W > 6 0 S W T X > H > > H 7 H 7 H > H Thus AB BA I > H I Therefore A nd B re inverses of ech other. Eercise.. Find the djoint of the mtri A G. If A > H then verify tht A ( dj A) A I nd lso find A. Find the inverse of ech of the following mtrices (i) 5 G (ii) G (iii) > 0 H (iv) If A G nd B 6 G, then verify dj (AB) (dj B)(dj A) 5. If A > 0H then, show tht (dj A)A O If A > H then, show tht the inverse of A is A itself If A > H then, find A. > H M t ri ces nd D et erm i n nt s 9

28 R V S W S W 8. Show tht the mtrices A > H nd B S W S W other. S W T X If A G nd 5 B G then, verify tht ( AB ) B A 0. Find m if the mtri 9 m 7 > H hs no inverse. re inverses of ech 8. If X > 5 Hnd Y th Std. Business Mthemtics p q..5 of system of liner equtions > H then, find p, q if Y X Consider system of n liner non homogeneous equtions with n unknowns,,... n s nn b nn b b m m mn n m This cn be written in mtri form s R V R V R V S... n W SW Sb W S... W S W S n b W S W S W S W S W S. W S. W S W S. W S. W Sm m... W S mn W S n b W m T X T X T X Thus we get the mtri eqution A X B ( ) R V R... V R S n W S W b V S W S... W S n W Sb W S W S W S W where A S W ; X S. W nd B S. W S W S. W S W Sm m... W S. W mn S T X W T n Sb X m W T X From () solution is X A B

29 Emple.0 Solve by using mtri inversion method: + 5y + y 7 The given system cn be written s 5 G G y G 7 i.e., AX B X A B where A 5 G, X y G nd B 7 G A 5 ] 0 A eists. dj A A 5 G A dja 5 G X A B 5 G G 7 G G G G y nd y. M t ri ces nd D et erm i n nt s

30 Emple. Solve by using mtri inversion method: y+ z 8; + y z ; y+ z The given system cn be written s y z 8 > H > H > H i.e., AX B X A B Here A > H, X A 7 ] 0 y z 8 > H nd B > H A eists A A 8 A 0 A 5 A 6 A A A 9 A 7 [A ij ] > H dj A [A ij ] T th Std. Business Mthemtics A A dja 9 7 > H

31 7 X A B 7 7 > yh > H z, y nd z. Emple > H > H > H 7 5 > H > H The cost of kg onion, kg whet nd kg rice is `0. The cost of kg onion, kg whet nd 6 kg rice is `560. The cost of 6 kg onion, kg whet nd kg rice is `80. Find the cost of ech item per kg by mtri inversion method. Let, y nd z be the cost of onion, whet nd rice per kg respectively. Then, + y + z 0 It cn be written s 6 6 y z + y + 6z y + z > H > H > H AX B where A > 6H, X 6 A ] 0 y z > H nd B > H M t ri ces nd D et erm i n nt s

32 A eist A 0 A 0 A 0 A 5 A 0 A 0 A 0 A 0 A 0 [A ij ] > H dj A [A ij ] T > H A A dja X A B y z > H > H > H > H > H > H > > H > H , y 0 nd z 60. H Cost of kg onion is `0, cost of kg whet is `0 nd cost of kg rice is ` th Std. Business Mthemtics

33 ICT Corner Epected finl outcomes Step Open the Browser nd type the URL given (or) Scn the QR Code. GeoGebr Work book clled th BUSINESS MATHS will pper. In this severl work sheets for Business Mths re given, Open the worksheet nmed Simultneous Equtions Step Type the prmeters for three simultneous equtions in the respective boes. You will get the solutions for the equtions. Also, you cn see the plnes for the equtions nd the intersection point() on the righthnd side. Solve yourself nd check the nswer. If the plnes do not intersect then there is no solution. St e p St e p St e p Browse in the link th Business Mths: (or) scn the QR Code M t ri ces nd D et erm i n nt s 5

34 Eercise.. Solve by mtri inversion method : + y 5 0 ; y + 0. Solve by mtri inversion method : (i) y + z ; + y z ; + y 5z 8 (ii) y + z ; + z ; + y + z. (iii) z 0 ; 5 + y ; y + z 5. A sles person Rvi hs the following record of sles for the month of Jnury, Februry nd Mrch 009 for three products A, B nd C He hs been pid commission t fied rte per unit but t vrying rtes for products A, B nd C Months Sles in Units Commission A B C Jnury Februry Mrch Find the rte of commission pyble on A, B nd C per unit sold using mtri inversion method.. The prices of three commodities A, B nd C re `, `y nd `z per unit respectively. P purchses units of C nd sells units of A nd 5 units of B. Q purchses units of B nd sells units of A nd unit of C. R purchses unit of A nd sells units of B nd 6 units of C. In the process P, Q nd R ern `6,000, `5,000 nd `,000 respectively. By using mtri inversion method, find the prices per unit of A, B nd C. 5. The sum of three numbers is 0. If we multiply the first by nd dd the second number nd subtrct the third we get. If we multiply the first by nd dd second nd third to it, we get 6. By using mtri inversion method find the numbers. 6. Weekly ependiture in n office for three weeks is given s follows. Assuming tht the slry in ll the three weeks of different ctegories of stff did not vry, clculte the slry for ech type of stff, using mtri inversion method. week Number of employees Totl weekly Slry A B C (in rupees) st week 900 nd week 500 rd week th Std. Business Mthemtics

35 Input output Anlysis Input Output nlysis is technique which ws invented by Prof. Wssily W.Leontief. Input Output nlysis is form of economic nlysis bsed on the interdependencies between economic sectors. The method is most commonly used for estimting the impcts of positive or negtive economic shocks nd nlyzing the ripple effects throughout n economy. The foundtion of Input Output nlysis involves input output tbles. Such tbles include series of rows nd columns of dt tht quntify the supply chin for sectors of the economy. Industries re listed in the heds of ech row nd ech column. The dt in ech column corresponds to the level of inputs used in tht industry s production function. For emple the column for uto mnufcturing shows the resources required for building utomobiles (ie., requirement of steel, luminum, plstic, electronic etc.,). Input Output models typiclly includes seprte tbles showing the mount of lbour required per rupee unit of investment or production. Consider simple economic model consisting of two industries A nd A where ech produces only one type of product. Assume tht ech industry consumes prt of its own output nd rest from the other industry for its opertion. The industries re thus interdependent. Further ssume tht whtever is produced tht is consumed. Tht is the totl output of ech industry must be such s to meet its own demnd, the demnd of the other industry nd the eternl demnd (finl demnd). Our im is to determine the output levels of ech of the two industries in order to meet chnge in finl demnd, bsed on knowledge of the current outputs of the two industries, of course under the ssumption tht the structure of the economy does not chnge. Let ij be the rupee vlue of the output of A i consumed by A j, i, j, Let nd be the rupee vlue of the current outputs of A nd A respectively. Let d nd d be the rupee vlue of the finl demnds for the outputs of A nd A respectively. The ssumptions led us to frme the two equtions + + d ; + + d... () ij Let b ij i, j, j b ; b ; b ; b M t ri ces nd D et erm i n nt s 7

36 The equtions () tke the form b + b + d b + b + d The bove equtions cn be rerrnged s ^ bh b d b+ ^ bh d The mtri form of the bove equtions is b b d G G G b b d 0 b b d ) e o e o G G 0 b b d ( B)X D b b where B G, I b b 0 0 G X nd D d G G d By solving we get X(I B) D The mtri B is known s the Technology mtri... The Hwkins Simon conditions Hwkins Simon conditions ensure the vibility of the system. If B is the technology mtri then Hwkins Simon conditions re Emple. (i) the min digonl elements in I B must be positive nd (ii) I B must be positive. The technology mtri of n economic system of two industries is G. Test whether the system is vible s per Hwkins Simon conditions th Std. Business Mthemtics

37 B G Emple. I B G G G I B (0.)(0.) ( 0.)( 0.9) < 0 Since I B is negtive, Hwkins Simon conditions re not stisfied. Therefore, the given system is not vible. The following inter industry trnsctions tble ws constructed for n economy of the yer 06. Industry Finl consumption Totl output 500,600 00,500,750,600,650 8,000 Lbours 50,800 Construct technology coefficient mtri showing direct requirements. Does solution eist for this system b ; b M t ri ces nd D et erm i n nt s 9

38 b ; b The technology mtri is G I B 0 0 G G G, elements of min digonl re positive Now, I B (0.8)(0.8) ( 0.7)( 0.) >0 Since digonl elements of re positive nd I B is positive, Hwkins Simon conditions re stisfied Hence this system hs solution. Emple.5 In n economy there re two industries P nd P nd the following tble gives the supply nd the demnd position in crores of rupees. Production sector Consumption sector P P Finl demnd Gross output P P Determine the outputs when the finl demnd chnges to 5 for P nd for P b b ; b ; b 0 th Std. Business Mthemtics

39 R V S 5 The technology mtri is B 5 W S W S 5 W T XR 0 V S 5 I B 0 G 5 W S W S 5 W R V S T X 5 W S 5 W, elements of min digonl re positive S 5 W T X 5 Now, I B Min digonl elements of I B re positive nd I B is positive The problem hs solution R V S 5 dj (I B) W S W S 5 5 W T X (I B) ( )( Since 0,( ) eist) I B dj I B I B! I B dj( I B) 0 7 R V S W S W 5 S 7 > W H T X X (I B) D, where D 5 G > H G G 0 The output of industry P should be `50 crores nd P should be `0 crores. Emple.6 An economy produces only col nd steel. These two commodities serve s intermedite inputs in ech other s production. 0. tonne of steel nd 0.7 tonne of col re needed to produce tonne of steel. Similrly 0. tonne of steel nd 0.6 tonne of col re required to produce tonne of col. No cpitl inputs re needed. Do you think tht M t ri ces nd D et erm i n nt s

40 the system is vible? nd 5 lbour dys re required to produce tonne s of col nd steel respectively. If economy needs 00 tonnes of col nd 50 tonnes of steel, clculte the gross output of the two commodities nd the totl lbour dys required. Here the technology mtri is given under Steel Col Finl demnd Steel Col Lbour dys The technology mtri is B G I B 0 0 G G G I B (0.6)(0.) ( 0.7)( 0.) Since the digonl elements of I B re positive nd vlue of I B is positive, the system is vible dj (I B) G (I B) ( ) I B dj I B G X (I B) D, where D 50 G G G G G th Std. Business Mthemtics

41 Steel output 76.5 tonnes Col output tonnes Totl lbour dys required 5(steel output) + ( col output) 5(76.5) + (558.8) lbour dys. Eercise The technology mtri of n economic system of two industries is G Test whether the system is vible s per Hwkins Simon conditions The technology mtri of n economic system of two industries is G. Test whether the system is vible s per HwkinsSimon conditions The technology mtri of n economic system of two industries is G. Test whether the system is vible s per HwkinsSimon conditions.. Two commodities A nd B re produced such tht 0. tonne of A nd 0.7 tonne of B re required to produce tonne of A. Similrly 0. tonne of A nd 0.7 tonne of B re needed to produce tonne of B. Write down the technology mtri. If 6.8 tonnes of A nd 0. tonnes of B re required, find the gross production of both of them. 5. Suppose the interindustry flow of the product of two industries re given s under. Production sector X Consumption sector Y Domestic demnd Totl output X Y Determine the technology mtri nd test Hwkin s Simon conditions for the vibility of the system. If the domestic demnd chnges to 80 nd 0 units respectively, wht should be the gross output of ech sector in order to meet the new demnds. M t ri ces nd D et erm i n nt s

42 6. You re given the following trnsction mtri for two sector economy. Sector Sles Finl demnd Gross output 0 5 (i) Write the technology mtri (ii) Determine the output when the finl demnd for the output sector lone increses to units. 7. Suppose the interindustry flow of the product of two sectors X nd Y re given s under. Production sector X Consumption Sector Y Domestic demnd Gross output X Y Find the gross output when the domestic demnd chnges to for X nd 8 for Y. Choose the correct nswer Eercise The vlue of if 0 is () 0, (b) 0, (c), (d),. The vlue of + y y+ z z+ y z y z is () y z (b) + y + z (c) + y + z (d) 0. The cofctor of 7 in the determinnt is () 8 (b) 8 (c) 7 (d) 7 th Std. Business Mthemtics

43 If D then is () (b) (c) (d) 5. The vlue of the determinnt b 0 () bc (b) 0 (c) b c (d) bc 6. If A is squre mtri of order then ka is 0 0 c is () k A (b) k A (c) k A 7. dj] ABg is equl to (d) k A T T () dja djb (b) dja djb (c) djb dja T T (d) djb dja 8. The inverse mtri of f p is 7 5 () 0 f 5 5 p (b) f 5 5 p (c) f 5 5 p (d) f 5 5 p 9. If A e b c d o such tht d bc! 0 then A is () d b e o d b d bc (b) e o c d bc c d b (c) e o d b d bc (d) e o c d bc c 0. The number of Hwkins Simon conditions for the vibility of n inputoutput nlysis is () (b) (c) (d). The inventor of input output nlysis is () Sir Frncis Glton (c) Prof. Wssily W. Leontief (b) Fisher (d) Arthur Cyly M t ri ces nd D et erm i n nt s 5

44 Which of the following mtri hs no inverse () (c) e o (b) cos sin e o (d) sin cos e e o sin cos sin o cos. The inverse mtri of e o is 5 () e o (b) 5 (c) e o (d) 5 e e 5 o 5 o. If A e o then A^djAh is () e o (b) e o (c) e 0 0 o (d) 0 e o 0 5. If A nd B nonsingulr mtri then, which of the following is incorrect? () A I implies A A (b) I I (c) If AX B then X B A (d) If A is squre mtri of order then dj A A 6. The vlue of 5 5 y 5 z is y z () 5 (b) (c) 0 (d) 7. If A is n invertible mtri of order then det^a h be equl to () de ( ta ) (b) det ] Ag (c) (d) 0 8. If A is mtri nd A then A is equl to () (b) 6 (c) (d) 9. If A is squre mtri of order nd A then dja is equl to () 8 (b) 7 (c) (d) 9 6 th Std. Business Mthemtics

45 0. The vlue of y y z yz z y is () (b) 0 (c) (d) yz. If A cos i sin i G, then A is equl to sin i cos i (). If cos i (b) (c) (d) D nd A ij is cofctor of ij, then vlue of D is given by () A + A + A (b) A + A + A (c) A + A + A (d) A + A + A. If (). If then the vlue of is (b) 6 5 then the vlue of is (c) 5 6 (d) 6 5 () 5 (b) 5 (c) 5 (d) 0 5. If ny three rows or columns of determinnt re identicl then the vlue of the determinnt is () 0 (b) (c) (d) Miscellneous Problems. Solve: 5 0. Evlute Without ctul epnsion show tht the vlue of the determinnt 0. Show tht b c b 0 bc c bc 0 bc is zero. M t ri ces nd D et erm i n nt s 7

46 R V 5. If A 7 verify tht A^djAh ^ djah] Ag A I. S W T X R V 6. If A S W then, find the Inverse of A. S 5 W T X 7. If A > H show tht A A+ 5I O nd lso find A 8. Solve by using mtri inversion method : y+ z, y 0, y z. 9. The cost of Kg of Whet nd Kg of Sugr is `70. The cost of Kg of Whet nd Kg of Rice is `70. The cost of Kg of Whet, Kg of Sugr nd Kg of rice is `70. Find the cost of per kg ech item using mtri inversion method. 0. The dt re bout n economy of two industries A nd B. The vlues re in crores of rupees. Producer A User B Finl demnd Totl output A B Find the output when the finl demnd chnges to 00 for A nd 600 for B Summry Determinnt is number ssocited to squre mtri. If A then its A Minor of n rbitrry element ij of the determinnt of the mtri is the determinnt obtined by deleting the i th row nd j th column in which the element ij stnds. The minor of ij is denoted by M ij. The cofctor is signed minor. The cofctor of i j is denoted by A ij nd is defined s i+ j A ( ) M. ij Let ij, then Δ A + A + A or M M + M 8 th Std. Business Mthemtics

47 dj A A ij T 7 A where [A ij ] is the cofctor mtri of the given mtri. dja A n, where n is the order of the mtri A. A^djAh ^ djah] Ag A I. dji I, I is the unit Mtri. dj(ab) (dj B)(dj A). A squre mtri A is sid to be singulr, if A 0. A squre mtri A is sid to be nonsingulr if A! 0 Let A be ny squre mtri of order n nd I be the identity mtri of order n. If there eists squre mtri B of order n such tht AB BA I then, B is clled the inverse of A nd is denoted by A Inverse of A ( if it eists ) is A A dja Hwkins Simon conditions ensure the vibility of the system. If B is the technology mtri then HwkinsSimon conditions re (i) the min digonl elements in I (ii) I B must be positive.. B must be positive nd Adjoint Mtri Anlysis Cofctor Determinnt Digonl Mtri Input Inverse Mtri Minors NonSingulr Mtri Output Sclr Singulr Mtri Trnspose of Mtri Tringulr Mtri GLOSSARY ச ர ப ப அண பக ப ப ய வ இமணக க ரண அண க சக மவ ம ம வ ட ட அண உள ள ட சநர ற அண றறண க சக மவகள ப ச ய றறக சக மவ அண வவள ய ட த ம ய ல ப ச யக சக மவ அண ந மர ந ரல றற அண ம க சக ண அண M t ri ces nd D et erm i n nt s 9

48 Chpter ALGEBRA Lerning Objectives After studying this chpter, the students will be ble to understnd definition of prtil frctions the vrious techniques to resolve into prtil frction different cses of permuttion such s word formtion different cses of combintion such s word formtion difference between permuttion nd combintion the concept nd principle of mthemticl induction Binomil epnsion techniques Introduction Algebr is mjor component of mthemtics tht is used to unify mthemtics concepts. Algebr is built on eperiences with numbers nd opertions long with geometry nd dt nlysis. The word lgebr is derived from the Arbic word ljbr. The Arbic mthemticin AlKhwrizmi hs trditionlly been known s the Fther of lgebr. Algebr is used to find perimeter, re, volume of ny plne figure nd solid.. Prtil Frctions A l K h w r iz m i Rtionl Epression p] g An epression of the form q ] g rtionl lgebric epression. n n g + m m b + b + g + b 0 n m,[q() 0] is clled If the degree of the numertor p () is less thn tht of the denomintor q () then p] g q] g is clled proper frction. If the degree of p () is greter thn or equl to tht of p] g q () then q ] g is clled n improper frction. 0 th Std. Business Mthemtics

49 An improper frction cn be epressed s the sum of n integrl function nd proper frction. Tht is, For emple, Prtil Frctions p] g q] g f r] g ] g + q] g + + We cn epress the sum or difference of two frctions nd frction. i.e., ] g] g 5 8 ] g] g + s single Process of writing single frction s sum or difference of two or more simpler frctions is clled splitting up into prtil frctions. Generlly if p () nd q () re two rtionl integrl lgebric functions of nd p] g the frction q ] g be epressed s the lgebric sum (or difference) of simpler frctions p] g ccording to certin specified rules, then the frction q ] g is sid to be resolved into prtil frctions... Denomintor contins nonrepeted Liner fctors p] g In the rtionl frction q ] g, if q ] g is the product of nonrepeted liner fctors p] g of the form ] + bg] c+ dg, then q ] g cn be resolved into prtil frction of the form A B + b + c + d, where A nd B re constnts to be determined. + 7 For emple ] g] g ] A g + ] B g, the vlues of A nd B re to be determined. Emple. Find the vlues of A nd B if A B + ^ h + Let ^ h A + + B Multiplying both sides by ] g] + g, we get A l g eb r

50 Put in () we get, A] g A] + g+ B] g... () ` A Put in () we get, A(0)+B( ) ` B Emple. Resolve into prtil frctions : Write the denomintor into the product of liner fctors. Here 5+ 6 ] g] g 7 Let ] A g + ] B g () Multiplying both the sides of () by ()(), we get 7 A] g+ B] g..() Put in (),we get B () & B 0 Put in (),we get A] g & A Substituting the vlues of A nd B in (), we get ] g + ] g th Std. Business Mthemtics

51 ] 7 g ] 7 g < F ] + < F g ] g ] g ] g 0 ] g + ] g Emple. Resolve into prtil frction : + ^ h] + g Write the denomintor into the product of liner fctors Here ^ h] + g ] g] + g] + g + ` ^ h] + g ] A g + ] + B g + ] C + g Multiplying both the sides by ( )(+)(+), we get...() + A] + g] + g+ B] g] + g+ C] g] + g... () Put in (), we get + A] 0g + B] g] g+ C] 0g ` B Put in (), we get + A] g] g+ B] 0g+ C] 0g ` A Put in (), we get + A] 0g+ B] 0g+ C] g] g A l g eb r

52 ` C Substituting the vlues of A, B nd C in (), we get + ^ h] + g ] g + ( + ) +.. Denomintor contins Liner fctors, repeted n times p () In the rtionl frction q (), if q () is of the form ] + bg n [the liner fctor is p () the product of (+ b), n times], then q () cn be resolved into prtil frction of the form. A A A An n + b g + ] g ] + bg ] + bg ] + bg For emple, ] + g] + g A B C ] + g + ] + g + ] + g Emple. Find the vlues of A, B nd C if ] g] + g ] g] + g A B C ] + g A B C ] + g A] + g + B] g] + g+ C] g... () Put in () A] + g + B] 0g+ C] 0g ` A Put in () A] 0g+ B] 0g+ C] g ` C Equting the constnt term on both sides of (),we get AB C 0 & B A C ` B th Std. Business Mthemtics

53 Emple.5 + Resolve into prtil frction ] g ] + g + Let ] g ] + g A B ] g + C + ] g ] + g Multiplying both the sides by ] g ] + g, we get... () + A] g] + g+ B] + g+ C] g () Put in (), we hve + A] 0g+ B] 5g+ C] 0g ` B Put 5 in (), we hve + A] 0g+ B] 0g+ C] g ` C 5 Equting the coefficient of on both the sides of (), we get 0 A+ C Emple.6 A ` A C Substituting the vlues of A, B nd C in (), + ] g ] + g Resolve into prtil frction 9 ] g] + g 5 5] g + 5] g 5] + g. 9 ] g] + g. Multiplying both sides by ] g] + g A B C ] g + ] + g + ] + g () ` 9 A] + g + B] g] + g+ C] g () Put in () A l g eb r 5

54 9 A] 0g+ B] 0g+ C] g ` C Put in () 9 A] g + B] 0g+ C] 0g ` A Equting the coefficient of on both the sides of (), we get 0 A+ B B A ` B Substituting the vlues of A, B nd C in (), we get 9 ] g] + g + ] + g.. Denomintor contins qudrtic fctor, which cnnot be fctorized into liner fctors p] g In the rtionl frction q] g, if q ] g is the qudrtic fctor + b + c nd cnnot p] g be fctorized into liner fctors, then q ] g cn be resolved into prtil frction of the A + B form. + b + c Emple.7 Resolve into prtil frctions: + ] g^ + h Here + cnnot be fctorized into liner fctors. + Let ] g^ + h A B c Multiplying both sides by ] g^ + h... () + A^ + h + ] B + Cg] g... () Put in () 6 th Std. Business Mthemtics

55 + A] + g+ 0 ` A Put 0 in () 0+ A] 0+ g + ] 0+ Cg] g A C ` C Equting the coefficient of on both the sides of (), we get A + B 0 B ` B A Substituting the vlues of A, B nd C in (), we get + ] g^ + h ] g ^ + h Eercise. Resolve into prtil frctions for the following: ] g] + g. ] g] + g ] + g] g ] g ] + 9. g ] + g^ + h ] g] + g 0. ^ + h] + g. Permuttions.. Fctoril For ny nturl number n, n fctoril is the product of the first n nturl numbers nd is denoted by n! or n. For ny nturl number n, n! n] ng] ng # # A l g eb r 7

56 For emples, 5! 5#### 0! ###! ##6! #! NOTE 0! For ny nturl number n n! n] ng] ng # # n] n g! n] ng] ng! In prticulr, 8! 8 (7 6 5 ) 8 7! Emple.8 Evlute the following: 7! (i) 6! (ii) 5 8!! (iii)! 6 9!! (i) 7! 7# 6 6! 6!! 7. (ii) 5 8!! 8# 7# 6# 5! 5! 6. (iii)! 6 9!! 9# 8# 7# 6! 6! #! 9# 8# 7 # # 8. Emple.9 Rewrite 7! in terms of 5! 7! 7 6! 7 6 5! 5! 8 th Std. Business Mthemtics

57 Emple.0 Find n, if n 9! + 0!! 9! + 0! n! 9! + 0 # 9! n! 9! : + 0 D n! 9! # 0 n!! n 9! # # 0! # 0! # 0! # 0!... Fundmentl principle of counting Multipliction principle of counting: Consider the following sitution in n uditorium which hs three entrnce doors nd two eit doors. Our objective is to find the number of wys person cn enter the hll nd then come out. Assume tht, P, P nd P re the three entrnce doors nd S nd S re the two eit doors. A person cn enter the hll through ny one of the doors P, P or P in wys. After entering the hll, the person cn come out through ny of the two eit doors S or S in wys. S P P S S P S S P S P S P S Hence the totl number of wys of entering the hll nd coming out is P S P S # 6 wys. S P S These possibilities re eplined in the given flow chrt (Fig..). Fig.. The bove problem cn be solved by pplying multipliction principle of counting. A l g eb r 9

58 Emple. Definition. There re two jobs. Of which one job cn be completed in m wys, when it hs completed in ny one of these m wys, second job cn be completed in n wys, then the two jobs in succession cn be completed in m # n wys. This is clled multipliction principle of counting. Find the number of letter words, with or without mening, which cn be formed out of the letters of the word NOTE, where the repetition of the letters is not llowed. Let us llot four vcnt plces (boes) to represent the four letters. N,O,T,E of the given word NOTE. 50 th Std. Business Mthemtics Fig :. The first plce cn be filled by ny one of the letters in wys. Since repetition is not llowed, the number of wys of filling the second vcnt plce by ny of the remining letters in wys. In the similr wy the third bo cn be filled in wys nd tht of lst bo cn be filled in wy. Hence by multipliction principle of counting, totl number of words formed is # # # words. Emple. If ech objective type questions hving choices, then find the totl number of wys of nswering the questions. Since ech question cn be nswered in wys, the totl number of wys of nswering questions # # # 56 wys. Emple. How mny digits numbers cn be formed if the repetition of digits is not llowed? Let us llot boes to represent the digits of the digit number

59 00 s plce 0 s plce unit s plce Fig :. Three digit number never begins with 0. ` 00 s plce cn be filled with ny one of the digits from to 9 in 9 wys. Since repetition is not llowed, 0 s plce cn be filled by the remining 9 numbers (including 0) in 9 wys nd the unit plce cn be filled by the remining 8 numbers in 8 wys. Then by multipliction principle of counting, number of digit numbers 9#9# Addition principle of counting Let us consider the following sitution in clss, there re 0 boys nd 8 girls. The clss techer wnts to select either boy or girl to represent the clss in function. Our objective is to find the number of wys tht the techer cn select the student. Here the techer hs to perform either of the following two jobs. One student cn be selected in 0 wys mong 0 boys. One student cn be selected in 8 wys mong 8 girls Hence the jobs cn be performed in wys This problem cn be solved by pplying Addition principle of counting. Emple. Definition. If there re two jobs, ech of which cn be performed independently in m nd n wys respectively, then either of the two jobs cn be performed in (m+n) wys. This is clled ddition principle of counting. There re 6 books on commerce nd 5 books on ccountncy in book shop. In how mny wys cn student purchse either book on commerce or book on ccountncy? Out of 6 commerce books, book cn be purchsed in 6 wys. A l g eb r 5

60 Out of 5 ccountncy books, book cn be purchsed in 5 wys. Then by ddition principle counting the totl number of wys of purchsing ny one book is 5+6 wys. Eercise.. Find if. Evlute 6! + 7! 8!. n! r!( n r)! when n 5 nd r.. If (n+)! 60[(n )!], find n. How mny five digits telephone numbers cn be constructed using the digits 0 to 9 if ech number strts with 67 with no digit ppers more thn once? 5. How mny numbers lesser thn 000 cn be formed using the digits 5, 6, 7, 8 nd 9 if no digit is repeted?.. Permuttion Suppose we hve fruit sld with combintion of APPLES, GRAPES & BANANAS. We don t cre wht order the fruits re in. They could lso be bnns, grpes nd pples or grpes, pples nd bnns. Here the order of miing is not importnt. Any how, we will hve the sme fruit sld. But, consider number lock with the number code 95. If we mke more thn three ttempts wrongly, then it locked permnently. In tht sitution we do cre bout the order 95. The lock will not work if we input 59 or 95. The number lock will work if it is ectly 95. We hve so mny such situtions in our prcticl life. So we should know the order of rrngement, clled Permuttion. Definition. The number of rrngements tht cn be mde out of n things tking r t time is clled the number of permuttion of n things tking r t time. For emple, the number of three digit numbers formed using the digits,, tking ll t time is 6 6 three digit numbers re,,,,, 5 th Std. Business Mthemtics

61 Nottions: If n nd r re two positive integers such tht 0 < r < n, then the number of ll permuttions of n distinct things tken r t time is denoted by P(n,r ) (or) npr We hve the following theorem without proof. Theorem : npr Emple.5 Evlute : 5P nd P(8, 5) n! ( n r)! 5P 5! ] 5 g! 5!! 60. P(8, 5) 8P 5 8! ] 8 5g! 8!! !! 670 Results (i) 0! (ii) np 0! ] n n 0g! n n!! (iii) np!! n n n] n g ] g! ] n g! (iv) np n n! n ] n ng! 0!! n! n (v) np r n(n ) (n )... [n (r )] Emple.6 Evlute: (i) 8P (ii) 5P (i) 8P 8# 7# 6 6. (ii) 5P 5# # # 0. A l g eb r 5

62 Emple.7 In how mny wys 7 pictures cn be hung from 5 picture nils on wll? The number of wys in which 7 pictures cn be hung from 5 picture nils on wll 7! is nothing but the number of permuttion of 7 things tken 5 t time 7P 5 ] 7 5g! 50 wys. Emple.8 Find how mny four letter words cn be formed from the letters of the word LOGARITHMS ( words re with or without menings) There re 0 letters in the word LOGARITHMS Therefore n 0 Since we hve to find four letter words, r Hence the required number of four letter words np r 0 p Emple.9 If np r 60, find n nd r. np r Therefore n 6 nd r. Permuttion of repeted things: 6 5 6P The number of permuttion of n different things tken r t time, when repetition is llowed is n r. 5 th Std. Business Mthemtics

63 Emple.0 Using 9 digits from,,, 9, tking digits t time, how mny digits numbers cn be formed when repetition is llowed? Here, n 9 nd r ` Number of digit numbers n r 9 79 numbers Permuttions when ll the objects re not distinct: The number of permuttions of n things tken ll t time, of which p things re of n! one kind nd q things re of nother kind, such tht p + q n is pq!!. In generl, the number of permuttion of n objects tken ll t time, p re of one kind, p re of nother kind, p re of third kind, p k re of k th kind such tht p n! + p + +p k n is. p! p!... p! Emple. k How mny distinct words cn be formed using ll the letters of the following words. (i) MISSISSIPPI (ii) MATHEMATICS. (i) There re letters in the word MISSISSIPPI In this word M occurs once I S occurs times occurs times P occurs twice. Therefore required number of permuttion (ii) In the word MATHEMATICS!!!! There re letters Here M occurs twice A l g eb r 55

64 T occurs twice A occurs twice nd the rest re ll different. Therefore number of permuttions!!!!..5 Circulr permuttion In the lst section, we hve studied permuttion of n different things tken ll together is n!. Ech permuttion is different rrngement of n things in row or on stright line. These re clled Liner permuttion. Now we consider the permuttion of n things long circle, clled circulr permuttion. Consider the four letters A, B, C, D nd the number of row rrngement of these letters cn be done in! Wys. Of those! rrngements, the rrngements ABCD, BCDA, CDAB, DABC re one nd the sme when they re rrnged long circle. C D A B D C( g, f) B A C( g, f) C B C( g, f) D C C( g, f) A A B C D Fig..! So, the number of permuttion of things long circle is! In generl, circulr permuttion of n different things tken ll t the time (n )! NOTE If clock wise nd nticlockwise circulr permuttions re considered to be sme (identicl), the number of circulr permuttion of n objects tken ll t time ] n g! is Emple. In how mny wys 8 students cn be rrnged (i) in line (ii) long circle 56 th Std. Business Mthemtics

65 (i) Number of permuttions of 8 students long line 8P 8 8! (ii) When the students re rrnged long circle, then the number of permuttion is (8 )! 7! Emple. In how mny wys 0 identicl keys cn be rrnged in ring? Since keys re identicl, both clock wise nd nticlockwise circulr permuttion re sme. The number of permuttion is ] n g! ] 0 g! 9! Emple. Find the rnk of the word RANK in dictionry. s Here mimum number of the word in dictionry formed by using the letters of the word RANK is! The letters of the word RANK in lphbeticl order re A, K, N, R Number of words strting with A! 6 Number of words strting with K! 6 Number of words strting with N! 6 Number of words strting with RAK! Number of words strting with RANK 0! ` Rnk of the word RANK is Rnk of word in dictionry The rnk of word in dictionry is to rrnge the words in lphbeticl order. In this dictionry the word my or my not be meningful. A l g eb r 57

66 Eercise.. If np (np ), find n.. In how mny wys 5 boys nd girls cn be seted in row, so tht no two girls re together?. How mny 6 digit telephone numbers cn be constructed with the digits 0,,,,,5,6,7,8,9 if ech numbers strts with 5 nd no digit pper more thn once?. Find the number of rrngements tht cn be mde out of the letters of the word ASSASSINATION 5. () In how mny wys cn 8 identicl beds be strung on necklce? (b) In how mny wys cn 8 boys form ring? 6. Find the rnk of the word CHAT in dictionry.. Combintions A combintion is selection of items from collection such tht the order of selection does not mtter. Tht is the ct of combining the elements irrespective of their order. Let us suppose tht in n interview two office ssistnts re to be selected out of persons nmely A, B, C nd D. The selection my be of AB, AC, AD, BC, BD, CD (we cnnot write BA, since AB nd BA re of the sme selection). Number of wys of selecting persons out of persons is 6.It is represented s C 6. The process of different selection without repetition is clled Combintion. Definition. Combintion is the selection of n things tken r t time without repetition. Out of n things, r things cn be selected in nc r wys. nc r n! r! ] n r g, 0 < r < n! Here n 0 but r my be 0. For emple, out of 5 blls blls cn be selected in 5C wys. 5C 5!! ] 5 g!! 5 5!! b l 0 wys. 58 th Std. Business Mthemtics

67 Emple.5 Find 8C 8 7 8C 8 Permuttions re for lists (order mtters) nd combintions re for groups (order doesn t mtter) Properties (i) nc 0 nc n. (ii) nc n. (iii) nc n] n g! (iv) nc nc y, then either y or + y n. (v) nc r nc n r. (vi) nc r + nc r (n + )C r. (vii) nc r np r r! Emple.6 If nc nc 6, find C n. If nc ncy, then + y n. Here nc nc 6 ` n C n C 0 Emple.7 C 66 If np r 70 ; nc r 0, find r A l g eb r 59

68 We know tht nc r Emple.8 0 r! ( r. If 5C r 5C r+, find r 5C r 5C r+ npr r! 70 r! ! Then by the property, nc nc y ( + y n, we hve Emple.9 60 th Std. Business Mthemtics r + r + 5 ( r. From clss of students, students re to be chosen for competition. In how mny wys cn this be done? The number of combintion C Emple.0!! ] g!!! ] 8 g.! A question pper hs two prts nmely Prt A nd Prt B. Ech prt contins 0 questions. If the student hs to choose 8 from prt A nd 5 from prt B, in how mny wys cn he choose the questions? In prt A, out of 0 questions 8 cn be selected in 0C 8 wys. In prt B out of 0 questions 5 cn be selected in 0C 5. Therefore by multipliction principle the totl number of selection is

69 0C 8 0C 5 0C 0C Emple. A Cricket tem of plyers is to be formed from 6 plyers including bowlers nd wicketkeepers. In how mny different wys cn tem be formed so tht the tem contins t lest bowlers nd t lest one wicketkeeper? A Cricket tem of plyers cn be formed in the following wys: (i) bowlers, wicket keeper nd 7 other plyers cn be selected in C C 0C7 wys C C 0C7 C C 0C 960 wys (ii) bowlers, wicket keepers nd 6 other plyers cn be selected in C C 0C6 wys C C 0C6 C C 0C 80 wys (iii) bowlers nd wicket keeper nd 6 other plyers C # C # 0C C # C # 0C 0 wys 6 bowlers, wicket keepers nd 5 other plyers cn be selected in C # C # 0C wys 5 C # C # 0C 5 wys 5 By ddition principle of counting. Totl number of wys A l g eb r 6

70 Emple. If ] ncg ] n + g C, find n ( nc ) ] n + g C nn ( ) # ] n+ g] n+ g] ng # # ] n g (n+) (n+) ] n g ^n + n+ h n 9n+ 0 ] ng] n7g 0 & n, n 7 Emple. If ] n + g Cn 5, find n ] n + g Cn 5 + n+ n ] n g C 5 ] n + g C 5 ( n + ) ] n + g 5 n + n 88 0 ] n+ g] n8g 0 n is not possible n, 8 ` n 8 6 th Std. Business Mthemtics

71 ICT Corner Epected finl outcomes Step Open the Browser nd type the URL given (or) Scn the QR Code. GeoGebr Work book clled th BUSINESS MATHS will pper. In this severl work sheets for Business Mths re given, Open the worksheet nmed Combintion Eercise Step Combintion prctice eercise will open. You cn generte s mny questions you like by clicking New question Solve the problem yourself nd check your nswer by clicking on the Show solution bo. Also click Short method nd follow this method. St e p St e p St e p B r o w s e i n t h e l i n k th Business Mths: (or) scn the QR Code A l g eb r 6

72 Eercise.. If np r 680 nd nc r 70, find n nd r.. Verify tht 8C + 8C 9C.. How mny chords cn be drwn through points on circle?. How mny tringles cn be formed by joining the vertices of hegon? 5. Out of 7 consonnts nd vowels, how mny words of consonnts nd vowels cn be formed? 6. If four dice re rolled, find the number of possible outcomes in which tlest one die shows. 7. There re 8 guests t dinner prty. They hve to sit 9 guests on either side of long tble, three prticulr persons decide to sit on one prticulr side nd two others on the other side. In how mny wys cn the guests to be seted? 8. If polygon hs digonls, find the number of its sides. 9. In how mny wys cn cricket tem of plyers be chosen out of btch of 5 plyers? (i) There is no restriction on the selection. (ii) A prticulr plyer is lwys chosen. (iii) A prticulr plyer is never chosen. 0. A Committee of 5 is to be formed out of 6 gents nd ldies. In how mny wys this cn be done when (i) tlest two ldies re included (ii) tmost two ldies re included. Mthemticl induction Mthemticl induction is one of the techniques which cn be used to prove vriety of mthemticl sttements which re formulted in terms of n, where n is positive integer. 6 th Std. Business Mthemtics

73 Mthemticl Induction is used in the brnches of Algebr, Geometry nd Anlysis where it turns out to be necessry to prove some truths of propositions. The principle of mthemticl induction: Emple. Let P(n) be given sttement for n! N. (i) Initil step: Let the sttement is true for n i.e., P() is true nd (ii) Inductive step: If the sttement is true for n k (where k is prticulr but rbitrry nturl number) then the sttement is true for n k +. i.e., truth of P(k) implies the truth of P(k+). Then P(n) is true for ll nturl numbers n. Using mthemticl induction method, Prove tht n n ] n + g, n! N. Let the given sttement P(n) be defined s +++.n n ] n + g for n! N. Step : Put n LHS P() RHS ] + g P() LHS RHS for n ` P() is true Step : Let us ssume tht the sttement is true for n k. i.e., P(k) is true k] k + g k is true Step : To prove tht P(k + ) is true P(k + ) k + (k +) P(k) + k + A l g eb r 65

74 ` P(k +) is true k ] k + g + k + k ] k+ g+ ] k+ g ] k+ g] k+ g Thus if P(k) is true, then P(k +) is lso true. ` P(n) is true for ll n! N n] n + g Hence n, n! N Emple.5 for ll n By the principle of Mthemticl Induction, prove tht (n ) n,! N. Let P(n) denote the sttement (n ) n Put n ` P() is true. Assume tht P(k) is true. LHS RHS LHS RHS i.e., (k ) k To prove: P(k + ) is true. ` (k ) + (k+) P] kg + ] k + g k + k+ ] k + g 66 th Std. Business Mthemtics

75 P(k + ) is true whenever P(k) is true. ` P(n) is true for ll n Emple.6! N. By Mthemticl Induction, prove tht n n ] n+ g] n+ g 6, for ll n! N. Let P(n) denote the sttement: n Put n ` LHS n] n+ g] n+ g 6 ] + g] + g RHS 6 LHS RHS ` P() is true. Assume tht P(k) is true. P(k) k Now, k] k+ g] k+ g k + (k+) p] kg + ] k + g k] k+ g] k+ g k 6 + ] + g k] k+ g] k+ g+ 6] k + g 6 ] k + g6 k] k+ g+ 6] k+ g@ 6 ] k + g^k + 7k+ 6h 6 ] k+ g] k+ g] k + g 6 ` P(k + ) is true whenever P(k) is true ` P(n) is true for lln! N. A l g eb r 67

76 Emple.7 Show by the principle of mthemticl induction tht n is divisible by 7, for ll n! N. Let the given sttement P(n) be defined s P(n) n Step : Put n ` P() 7 is divisible by 7 i.e., P() is true. Step : Let us ssume tht the sttement is true for n k i.e., P(k) is true. We ssume k is divisible by 7 k & 7m Step : To prove tht P( + ) is true Emple.8 68 th Std. Business Mthemtics P(k +) k which is divisible by 7 ] + g ] k + g k $ k $ 8 k $ ] 7+ g k k $ 7+ k k $ 7+ 7m 7( + m) P(k +) is true whenever p(k) is true ` P(n) is true for ll n Hence the proof.! N. By the principle of mthemticl induction prove tht n + n is n even number, for ll n! N.

77 Let P(n) denote the sttement tht, n + n is n even number. Put n Let us ssume tht P(k) is true. ` + +, n even number. ` k + k is n even number is true ` tke k + k m () To prove P(k + ) is true ` ] k+ g + ] k+ g k + k+ + k + k + k+ k + ` (k + ) + (k + ) is n even number m + (k + ) by () ` P(k + ) is true whenever P(k) is true. P(n) is true for n! N. (m + k + ) ( multiple of ) Eercise.5 By the principle of mthemticl induction, prove the following n n ( n+ ) for ll n! N n(n + ) n ] n+ g] n+ g, for ll n! N n n(n+), for ll n! N (n ) n ] n g, for ll n! N. 5. n is divisible by 8, for ll n! N. n n 6. b is divisible by b, for ll n! N n is divisible by, for ll n! N. A l g eb r 69

78 8. n(n + ) (n+) is divisible by 6, for ll n! N. 9. n > n, for ll n! N..5 Binomil theorem An lgebric epression of sum or the difference of two terms is clled binomil. For emple y, 5 b 5 7 ^ + h ] g, c + y m, c p + p m, d + n y etc., re binomils. The Binomil theorem or Binomil Epression is result of epnding the powers of binomils. It is possible to epnd ( + y ) n into sum involving terms of the form b y c, eponents b nd c re nonnegtive integers with b + c n, the coefficient of ech term is positive integer clled binomil coefficient. Epnsion of ( + ) ws given by Greek mthemticin Euclid on th century nd there is n evidence tht the binomil theorem for cubes i.e., epnsion of (+) ws known by 6th century in Indi. The term Binomil coefficient ws first introduced by Michel Stifle in 5.Blise Pscl (9 June 6 to 9 August 66) French Mthemticin, Physicist, inventor, writer nd ctholic theologin. In his Tretise on Arithmeticl tringle of 65 described the convenient tbulr presenttion for Binomil coefficient now clled Pscl s tringle. Sir Issc Newton generlized the Binomil theorem nd mde it vlid for ny rtionl eponent. Now we study the Binomil theorem for ( + ) n Theorem(without proof) If nd re rel numbers, then for ll n! N ] + g n n 0 n n n r r nc + nc + nc nc nc + nc nc n n n n n 0 / r 0 r r n r r NOTE When n 0, (+) 0 When n, ] + g C+ C + 70 th Std. Business Mthemtics 0 When n, ] + g C + C + C When n, ] + g C + C+ C + C nd so on.

79 Given below is the Pscl s Tringle showing the co efficient of vrious terms in Binomil epnsion of ] + g n n 0 n n n n 6 n Fig..5 NOTE (i) Number of terms in the epnsion of ( + ) n is n+ (ii) Sum of the indices of nd in ech term in the epnsion is n (iii) nc0, nc, nc, nc,... ncr,... ncnre lso represented s C0, C, C, C,..., Cr,..., Cn, re clled Binomil coefficients. (iv) Since ncr ncn r,, for r 0,,,, n in the epnsion of ( + ) n, binomil coefficients of terms equidistnt from the beginning nd from the end of the epnsion re equl. (v) Sum of the coefficients in the epnsion of (+) n is equl of n (vi) In the epnsion of (+) n, the sum of the coefficients of odd terms the sum of the coefficients of the even terms n (vii) Generl term in the epnsion of (+) n is t r+ nc Middle term of ] + g n r n r r Cse (i) If n is even, then the number of terms n + is odd, then there is only one middle term, given by t n + Cse (ii) If n is odd, then the number of terms (n+) is even.. Therefore, we hve two middle terms given by t n + nd t n + A l g eb r 7

80 Sometimes we need prticulr term in the epnsion of ( + ) n. For this we first write the generl term tr +. The vlue of r cn be obtined by tking the term tr + s the required term. To find the term independent of (term without ), equte the power of in tr + to zero, we get the vlue of r. By substituting the vlue of r in tr +, we get the term independent of. NOTE ] + g n n 0 n n nc + nc + nc nc (i) To find ( ) n, replce by 0 r n r r n.() ] g n n 0 n n nc + nc + nc ] g ] g nc ] g ]g 0 r n r r n n 0 n nc nc n nc r n r... ( ) ncr r n... ( ) n 0 ] g+ ] g + + ] g + + ] g...() Note tht we hve the signs lterntively. (ii) (iii) If we put in (), we get n r ] + g + nc + nc nc nc n...() If we replce by in (), we get r n n r r n ] g nc + nc nc ] g nc ( ) r n n Emple.9 Epnd ( + y) 5 using binomil theorem. ] + g n n 0 n n n r r nc + nc + nc nc nc + nc n r n n 0 ^+ yh 5 5C 5 5C y 5C ] g + ] g ^ h + ] g ^ yh + 5C ] g ^yh + 0 5C ] g^yh + 5C ^yh y y ] g ^ h + ^ h_ i+ _ y i+ 5] g8y + y y+ 70 y y + 80 y + y n 7 th Std. Business Mthemtics

81 Emple.0 Using binomil theorem, epnd c Emple. + m c + m C ^ h + ^ h + C ^ h c m + C ^ hc m + C c m Using binomil theorem, evlute (0) 5 (0) 5 (00 + ) 5 (00) 5 + 5C (00) + 5C (00) + 5C (00) + 5C (00) + 5C ( ) + 0(000000) + 0(0000) + 5(00) Emple. Find the 5 th term in the epnsion of c m 0 Generl term in the epnsion of ] + g n is tr + ncr To find the 5 th term of For this tke r c m 0 n r r Then, t+ t5 0C 6 ] g c m (Here n 0,, ) 0C 6 ] g 8 Emple. 700 Find the middle term in the epnsion of b l 0 () A l g eb r 7

82 0 Compre b l with ] + g n n 0,, Since n 0, we hve terms (odd) ` 6 th term is the middle term. n r r The generl term is tr + ncr () To get t6, put r 5 t + t 0C ^ C h b l 5 ]g Emple. Find the middle term in the epnsion of b + 9yl 9 Compre b + 9yl 9 with ( + ) n Since n 9,we hve 0 terms(even) t + t + ` There re two middle terms nmely, Generl term in the epnsion of ( + ) n is tr + nc r n r r n n i.e., t, t s () Here t5 nd t6 re middle terms. put r in (), t+ t5 9C b l $ ^9yh 7 th Std. Business Mthemtics 9C 5 $ 9 y y 5 put r 5 in (), t5+ t6 9C 5 b l $ ^9yh 5 5 9C 5 $ 9 y 985 y 5 5 $ 5 $ 9 y 5

83 Emple.5 Find the Coefficient of 0 in the binomil epnsion of b l n r r Generl term of ( + ) n is tr + ncr Compre b l with ( + ) n r tr + Cr ^ h b l r () r ] r C g r r ] g b l C ]g $ r r r r r To find the coefficient of 0, tke r 0 r Put r in (), t5 C 0 C $ r 7 0 ` Coefficient of 0 is C 7. Emple.6 Find the term independent of in the epnsion of 9 c + m. n r r Generl term in the epnsion of ( + ) n is tr + ncr Compre c + ` tr 9 m w i t ( h + ) n + 9C 9 r r ] g c tr r r 9Cr r + 9Cr r $ m 9 9 r 9 r 9 r To get the term independent of, equting the power of s 0 9 r 0 ` r Put r in () t+ 9C 5 $ 9C 5 79 r... () A l g eb r 75

84 ICT Corner Epected finl outcomes Step Open the Browser nd type the URL given (or) Scn the QR Code. GeoGebr Work book clled th BUSINESS MATHS will pper. In this severl work sheets for Business Mths re given, Open the worksheet nmed Pscl s Tringle Step Pscl s tringle pge will open. Click on the check boes View Pscl Tringle nd View Pscl s Numbers to see the vlues. Now you cn move the sliders n nd r to move the red point to respective n C r. Now click on view Combintion to see the n C r clcultion. St e p St e p St e p B r o w s e i n t h e l i n k th Business Mths: (or) scn the QR Code 76 th Std. Business Mthemtics

85 Eercise.6. Epnd the following by using binomil theorem 7 (i) ] bg (ii) c + y m (iii) c + m 6. Evlute the following using binomil theorem: (i) (0) (ii) (999) 5. Find the 5 th term in the epnsion of ( y).. Find the middle terms in the epnsion of (i) b + l (ii) b + l (iii) c 8 m 0 5. Find the term independent of in the epnsion of (i) 9 5 b l (ii) c m (iii) b + l n 6. Prove tht the term independent of in the epnsion of b + l is $ $ 5..., ] n g n n!. n 7. Show tht the middle term in the epnsion of ] + g is 5.. ( n ) n n n! Eercise.7 Choose the correct nswer. If nc nc then the vlue of nc is () (b) (c) (d) 5. The vlue of n, when np 0 is () (b) 6 (c) 5 (d). The number of wys selecting plyers out of 5 is ()! (b) 0 (c) 5 (d) 5 A l g eb r 77

86 If npr 70] ncrg, then r is equl to () (b) 5 (c) 6 (d) 7 5. The possible out comes when coin is tossed five times () 5 (b) 5 (c) 0 (d) 5 6. The number of digonls in polygon of n sides is equl to () nc (b) nc (c) nc n (d) nc 7. The gretest positive integer which divide n] n+ g] n+ g] n+ g for ll n! N is () (b) 6 (c) 0 (d) 8. If n is positive integer, then the number of terms in the epnsion of ] + g n is () n (b) n + (c) n (d) n 9. For ll n > 0, nc + nc + nc + + nc n is equl to () n (b) n (c) n (d) n 0. The term contining in the epnsion of ^ yh 7 is () rd (b) th (c) 5 th (d) 6 th. The middle term in the epnsion of b + l is () 0C b l (b) 0C5 (c) 0C6 (d) C 78 th Std. Business Mthemtics 0 6. The constnt term in the epnsion of b + l is 0 7 () 56 (b) 65 (c) 6 (d) 60. The lst term in the epnsion of ^+ h 8 is () 8 (b) 6 (c) 8 (d) 7 k. If ] + g] g + + then k is equl to () 9 (b) (c) 5 (d) 7 5. The number of letter words tht cn be formed from the letters of the word number when the repetition is llowed re () 06 (b) (c) 6 (d) 00

87 6. The number of prllelogrms tht cn be formed from set of four prllel lines intersecting nother set of three prllel lines is () 8 (b) (c) 9 (d) 6 7. There re 0 true or flse questions in n emintion. Then these questions cn be nswered in () 0 wys (b) 0 wys (c) 0 wys (d) 00 wys 8. The vlue of ] 5C0+ 5Cg + ] 5C+ 5Cg + ] 5C+ 5Cg+ ] 5C+ 5Cg+ ] 5C+ 5C5g is () 6 (b) 5 (c) 8 (d) 7 9. The totl number of 9 digit number which hve ll different digit is () 0! (b) 9! (c) 9 9! (d) 0 0! 0. The number of wys to rrnge the letters of the word CHEESE () 0 (b) 0 (c) 70 (d) 6. Thirteen guests hs prticipted in dinner. The number of hndshkes hppened in the dinner is () 75 (b) 78 (c) 86 (d). Number of words with or without mening tht cn be formed using letters of the word EQUATION, with no repetition of letters is () 7! (b)! (c) 8! (d) 5!. Sum of Binomil coefficient in prticulr epnsion is 56, then number of terms in the epnsion is () 8 (b) 7 (c) 6 (d) 9. The number of permuttion of n different things tken r t time, when the repetition is llowed is () r n (b) n r n! n! (c) ] n rg (d)! ] n+ rg! 5. Sum of the binomil coefficients is () n (b) n (c) n (d) n+7 A l g eb r 79

88 Miscellneous Problems. Resolve into Prtil Frctions : ] g] + g. Resolve into Prtil Frctions: +. Decompose into Prtil Frctions:. Decompose into Prtil Frctions: 6 7 ] + g] g ] g^ + h 5. Evlute the following epression. 7! (i) 6! (ii) 5 8!! (iii)! 6 9!! 6. How mny code symbols cn be formed using 5 out of 6 letters A, B, C, D, E, F so tht the letters ) cnnot be repeted b) cn be repeted c) cnnot be repeted but must begin with E d) cnnot be repeted but end with CAB. 7. From 0 rffle tickets in ht, four tickets re to be selected in order. The holder of the first ticket wins cr, the second motor cycle, the third bicycle nd the fourth sktebord. In how mny different wys cn these prizes be wrded? 8. In how mny different wys, Mthemtics, Economics nd History books cn be selected from 9 Mthemtics, 8 Economics nd 7 History books? 9. Let there be red, yellow nd green signl flgs. How mny different signls re possible if we wish to mke signls by rrnging ll of them verticlly on stff? 0. Find the Coefficient of in the epnsion of c + 80 th Std. Business Mthemtics Summry For ny nturl number n, n fctoril is the product of the first n nturl numbers nd is denoted by n! or n. For ny nturl number n n! n(n ) (n ) 0! If there re two jobs, ech of which cn be performed independently in m nd n wys respectively, then either of the two jobs cn be performed in (m+n) wys. The number of rrngements tht cn be mde out of n things tking r t time is clled the number of permuttion of n things tking r t time. m 7

89 np np np np n! ] n rg!!! ] n n 0g! n n!!! n n nn ^ h ] g! ] n g! n! n ] n ng! 0!! n! r q n n npr nn ( )( n) [ n( r )] The number of permuttion of n different things tken r t time, when repetition is llowed is n r. The number of permuttions of n things tken ll t time, of which p things re n! of one kind nd q things re of nother kind, such tht p+q n is pq!! Circulr permuttion of n different things tken ll t the time (n )! The number of circulr permuttion of n identicl objects tken ll t time is ] n g! Combintion is the selection of n things tken r t time without repetition. Out of n things, r things cn be selected in ncr wys. n! ncr, 0 # r # n r! ^n rh! ncn. nc n. nn ^ h nc! nc ncy, then either y or + y n. ncr ncn r. nc + nc ^n+ hc r r r n ( ) nco n o nc n nc n n r r ncr + ncn + nc nc n 0 n n / r 0 r n r r Number of terms in the epnsion of (+) n is n+ n Sum of the indices of nd in ech term in the epnsion of (+) n is n A l g eb r 8

90 nc0, nc, nc, nc ncr ncnre lso represented s C, C, C, C Cr C clled Binomil coefficients. 0 n, SincenCr ncn r, for r 0,,, n in the epnsion of (+) n, Binomil co efficients which re equidistnt from either end re equl nc nc, nc nc, nc nc 0 n n n Sum of the coefficients is equl of n Sum of the coefficients of odd terms sum of the coefficients of even terms n Generl term in the epnsion of ^+ h n istr+ nc r n r r GLOSSARY Binomil ஈர ற ப ப Circulr Permuttion வட ட வ ம றறம Coefficient வகழ Combintion ச ர வ Fctoril க ரண யப வபர க கம Independent term ர உற ப ப Liner fctor ஒர பட க க ரண Mthemticl Induction கண தத வத க ததற தல Middle term ம ய உற ப ப Multipliction Principle of counting எண ண தல ன வபர க கல வக ள மக Prtil frction பக த ப ப னனஙகள Pscl s Tringle ப ஸகல ன ம க சக ணம Permuttions வ ம றறஙகள Principle of counting எண ண தல ன வக ள மக Rtionl Epression வ க தம ற சக மவ 8 th Std. Business Mthemtics

91 Chpter ANALYTICAL GEOMETRY Lerning Objectives Introduction After studying this chpter, the students will be ble to understnd the locus the ngle between two lines. the concept of concurrent lines. the pir of stright lines the generl eqution nd prmetric eqution of circle. the centre nd rdius of the generl eqution of circle. the eqution of circle when the etremities of dimeter re given. the eqution of tngent to the circle t given point. identifiction of conics. the stndrd eqution of prbol, its focus, directri, ltus rectum nd commercil pplictions The word Geometry is derived from the word geo mening erth nd metron mening mesuring. The need of mesuring lnd is the origin of geometry. Geometry is the study of points, lines, curves, surfce etc., nd their properties. The importnce of nlyticl geometry is tht it estblishes correspondence between geometric curves nd lgebric equtions. A systemtic study of geometry by the use of lgebr ws first crried out by celebrted French Philosopher nd Rene Descrtes (596650) mthemticin Rene Descrtes (596650), in his book L Geometry, published in 67. The resulting combintion of nlysis nd geometry is referred now s nlyticl geometry. He is known s the fther of Anlyticl geometry Anlyticl geometry is etremely useful in the ircrft industry, especilly when deling with the shpe of n irplne fuselge. n tic e met 8

92 Locus Definition. The pth trced by moving point under some specified geometricl condition is clled its locus... Eqution of locus Any reltion in nd y which is stisfied by every point on the locus is clled the eqution of the locus. For emple, (i) The locus of point P^, yh whose distnce from fied point C(h, k) is constnt, is circle. The fied point C is clled the centre. C Fig.. P P (ii) The locus of point whose distnces from two points A nd B re equl is the perpendiculr bisector of the line segment AB. A B Stright line is the locus of point which moves in the sme direction. Emple. Fig.. A point in the plne moves so tht its distnce from the origin is thrice its distnce from the y is. Find its locus. Let P^, yh be ny point on the locus nd A be the foot of the perpendiculr from P^, y h to the y is. Given tht OP AP OP 9AP 8 th Std. Business Mthemtics

93 ^ 0h + ^y 0h 9 + y 9 8 y 0 ` The locus of P^, y h is 8 y 0 Emple. Find the locus of the point which is equidistnt from (, ) nd (, ). Let A(, ) nd B (, ) be the given points Let P^, y h be ny point on the locus. Given tht PA PB. ^ PA PB h + ^y + h ^ h + ^y + h + + y + 6y y + 8y + 6 i.e., y 0 Emple. i.e., y 6 0 The locus of P^, y h is y 6 0. Find the locus of point, so tht the join of ( 5, ) nd (, ) subtends right ngle t the moving point. Let A( 5, ) nd B(, ) be the given points Let P^, y h be ny point on the locus. Given tht + APB 90 o. Tringle APB is right ngle tringle. BA PA +PB n tic e met 85

94 (+5) + ( ) ^ + 5h + ^y h + ^ h + ^y h y y y y + i.e., + y + 6y i.e., + y + y 0 The locus of P^, y h is + y y 0 Eercise.. Find the locus of point which is equidistnt from (, ) nd is.. A point moves so tht it is lwys t distnce of units from the point (, ). If the distnce of point from the points (,) nd (,) re in the rtio :, then find the locus of the point.. Find point on is which is equidistnt from the points (7, 6) nd (, ). 5. If A(, ) nd B(, ) re two fied points, then find the locus of point P so tht the re of tringle APB 8 sq.units.. System of stright lines.. Recll In lower clsses, we studied the bsic concept of coordinte geometry, like distnce formul, section formul, re of tringle nd slope of stright lines etc. We lso studied vrious form of equtions of lines in X std. Let us recll the equtions of stright lines. Which will help us for better understnding the new concept nd definitions of XI std coordinte geometry. Vrious forms of stright lines: (i) Slopeintercept form Eqution of stright line hving slope m nd yintercept c is y m+c (ii) Point slope form Eqution of Stright line pssing through the given point P^, yh nd hving slope m is y y m^ h 86 th Std. Business Mthemtics

95 (iii) T w o P oi n t f or m Eqution of stright line joining the given points A^, yh, B^, yh is y y y y. In determinnt form, eqution of stright line joining two given points A^, yh nd B^, y h is y y y 0 (iv) Intercept form Eqution of stright line whose nd y intercepts re nd b, is y + b. (v) Generl form Eqution of stright line in generl form is + by + c 0 where, b nd c re constnts nd, b re not simultneously zero... Angle between two stright lines Let l nd l be two stright lines represented by the equtions l : y m+ c nd l : y m+ c intersecting t P. m If i nd i re two ngles mde by l nd l with is then slope of the lines re tni nd m tn i. y From fig., if i is ngle between the lines l l & l then l i ii ` tn i tn^i i h tni tn i + tni tn i O P i i i tnθ m m + mm Fig.. θ tn m m + mm n tic e met 87

96 NOTE m m (i) If + mm is positive, then i, the ngle between l nd l is cute nd if it is negtive, then, i is the obtuse. (ii) We know tht two stright lines re prllel if nd only if their slopes re equl. (iii) We know tht two lines re perpendiculr if nd only if the product of their slopes is. (Here the slopes m nd m re finite.) The stright lines is nd yis re perpendiculr to ech other. But, the condition mm is not true becuse the slope of the is is zero nd the slope of the yis is not defined. Emple. Find the cute ngle between the lines y+ 0 nd + y+ 0 Let m nd m be the slopes of y+ 0 nd + y+ 0 Now m, m Let i be the ngle between the given lines m m tn i + mm 88 th Std. Business Mthemtics + + ^h i tn ^h.. Distnce of point from line (i) The length of the perpendiculr from point P(l, m) to the line + by + c 0 l + bm + c is d + b (ii) The length of the perpendiculr form the origin (0,0) to the line + by + c 0 c is d + b Emple.5 Show tht perpendiculr distnces of the line y+ 5 0from origin nd from the point P(, ) re equl.

97 Given line is y+ 5 0 Perpendiculr distnce of the given line from P(, ) is Distnce of (0,0) from the given line The given line is equidistnce from origin nd (, ) Emple.6 If the ngle between the two lines is r nd slope of one of the lines is, then find the slope of the other line. by We know tht the cute ngle i between two lines with slopes m m tn i + mm r Given m ndi r m ` tn + m m d + m n + m m & m Hence the slope of the other line is. mnd m is given.. Concurrence of three lines If two lines l nd l meet t common point P, then tht P is clled point of intersection of l nd l. This point of intersection is obtined by solving the equtions of l nd l. If three or more stright lines will hve point in common then they re sid to be concurrent. The lines pssing through the common point re clled concurrent lines nd the common point is clled concurrent point. n tic e met 89

98 Conditions for three given stright lines to be concurrent Let + by + c 0 " () + by + c 0 " () + by + c 0 " () be the equtions of three stright lines, then the condition tht these lines to be concurrent is Emple.7 b b b c c c 0 Show tht the given lines y 08, y nd y 7 0 re concurrent nd find the concurrent point. Given lines y 0... () 8 y... () y () Conditon for concurrent lines is b b b c c c 0 i.e., 8 7 (77 99) + ( 56+66) ( +) & Given lines re concurrent. To get the point of concurrency solve the equtions () nd () Eqution () # & 6 8y 6 Eqution () # & 6 9y y 5 90 th Std. Business Mthemtics

99 When y5 from () 8 88 Point of concurrency is (, 5) Emple.8 If the lines 5y 05, + y 7 0nd + ky 0 re concurrent, find the vlue of k. Given the lines re concurrent. Therefore b b b c c c k 0 Emple.9 (5+) k( +55) 0 & k 68. ` k. A privte compny ppointed clerk in the yer 0, his slry ws fied s `0,000. In 07 his slry rised to `5,000. (i) Epress the bove informtion s liner function in nd y where y represent the slry of the clerk nd represent the yer (ii) Wht will be his slry in 00? Let y represent the slry (in Rs) nd represent the yer yer () slry (y) 0( ) 0,000(y ) 07( ) 5,000(y ) 00? The eqution of stright line epressing the given informtion s liner eqution in nd y is n tic e met 9

100 y y y y y 0, 000 5, 000 0, y 0, In 00 his slry will be 5 0 y ,000 y 000 9,9,000 y 000(00) 9,9,000 y `8000 Eercise.. Find the ngle between the lines whose slopes re nd. Find the distnce of the point (,) from the line y+ 0. Show tht the stright lines + y 0, + 0nd y re concurrent.. Find the vlue of for which the stright lines + y ; 7y nd y 0 re concurrent. 5. A mnufcturer produces 80 TV sets t cost of `,0,000 nd 5 TV sets t cost of `,87,500. Assuming the cost curve to be liner, find the liner epression of the given informtion. Also estimte the cost of 95 TV sets.. Pir of stright lines.. Combined eqution of the pir of stright lines Let us consider the two individul equtions of stright lines l + my + n 0 nd l+ my + n 0 Then their combined eqution is ^l + my + nh ^l + my + nh 0 9 th Std. Business Mthemtics

101 ll + ^lm + lmhy + mmy + ^ln + lnh+ ^mn + mnhy+ n n 0 Hence the generl eqution of pir of stright lines cn be tken s + hy+ by + g+ fy+ c 0 where bc,,, f, g nd h re ll constnts... Pir of stright lines pssing through the origin The homogeneous eqution + hy+ by 0... () of second degree in nd y represents pir of stright lines pssing through the origin. Let y m nd y m be two stright lines pssing through the origin. Then their combined eqution is ^ymh^ym h 0 & mm m m ^ + h y + y 0... () () nd () represent the sme pir of stright lines ` mm h b m + ^ mh & mm b nd m m b h +. i.e., product of the slopes b nd sum of the slopes b h.. Angle between pir of stright lines pssing through the origin The eqution of the pir of stright lines pssing through the origin is + hy+ by 0 Let m nd m be the slopes of bove lines. Here m m b h + nd mm b. Let i be the ngle between the pir of stright lines. Then m m tn i + mm n tic e met 9

102 ± h b + b Let us tke i s cute ngle ` i tn h b < F + b NOTE (i) If i is the ngle between the pir of stright lines + hy+ by + g+ fy+ c 0, then i tn h b < F + b (ii) If the stright lines re prllel, then h b. (iii) If the stright lines re perpendiculr, then + b 0 i.e., cqefficient qf + cqefficient qf y 0.. The condition for generl second degree eqution to represent the pir of stright lines NOTE The condition for generl second degree eqution in, y nmely + hy+ by + g+ fy+ c 0 to represent pir of stright lines is bc+ fgh f bg ch 0. Emple.0 The condition in determinnt form is Find the combined eqution of the given stright lines whose seprte equtions re + y 0 nd + y 5 0. The combined eqution of the given stright lines is ^+ yh^+ y 5h 0 i.e., + y + y + y y0 5y+ 5 0 i.e., + 5y+ y 7y+ 5 0 h g h b f g f c 0 9 th Std. Business Mthemtics

103 Emple. Show tht the eqution + 5y+ y y+ 0 represents pir of stright lines. Also find the ngle between them. Compre the eqution + 5y+ y y+ 0 with + hy+ by + g+ fy+ c 0, we get 5 7, b, h, g, f nd c Condition for the given eqution to represent pir of stright lines is bc+ fgh f bg ch bc+ fgh f bg ch Hence the given eqution represents pir of stright lines. Let θ be the ngle between the lines. h b 5 Then i tn < F + tn b > 6 H 5 ` i tn b 5 l Emple. The slope of one of the stright lines + hy+ by 0 is twice tht of the other, show tht 8h 9b Let m nd m be the slopes of the pir of stright lines + hy+ by 0 ` m+ m b h nd mm b It is given tht one slope is twice the other, so let m ` m+ m b h nd m m b $ m n tic e met 95

104 ` h m b nd m & b h b l b & & 8h 9b b 8h 9b b Emple. Show tht the eqution + 7y+ y y+ 0 represent two stright lines nd find their seprte equtions. 96 th Std. Business Mthemtics Compre the eqution + 7y+ y y+ 0 with + hy+ by + g+ fy+ c 0, we get, 7 5 5, b, h, g, f, c 7 5 h g 7 5 Now h b f 5 5 g f c 5 b6 7 l b 5 9 l+ 5 b 5 5 l b l 7 $ + 5 $ Hence the given eqution represents pir of stright lines. Now consider, + 7y+ y + 6y+ y+ y ^+ yh+ y^+ yh ^+ yh^+ yh Let + 7y+ y y+ ^+ y+ lh^+ y+ mh Compring the coefficient of, l + m 5 () Compring the coefficient of y, l + m 5 () Solving () nd (), we get m nd l ` The seprte equtions re + y+ 0 nd + y+ 0.

105 Emple. Show tht the pir of stright lines y+ 9y + 8 7y+ 8 0 represents pir of prllel stright lines nd find their seprte equtions. The given eqution is y+ 9y + 8 7y+ 8 0 Here, b 9 nd h 6 h b Hence the given eqution represents pir of prllel stright lines. Now y+ 9y ^ yh Consider y+ 9y + 8 7y+ 8 0 & ^ yh + 9 ^ yh Put y z z + 9z+ 8 0 (z+)(z+8) 0 z+ 0 z+8 0 y+ 0 y+8 0 Hence the seprte equtions re Emple.5 y+ 0 nd y+8 0 Find the ngle between the stright lines + y+ y 0 The given eqution is + y+ y 0 Here, b nd h. n tic e met 97

106 If i is the ngle between the given stright lines, then θ tn h b < F + b tn < F tn ^ h Emple.6 i r For wht vlue of k does + 5y+ y y+ k 0 represent pir of stright lines. 5 Here, b, h 5, g, f 9, c k. The given line represents pir of stright lines if, bc+ fgh f bg ch i.e., k+ 6 k 0 & 6k k 0 & 9k 5 ` k 8 Eercise.. If the eqution + 5y 6y + + 5y+ c 0 represents pir of perpendiculr stright lines, find nd c.. Show tht the eqution 0y+ y + 5y+ 0 represents pir of stright lines nd lso find the seprte equtions of the stright lines.. Show tht the pir of stright lines + y+ 9y 6 9y+ 0 represents two prllel stright lines nd lso find the seprte equtions of the stright lines.. Find the ngle between the pir of stright lines 5y y + 7+ y th Std. Business Mthemtics

107 Circles Definition. A circle is the locus of point which moves in such wy tht its distnce from fied point is lwys constnt. The fied point is clled the centre of the circle nd the constnt distnce is the rdius of the circle... The eqution of circle when the centre nd rdius re given. Let C(h, k) be the centre nd r be the rdius of the circle Let P(, y) be ny point on the circle y C r R P(, y) CP r circle. CP r ( h) + (y k) r is the eqution of the O L Fig.. M In prticulr, if the centre is t the origin, the eqution of circle is + y r Emple.7 Find the eqution of the circle with centre t (, ) nd rdius is units. Eqution of circle is ^ hh + ^y kh r Here (h, k) (, ) nd r Eqution of circle is ^ h + ^y+ h y + y+ 6 + y 6+ y 6 0 Emple.8 Find the eqution of the circle with centre t origin nd rdius is units. n tic e met 99

108 Eqution of circle is + y r Here r i.e eqution of circle is + y 9.. Eqution of circle when the end points of dimeter re given Let A^, yh nd B^, yh be the end points of dimeter of circle nd P(, y) be ny point on the circle. P(, y) We know tht ngle in the semi circle is 90 o ` APB 90c ` (Slope of AP) (Slope of BP) y y d y y n# d n ^yy h^yy h ^ h^ h A(, y ) C B(, y ) Fig..5 & ^h^ h+ ^yyh^yyh 0 is the required eqution of circle. Emple.9 Find the eqution of the circle when the end points of the dimeter re (, ) nd (, ). Eqution of circle when the end points of the dimeter re given is ^h^ h+ ^yyh^yyh 0 Here ^, yh (, ) nd ^, yh (, ) ` ] g] g+ _ y i_ y+ i 0 + y 5y 0.. Generl eqution of circle The generl eqution of circle is + y + g + fy + c 0 where g, f nd c re constnts. 00 th Std. Business Mthemtics

109 i.e + y + g + fy c + g+ g g + y + fy + f f c ^+ gh g + ^y+ fh f c ^+ gh + ^y+ f h g + f c 7 ^ gha + 7y ^f ha 7 g + f ca Compring this with the circle^ hh + ^y kh r We get, centre is ^g, fh nd rdius is g + f c NOTE The generl second degree eqution + by + hy+ g+ fy+ c 0 represents circle if (i) b i.e., coefficient of coefficient of y. (ii) h 0 i.e., no y term. Emple.0 Find the centre nd rdius of the circle + y 8+ 6y 0 Eqution of circle is + y 8+ 6y 0 Here g, f nd c Centre C^g, fh C^, h nd Rdius: r g + f c Emple unit. For wht vlues of nd b does the eqution ] g + by + ] b g y + + y 0 represents circle? Write down the resulting eqution of the circle. n tic e met 0

110 The given eqution is ] g + by + ] b g y + + y 0 As per conditions noted bove, (i) coefficient of y 0 & b 0 ` b (ii) coefficient of Coefficient of y & b & ` Resulting eqution of circle is + y + + y 0 Emple. If the eqution of circle + y + + by 0 pssing through the points (, ) nd (, ), find the vlues of nd b The circle + y + + by 0 pssing through (, ) nd (, ) ` We hve b 0 nd b 0 & + b 5 () nd + b () solving () nd (), we get, b,. Emple. If the centre of the circle + y + 6y+ 0 lies on stright line + y + 0, then find the vlue of Centre C(, ) It lies on + y th Std. Business Mthemtics

111 Emple. Show tht the point (7, 5) lies on the circle + y 6+ y 0 nd find the coordintes of the other end of the dimeter through this point. Let A(7, 5) Eqution of circle is + y 6+ y 0 Substitute (7, 5) for (, y), we get + y 6+ y ] g 6] 7g + ] 5g ` (7, 5) lies on the circle Here g nd f ` Centre C(, ) Let the other end of the dimeter be B (, y) Midpoint of AB + y, C, 7 5 c m ^ h + 7 y 5 y Other end of the dimeter is (, ). Emple.5 Find the eqution of the circle pssing through the points (0,0), (, ) nd (,0). Let the eqution of the circle be + y + g + fy + c 0 The circle psses through the point (0, 0) c 0... () The circle psses through the point (, ) n tic e met 0

112 + + g() + f() + c 0 g + f + c 5... () The circle psses through the point (, 0) g() + f() 0 + c 0 Solving (), () nd (), we get g, f nd c 0 ` The eqution of the circle is + y + ( ) + b l y+ 0 0 ie., + y y 0.. Prmetric form of circle g + c... () Consider circle with rdius r nd centre t the origin. Let Py (, ) be ny point on the circle. Assume tht OP mkes n ngle i with the positive direction of is. Drw PM perpendiculr to is. Y From the figure, cosi r & r cos i sini r & y rsin i The equtions r cos i, y r sin i re clled the prmetric equtions of the circle + y r. Here i is clled the prmeter nd 0 # i # r. Emple.6 r i O Fig..6 M P(, y) X Find the prmetric equtions of the circle + y 5 Here r 5 & r 5 Prmetric equtions re r cos i, y r sin i & 5cosi, y 5sin i, 0 # i # r 0 th Std. Business Mthemtics

113 Eercise.. Find the eqution of the following circles hving (i) the centre (,5) nd rdius 5 units (ii) the centre (0,0) nd rdius units. Find the centre nd rdius of the circle (i) + y 6 (ii) + y y+ 5 0 (iii) 5 + 5y + 8y 6 0 (iv) ] + g] 5g+ ^yh^y h 0. Find the eqution of the circle whose centre is (, ) nd hving circumference 6r. Find the eqution of the circle whose centre is (,) nd which psses through (,) 5. Find the eqution of the circle pssing through the points (0, ),(, ) nd (, ). 6. Find the eqution of the circle on the line joining the points (,0), (0,) nd hving its centre on the line + y 7. If the lines + y 6 nd + y re dimeters of the circle, nd the circle psses through the point (, 6) then find its eqution. 8. Find the eqution of the circle hving (, 7) nd (, 5) s the etremities of dimeter. 9. Find the Crtesin eqution of the circle whose prmetric equtions re cos i, y sin i, 0 # i # r...5 Tngents The eqution of the tngent to the circle + y + g + fy + c 0 t ^, yh is + yy + g ( + ) + f( y+ y ) + c 0. Corollry: The eqution of the tngent t ^, yh to the circle + y is + yy C( g, f) P(, y ) Fig..7 n tic e met T 05

114 NOTE To get the eqution of the tngent t ^, yh to the circle Emple.7 replce by, y + y + g + fy + c 0... () by yy, by + nd y by y + y in eqution (). Find the eqution of tngent t the point (, 5) on the circle + y + 8y The eqution of the tngent t, y ^ h to the given circle + y + 8y+ 7 0 is yy + + # y y ^ + h 8 # + ^ h Here ^, yh (, 5) + 5y+ ( ) ( y+ 5) y+ y y+ 68y y+ 0 is the required eqution. Length of the tngent to the circle Length of the tngent to the circle + y + g + fy + c 0 from point P ^, yh is PT + y + g + fy + c T C( g, f) P(, y ) 06 th Std. Business Mthemtics Fig..8

115 Emple.8 NOTE (i) If the point P is on the circle then PT 0 (ii) If the point P is outside the circle then PT > 0 (iii) If the point P is inside the circle then PT < 0 Find the length of the tngent from the point (,) to the circle + y + 8+ y+ 8 0 The length of the tngent to the circle + y + g + fy + c 0 from point ^, yh is + y + g + fy + c Emple.9 Length of the tngent + y y () + () + 8 [Here ^, y h (, )] 9 Length of the tngent 7 units Determine whether the points P(0,), Q(5,9), R(, ) nd S(, ) lie outside the circle, on the circle or inside the circle + y + y 8 0 The eqution of the circle is + y + y 8 0 PT + y + y 8 At P(0, ) PT < 0 At Q(5, 9) QT > 0 At R(, ) RT > 0 At S(, ) ST ` The point P lies inside the circle. The points Q nd R lie outside the circle nd the point S lies on the circle. n tic e met 07

116 Result Condition for the stright line y m + c to be tngent to the circle + y is c ( + m ) Emple.0 Find the vlue of k so tht the line + y k 0 is tngent to + y 6 0 The given equtions re + y 6 0 nd + y k 0 The condition for the tngency is c ( + m ) Here 6, m nd c k k c ( + m ) & 6 6b l k 6 5 k ± 0 08 th Std. Business Mthemtics Eercise.5. Find the eqution of the tngent to the circle + y + y 8 0 t (, ). Determine whether the points P(, 0), Q(, ) nd R(, ) lie outside the circle, on the circle or inside the circle + y 6y Find the length of the tngent from (, ) to the circle + y + y Find the vlue of P if the line + y P 0 is tngent to the circle + y 6.5 Conics l Definition. If point moves in plne such tht its M P(Moving point) distnce from fied point bers constnt rtio to its perpendiculr distnce from fied stright line, then the pth described by the moving point is clled conic. F(Fied point) In figure, the fied point F is clled focus, the fied stright line l is clled directri nd P is the Fig..9 moving point such tht PM e, constnt. Here the locus of P is clled conic nd the constnt e is clled the eccentricity of the conic. Fied line (directri)

117 Bsed on the vlue of eccentricity we cn clssify the conics nmely, ) If e, then, the conic is clled prbol b) If e <, then, the conic is clled n ellipse c) If e >, then, the conic is clled hyperbol. The generl second degree eqution + hy+ by + g+ fy+ c 0 represents, (i) pir of stright lines if bc+ fgh f bg ch 0 (ii) circle if b nd h 0 If the bove two conditions re not stisfied, then + hy+ by + g+ fy+ c 0 represents, (i) prbol if h b 0 (ii) n ellipse if h b < 0 (iii) hyperbol if h b > 0.5. Prbol In this chpter, we study bout prbol only. Definition. The locus of point whose distnce from fied point is equl to its distnce from fied line is clled prbol. y i s t he s t nd r d e qu t i on of t he p r bol. I t i s ope n r i ght w.5. Definitions regrding prbol: y M(, y) y P(, y) Z(, 0) V F(, 0) Fig..0 n tic e met 09

118 Focus The fied point used to drw the prbol is clled the focus (F). Here, the focus is F(, 0). Directri The fied line used to drw prbol is clled the directri of the prbol. Here, the eqution of the directri is. Ais The is of the prbol is the is of symmetry. The curve y is symmetricl bout is nd hence is or y 0 is the is of the prbol y. Note tht the is of the prbol psses through the focus nd perpendiculr to the directri. Verte The point of intersection of the prbol with its is is clled its verte. Here, the verte is V(0, 0). Focl distnce The distnce between point on the prbol nd its focus is clled focl distnce Focl chord A chord which psses through the focus of the prbol is clled the focl chord of the prbol. Ltus rectum It is focl chord perpendiculr to the is of the prbol. Here, the eqution of the ltus rectum is. Length of the ltus rectrum is..5. Other stndrd prbols :. Open leftwrd : y 6 0@ If > 0, then y become imginry. i.e., the curve eist for 0. y y F(, 0) V Fig... Open upwrd : y 6 0@ 0 th Std. Business Mthemtics

119 If y < 0, then becomes imginry. i.e., the curve eist for y 0. y y F(0, ) V y Fig... Open downwrd : y 6 0@ If y > 0, then becomes imginry. i.e., the curve eist for y 0. y V y y F(0, ) Fig.. Equtions y y y y Ais y 0 y Verte V(0, 0) V(0, 0) V(0, 0) V(0, 0) Focus F(, 0) F(, 0) F(0, ) F(0, ) Eqution of directri y y Length of Ltus rectum Eqution of Ltus rectum y y The process of shifting the origin or trnsltion of es. Consider the oy system. Drw line prllel to is (sy X is) nd drw line prllel to y is (sy Y is). Let P^, yhbe point with respect to oy system nd P^X, Yh be the sme point with respect to XOlY system. n tic e met

120 y Y (X, Y) P(, y) O Y O h (0, 0) k M L X Fig.. Let the coordintes of Ol with respect to oy system be (h, k) The coordinte of P with respect to oy system: OL OM + ML h + X i.e) X + h similrly y Y + k ` The new coordintes of P with respect to XOl Y system. X h nd Y y k.5. Generl form of the stndrd eqution of prbol, which is open rightwrd (i.e., the verte other thn origin) : Consider prbol with verte V whose coordintes with respect to XOY system is (0, 0) nd with respect to oy system is (h, k). Since it is open rightwrd, the eqution of the prbol w.r.t. XOY system is Y X. By shifting the origin X h nd Y y k, the eqution of the prbol with respect to old oy system is^y kh ^ hh. This is the generl form of the stndrd eqution of the prbol, which is open rightwrd. Similrly the other generl forms re ^y kh ^hh (open leftwrds) th Std. Business Mthemtics

121 ^ hh ^y kh (open upwrds) ^ hh ^ykh (open downwrds) Emple. Find the eqution of the prbol whose focus is (,) nd whose directri is y+ 0. F is (, ) nd directri is y+ 0 y+ 0 M P(, y) let P^, yh be ny point on the prbol. FP For prbol PM FP PM FP ^ h + ^y h y + 0 F(, ) + + y 6y+ 9 + y 6y+ 0 ^ PM + y + h + ^ + y +h Fig..5 PM FP PM ^ y+ h + y + y y+ + y y+ 0 + y y y+ + The required eqution of the prbol is & + y + y 8 8y+ 6 0 Emple. Find the focus, the verte, the eqution of the directri, the is nd the length of the ltus rectum of the prbol y n tic e met

122 The given eqution is of the form y where,. Y The prbol is open left, its focus is F^, qh, F^0, h Its verte is Vhk ^, h V^00, h F(, 0) V(0,0) X The eqution of the directri is i.e., Fig..6 Its is is is whose eqution is y 0. Length of the ltus rectum Emple. Show tht the demnd function 0p0 p is prbol nd price is mimum t its verte. 0p0 p th Std. Business Mthemtics p 0 + 0p p + 0p 5 _ p 0p+ 5i ^p 5h Put X 5 nd P p5 X P i.e P X It is prbol open downwrd. ` At the verte p 5 when 5. i.e., price is mimum t the verte. Emple. Find the is, verte, focus, eqution of directri nd length of ltus rectum for the prbol + 6 y + 0

123 X +, Y y y + 6+ y ^ h + y ^ + h ^ + h (y ) X nd y Y+ X Y Compring with X Y, Referred to (X, Y) Referred to (, y) X, y Y+ Ais y 0 Y 0 y Verte V(0,0) V(0,0) V(,) Focus F(0, ) F(0, ) F(, ) Eqution of directri (y ) Y y Length of Ltus rectum () () Emple.5 The supply of commodity is relted to the price by the reltion 5P 5. Show tht the supply curve is prbol. The supply price reltion is given by 5p 5 & P Where X nd P p 5(p ) ` the supply curve is prbol whose verte is ( X 0, P 0) i.e., The supply curve is prbol whose verte is (0, ) n tic e met 5

124 ICT Corner Epected finl outcomes Step Open the Browser nd type the URL given (or) Scn the QR Code. GeoGebr Work book clled th BUSINESS MATHS will pper. In this severl work sheets for Business Mths re given, Open the worksheet nmed LocusPrbol Step LocusPrbol pge will open. Click on the check boes on the Lefthnd side to see the respective prbol types with Directri nd Focus. On righthnd side Locus for prbol is given. You cn ply/puse for the pth of the locus nd observe the condition for the locus. St e p St e p St e p B r o w s e i n t h e l i n k th Business Mths: (or) scn the QR Code 6 th Std. Business Mthemtics

125 Eercise.6. Find the eqution of the prbol whose focus is the point F^, hnd the directri is the line y+ 0.. The prbol y k psses through the point (, ). Find its ltus rectum nd focus.. Find the verte, focus, is, directri nd the length of ltus rectum of the prbol y 8y Find the coordintes of the focus, verte, eqution of the directri, is nd the length of ltus rectum of the prbol () y 0 (b) 8y (c) 6y 5. The verge vrible cost of monthly output of tonnes of firm producing vluble metl is ` Show tht the verge vrible cost curve is prbol. Also find the output nd the verge cost t the verte of the prbol. 6. The profit ` y ccumulted in thousnd in months is given by y Find the best time to end the project. Choose the correct nswer Eercise.7. If m nd m re the slopes of the pir of lines given by + hy+ by 0, then the vlue of m+ m is () b h (b) b h (c) h h (d). The ngle between the pir of stright lines 7y+ y 0 is () tn b l (b) tn b l (c) tn c m 5 (d) tn 5 d n. If the lines y 5 0 nd y re the dimeters of circle, then its centre is () (, ) (b) (,) (c) (, ) (d) (, ). The intercept of the stright line + y 0 is () (b) (c) (d) n tic e met 7

126 5. The slope of the line 7+ 5y 8 0 is () 5 7 (b) 5 7 (c) 7 5 (d) The locus of the point P which moves such tht P is t equidistnce from their coordinte es is () y (b) y (c) y (d) y 7. The locus of the point P which moves such tht P is lwys t equidistnce from the line + y+ 7 0 is () + y+ 0 (b) y+ 0 (c) y+ 0 (d) + y If k + y y 0 represent pir of lines which re perpendiculr then k is equl to () (b) (c) (d) 9. (, ) is the centre of the circle + y + + by 0, then its rdius () (b) (c) (d) 0. The length of the tngent from (,5) to the circle + y 6 is () (b) 5 (c) 6 (d) 5. The focus of the prbol 6y is () (,0) (b) (, 0) (c) (0, ) (d) (0, ). Length of the ltus rectum of the prbol y 5. () 5 (b) 5 (c) 5 (d) 5. The centre of the circle + y + y 9 0 is 8 th Std. Business Mthemtics () (,) (b) (, ) (c) (,) (d) (, ). The eqution of the circle with centre on the is nd pssing through the origin is () + y 0 (b) y y+ 0 (c) + y (d) y+ y 0

127 5. If the centre of the circle is (, b) nd rdius is b, then the eqution of circle is () + y + + by + b 0 (b) + y + + by b 0 (c) + y by b 0 (d) + y by + b 0 6. Combined eqution of coordinte es is () y 0 (b) + y 0 (c) y c (d) y y + y 0 represents pir of prllel lines then is () (b) (c) (d) 8. In the eqution of the circle + y 6 then y intercept is (re) () (b)6 (c) ± (d) ±6 9. If the perimeter of the circle is 8π units nd centre is (,) then the eqution of the circle is () ^ h + ^y h (b) ^ h + ^y h 6 (c) ^ h + ^y h (d) + y 0. The eqution of the circle with centre (, ) nd touches the is is () ^ h + ^y h (b) ^ h + ^y+ h 6 (c) ^ h + ^y h 6 (d) + y 6. If the circle touches is, y is nd the line 6 then the length of the dimeter of the circle is () 6 (b) (c) (d). The eccentricity of the prbol is () (b) (c)0 (d). The double ordinte pssing through the focus is () focl chord (b) ltus rectum (c) directri (d) is n tic e met 9

128 The distnce between directri nd focus of prbol y is () (b) (c) (d) 5. The eqution of directri of the prbol y is () + 0 (b) 0 (c) 0 (d) + 0 Miscellneous Problems. A point P moves so tht P nd the points (, ) nd (, 5) re lwys colliner. Find the locus of P.. As the number of units produced increses from 500 to 000 nd the totl cost of production increses from ` 6000 to `9000. Find the reltionship between the cost (y) nd the number of units produced () if the reltionship is liner.. Prove tht the lines + y 0, y 5 nd 5+ y 7re concurrent.. Find the vlue of p for which the stright lines 8p + ( py ) + 0 nd p + py 7 0 re perpendiculr to ech other. 5. If the slope of one of the stright lines + hy+ by 0 is thrice tht of the other, then show tht h b 6. Find the vlues of nd b if the eqution ] g + by + ] b 8g y + + y 0 represents circle. 7. Find whether the points ^, h, ^0, h nd ^, h lie bove, below or on the line + y If ^, h is one etremity of dimeter of the circle + y + 6y 5 0, find the other etremity. 9. Find the eqution of the prbol which is symmetricl bout is nd pssing through (, ). 0. Find the is, verte, focus, eqution of directri length of ltus rectum of the prbol^y h ^ h 0 th Std. Business Mthemtics

129 Summry Angle between the two intersecting lines y m + c nd m m y m+ c is i tn + mm Condition for the three stright lines + by + c 0, + by + c 0, nd b c + by + c 0 to be concurrent is b c 0 b c The condition for generl second degree eqution + hy+ by + g+ fy+ c 0 to represent pir of stright lines is bc+ fgh f bg ch 0. Pir of stright lines pssing through the origin is + hy+ by 0. Let + hy+ by 0 be the stright lines pssing through the origin then the product of the slopes is b nd sum of the slopes is b h. If i is the ngle between the pir of stright lines + hy+ by + g+ fy+ c 0, then i tn! h b < + F. b The eqution of circle with centre (h, k) nd the rdius r is^ hh + ^y kh r The eqution of circle with centre t origin nd the rdius r is + y r. The eqution of the circle with ^, yh nd ^, yh s end points of dimeter is ^h^ h+ ^yyh^yyh 0. The prmetric equtions of the circle + y r re r cos i, y r sin i 0 # i # r The eqution of the tngent to the circle + y + g + fy + c 0 t ^, yh is + yy + g ( + ) + f( y+ y ) + c 0 The eqution of the tngent t, y ^ h to the circle + y is + yy. Length of the tngent to the circle + y + g + fy + c 0 from point ^, yh is + y + g + fy + c. Condition for the stright line y m+ c to be tngent to the circle + y is c ^ + m h. The stndrd eqution of the prbol is y n tic e met

130 GLOSSARY C e nt r e ம யம C hor d ந ண C i r c l e வட டம C onc ur r e nt L i ne ஒர ப ள ள வழ க சக ட C oni c s க ம ப வவடட கள D i m e t e r வ ட டம D i r e c t i இயக க வமர E qu t i on னப ட F oc l di s t nc e க வ யதத ரம F oc us க வ யம L t us r e c t um வ வவக ம L e ngt h of t he t nge ntவத ட சக டட ன ந ளம L oc us ந ய ப ப மத அல த இயஙக வமர O r i gi n ஆத P i r of s t r i ght l i ne இரடம ட சநர சக ட P r bol பரவமளயம P r l l e l l i ne இமண சக ட P r m e t e r த மணய க P e r pe ndi c ul r l i ne வ ஙக தத சக ட P oi nt of c onc ur r e nc yஒர ஙக மணவ ப ப ள ள P oi nt of i nt e r s e c t i on வவடட ம ப ள ள R di us ஆரம S t r i ght l i ne சநர சக ட T nge nt வத ட சக ட V e r t e ம மன th Std. Business Mthemtics

131 Chpter TRIGONOMETRY Lerning Objectives After studying this chpter, the students will be ble to understnd trigonometric rtio of ngles ddition formule, multiple nd submultiple ngles trnsformtion of sums into products nd vice vers bsic concepts of inverse trigonometric functions properties of inverse trigonometric functions Introduction The word trigonometry is derived from the Greek word tri (mening three), gon (mening sides) nd metron (mening mesure). In fct, trigonometry is the study of reltionships between the sides nd ngles of tringle. Around second century A.D. George Rheticus ws the first to define the trigonometric functions in terms of right ngles. The study of trigonometry ws first strted in Indi. The ncient Indin Mthemticin, Arybhtt, Brhmgupt, Bhskr I nd Bhskr II obtined importnt results. Brhmgupt Bhskr I gve formule to find the vlues of sine functions for ngles more thn 90 degrees.the erliest pplictions of trigonometry were in the fields of nvigtion, surveying nd stronomy. Currently, trigonometry is used in mny res such s electric circuits, describing the stte of n tom, predicting the heights of tides in the ocen, nlyzing musicl tone. T ri g onom et ry

132 Recll. sinθ side opposite to ngle i Hypotenuse side opposite to ngle i. tnθ Adjcent side to ngle i Hypotenuse 5. secθ Adjcent side to ngle i. cosθ Adjcent side tongle i Hypotenuse Adjcent side to ngle i. cotθ side opposite to ngel i Hypotenuse 6. cosecθ side opposite to ngle i Reltions between trigonometric rtios. sinθ cosec i. cosθ sec i. tnθ sin i cos i. tnθ cot i or cosecθ sin i or secθ cos i cos i or cotθ sin i or cotθ tn i Trigonometric Identities. sin i+ cos i i. + tn sec θ i. + cot cosec θ. Angle Angle is mesure of rottion of given ry bout its initil point. The ry OA is clled the initil side nd the finl position of the ry OB fter rottion is clled the terminl side of the ngle. The point O of rottion is clled the verte. O Initil side Angle AOB If the direction of rottion is nticlockwise, then ngle is sid to be positive nd if the Fig.. direction of rottion is clockwise, then ngle is sid to be negtive. Terminl side B A Mesurement of n ngle Two types of units of mesurement of n ngle which re most commonly used nmely degree mesure nd rdin mesure. th Std. Business Mthemtics

133 Degree mesure th If rottion from the initil position to the terminl position b 60 l of the revolution, the ngle is sid to hve mesure of one degree nd written s o. A degree is divided into minutes nd minute is divided into seconds. One degree 60 minutes 60l One minute 60 seconds (60m ) Rdin mesure r B r The ngle subtended t the centre of the circle by n rc equl to the length of the rdius of the circle is clled rdin, nd it is denoted by c O c r A NOTE Fig.. The number of rdins in n ngle subtended by n rc of circle t the centre of circle is length of the rc rdius i.e., i r s, where θ is the ngle subtended t the centre of circle of rdius r by n rc of length s. Reltion between degrees nd rdins We know tht the circumference of circle of rdius unit is π. One complete revolution of the rdius of unit circle subtends π rdins. A circle subtends t the centre n ngle whose degree mesure is60c. Emple. π rdins 60c π rdins 80c rdin 80c r Convert (i) 60c into rdins (ii) degree r 5 rdins into degree (iii) rdins into T ri g onom et ry 5

134 i) 60c 60 # 80 r 8 rdins 9 r r 80 ii) 5 rdins 5 # c r r c 80 iii) rdin r # 80 # 7 c9lll 5. Trigonometric rtios.. Qudrnts: Y Let XOX l nd Yl OY be two lines t right ngles to ech other s in the figure. We cll XOX l nd Yl OY s X is nd Y is respectively. Clerly these es divided the entire plne into four equl prts clled Qudrnts. X II III O I IV X The prts XOY, YOX ', XOYl ' nd YlOX re known s the first, second, third nd the fourth qudrnt respectively. Emple. Y Fig.. (i) Find the qudrnts in which the terminl sides of the following ngles lie. (i) 70c (ii) 0c (iii) 5c. (i) The terminl side of 70c lies in IV qudrnt. Y X O X P 6 th Std. Business Mthemtics Y Fig..

135 (ii) The terminl side 0cof lies in I qudrnt. Y P X O X Y Fig..5 (iii) 5c ] # 60g + 80c+ 65c the terminl side lies in III qudrnt. Y X X P Y Fig..6.. Signs of the trigonometric rtios of n ngle θ s it vries from 0º to 60º In the first qudrnt both nd y re positive. So ll trigonometric rtios re positive. In the second qudrnt ] 90c i 80cg is negtive nd y is positive. So trigonometric rtios sin θ nd cosec θ re positive. In the third qudrnt] 80c i 70cg both nd y re negtive. So trigonometric rtios tn θ nd cot θ re positive. In the fourth qudrnt ] 70c < i < 60cg is positive nd y is negtive. So trigonometric rtios cos θ nd sec θ re positive. A function f () is sid to be odd function if f( ) f. () sini, tn i, cotind cosec i re ll odd function. A function f () is sid to be even function if f( ) f. () cosind sec i re even function. T ri g onom et ry 7

136 Y II Qudrnt sin & cosec I Qudrnt All X III Qudrnt tn & cot O IV Qudrnt cos & sec X Y ASTC : All Sin Tn Cos Fig..7.. Trigonometric rtios of llied ngles: Two ngles re sid to be llied ngles when their sum or difference is either zero or multiple of 90c. The ngles i, 90c! i, 80c! i, 60c! i etc.,re ngles llied to the ngle θ. Using trigonometric rtios of the llied ngles we cn find the trigonometric rtios of ny ngle. Angle/ Function Sine i 90c i or r i 90c + i or r + i 80c i 80c + i 70c i 70c + i 60c i 60c + i or or or or or or r r r i r+ i i + i r i r+ i sin i cos i cos i sin i sin i cos i cos i sin i sin i Cosine cos i sin i sin i cos i cos i sin i sin i cos i cos i tngent cotngent tn i cot i cot i tn i tn i cot i cot i tn i tn i cot i tn i tn i cot i cot i tn i tn i cot i cot i secnt sec i cosec i cosec i sec i sec i cosec i cosec i sec i sec i cosecnt cosec i sec i sec i cosec i cosec i sec i sec i cosec i cosec i Emple. Tble :. Find the vlues of ech of the following trigonometric rtios. (i) sin 50c (ii) cos( 0c) (iii) cosec 90c (iv) tn( 5c) (v) sec85c 8 th Std. Business Mthemtics

137 (i) sin50c sin( 90c+ 60c) Since 50c lies in the second qudrnt, we hve sin50c sin( 90c+ 60c) cos60c (ii) we hve cos( 0 o ) cos0 o Since 0 lies in the third qudrnt, we hve cos0 o cos(80 o + 0 o ) cos0 o (iii) cosec90 o cosec(60 o + 0 o ) cosec0 o (iv) tn( 5 o ) tn(5 o ) tn( 60 o + 5 o ) tn5 o tn(90 o + 5 o ) ( cot5 o ) (v) sec85 o sec( 60 o + 5 o ) sec5 o Emple. Prove tht sin600 o cos 90 o + cos 80 o sin 50 o sin600 o sin(60 o +0 o ) sin0 o sin(80 o + 60 o ) sin 60 o cos90 o cos(60 o + 0 o ) cos0 o cos80 o cos(60 o + 0) cos0 o cos(80 o 60 o ) cos60 o sin50 o sin(80 o 0 o ) sin0 o Now sin 600 o cos90 o + cos80 o sin50 o b lb l+ b lb l T ri g onom et ry 9

138 Emple.5 Prove tht sin] igtn] 90cigsec] 80cig sin( 80 + i) cot( 60 i) cosec] 90c ig L.H.S sin] igtn] 90cigsec] 80cig sin( 80 + i) cot( 60 i) cosec] 90c ig ]sin igcot i] sec ig ]sinig]cot igsec i 0 th Std. Business Mthemtics Eercise.. Convert the following degree mesure into rdin mesure (i) 60 o (ii) 50 o (iii) 0 o (iv) 0 o. Find the degree mesure corresponding to the following rdin mesure. (i) r 8 (ii) 9r 5 (iii). Determine the qudrnts in which the following degree lie. (i) 80 o (ii) 0 o (iii) 95 o. Find the vlues of ech of the following trigonometric rtios. (iv) r 8 (i) sin00 o (ii) cos( 0 o ) (iii) sec90 o (iv) tn( 855 o ) (v) cosec5 o 5. Prove tht: (i) tn( 5 o ) cot( 05 o ) tn( 765 o ) cot(675 o ) 0 (ii) r sin 6 cosec r r + 6 cos (iii) r 5r 5r 5r secb i lsecb i l+ tnb + i ltnb i l 6. If A, B, C, D re ngles of cyclic qudrilterl, prove tht : cosa + cosb + cosc + cosd 0 7. Prove tht : (i) sin] 80c igcos] 90c+ igtn] 70 igcot] 60 ig sin] 60 igcos] 60 + igsin] 70 igcosec] ig (ii) sini$ cos i ' sin b r r il $ cosec i + cos b il $ sec i

139 8. Prove tht : cos50 o cos0 o + sin90 o cos0 o 9. Prove tht: (i) tn] r+ gcot] rg cos] r gcos] r+ g sin (ii) sin] 80c+ Agcos] 90cAgtn] 70c Ag sin cos sin 50c A cos 60c+ A cosec 70c + A A A ] g ] g ] g 0. If sin 5, tn r r i { nd < i < r < { < 8tni 5sec {. Trigonometric rtios of compound ngles.. Compound ngles, then find the vlue of When we dd or subtrct ngles, the result is clled compound ngle. i.e.,the lgebric sum of two or more ngles re clled compound ngles For emple, If A, B, C re three ngles then A ± B, A + B + C, A B + C etc., re compound ngles... Sum nd difference formule of sine, cosine nd tngent (i) sin] A+ Bg sina cosb + cosa sinb (ii) sin] A Bg sina cosb cosa sinb (iii) cos] A+ Bg cosa cosb sina sinb (iv) cos] A Bg cosa cosb + sina sinb Emple.6 (v) tn(a+b) (vi) tn(a B) tna+ tn B tnatn B tna tn B + tnatn B If cosa r 5 nd cosb, < A, B < r, find the vlue of (i) sin(a B) (ii) cos(a+b). Since r < A, B < r, both A nd B lie in the fourth qudrnt, ` sina nd sinb re negtive. Given cosa 5 nd cosb, T ri g onom et ry

140 Therefore, sina (i) (ii) Emple.7 cos A Sin B cos B cos(a + B) cosa cosb sina sinb 5 # b 5 l# b 5 l sin(a B) sina cosb cosa sinb 5 5 b lb l b lb 5 l Find the vlues of ech of the following trigonometric rtios. (i) sin5 o (ii) cos( 05 o ) (iii) tn75 o (iv) sec65 o (i) sin5 o sin(5 o 0 o ) sin5ccos 0c cos5csin 0c # # (ii) cos( 05 o ) cos05 o cos(60 o + 5 o ) cos60 o cos5 o sin60 o sin5 o # # (iii) tn75 o tn(5 o + 0 o ) th Std. Business Mthemtics

141 tn5c+ tn 0c tn5ctn 0c (iv) cos 65º cos(80º 5º) Emple.8 cos 5º cos(60º 5º) (cos60º cos5º + sin60º sin 5º) e # + # e + o ` sec65º + # + ^ h sin] A If tna m tnb, prove tht sin ] A + o Bg B g m m + Given tna m tnb sin A cos A sin B m cos B sinacos B cosasin B m Componendo nd dividendo rule: If b d c + b c+ d, then b c d Applying componendo nd dividendo rule, we get sinacos B+ cosasin B m + sinacos B cosasin B m sin] A+ Bg sin] A B g m m +, which completes the proof. T ri g onom et ry

142 Trigonometric rtios of multiple ngles Identities involving sina, cosa, tna, sina etc., re clled multiple ngle identities. (i) sin A sin^a+ Ah sinacos A+ cos Asin A sin Acos A. (ii) cos A cos( A+ A) cos AcosA sin AsinA cos A sin A (iii) sin A sin(a+a) sin AcosA+ cos Asin A ^sin AcosAhcos A+ ^ sin Ahsin A sin Acos A+ sin Asin A sin A^ sin Ah+ sin A sin A sina sin A Thus we hve the following multiple ngles formule. (i) sina sina cosa (ii) tn A sina + tn A. (i) cosa cos A sin A (ii) cosa cos A (iii) cosa sin A (iv) tn A cosa + tn A. tn A tna tn A. sina sina sin A 5. cosa cos A cos A 6. tna tna tn A tn A NOTE (i) cos A + cos A ] g (or) cos A! (ii) sin A cos A ] g (or) sin A! + cos A cos A th Std. Business Mthemtics

143 Emple.9 Show tht sin i + cos i Emple.0 sin i tn i + cos i sinicos i sin i tn i cos i cos i Using multiple ngle identity, find tn60 o tn A tn A tn A Put A 0 o in the bove identity, we get tn60 o tn 0c tn 0c # Emple. If tna 7 nd tnb, show tht cosa sinb Now cosa + tn A tn A sinb sinb cosb # () + tn B tn B # + tn B + tn B 5 () From() nd () we get, cosa sinb. Emple. If tna sin cos B B then prove tht tna tnb T ri g onom et ry 5

144 Consider cos B sin B tna B sin B B sin cos sin cos B B B sin B B tn cos tn B A B A B ` tna tnb Emple. If tn nd tn b 7 then prove tht ^+ bh r. tn+ tn b tn^+ bh tn$ tn b tn # tn tn 9 5 tn^+ bh # b tn r r 6 th Std. Business Mthemtics Eercise.. Find the vlues of the following : (i) cosec5º (ii) sin(05º) (iii) cot75º. Find the vlues of the following r r r r (i) sin76º cos6º cos76º sin6º (ii) sin cos + cos sin (iii) cos70º cos0º sin70º sin0º (iv) cos 5º sin 5º. If sin A 5, 0 < A< r nd cos B, A r r < < find the vlues of the following : (i) cos(a+b) (ii) sin(a B) (iii) tn(a B)

145 If cosa r nd cosb 7 where A, B re cute ngles prove tht A B 5. Prove tht tn80º tn85º tn 5º 6. If cot α, sec β 5, where π < α < tn(α + β). Stte the qudrnt in which α + β termintes. r r nd < b < r, find the vlue of 7. If A+B 5º, prove tht (+tna)(+tnb) nd hence deduce the vlue of tn c 8. Prove tht (i) sin] A+ 60cg+ sin] A 60cg sin A (ii) tnatn AtnA+ tn A+ tna tn A 0 9. (i) If tnθ find tnθ (ii) If sin A, find sina 0. If sina 5, find the vlues of cosa nd tna. Prove tht sin] B Cg sin] C Ag sin] A Bg cosbcos C + cosc cos A + cosacos B 0.. If tna tn B nd cotb cot A y prove tht cot] A Bg + y.. If sin+ sin b nd cos+ cos b b, then prove tht cos(α β). Find the vlue of tn 8 r. + b. 5. If tn 7, sin b Prove tht + b r 0 where 0 < < r 0 < b <.. Trnsformtion formule The two sets of trnsformtion formule re described below. r nd.. Trnsformtion of the products into sum or difference sinacos B+ cos Asin B sin] A+ Bg () T ri g onom et ry 7

146 sinacos B cosasin B sin] A Bg () cosacos B sinasin B cos] A+ Bg () cosacos B+ sinasin B cos] A Bg () Adding () nd (), we get sinacos B sin] A+ Bg+ sin] A Bg Subtrcting () from (), we get cosa sinb sin] A+ Bg sin] A Bg Adding () nd (), we get cosacos B cos] A+ Bg+ cos] A Bg Subtrcting () from (), we get sina sinb cos] A Bg cos] A+ Bg Thus, we hve the following formule; sina cosb sin] A+ Bg+ sin] A Bg cosa sinb sin] A+ Bg sin] A Bg cosacos B cos] A+ Bg+ cos] A Bg sinasin B cos] A Bg cos] A+ Bg Also sin A sin B sin] A+ Bgsin] ABg cos A sin B cos] A+ Bgcos] ABg.. Trnsformtion of sum or difference into product C D C D sinc + sind sinb + lcosb l C D C D sinc sind cosb + lsinb l C D C D cosc + cosd cosb + lcosb l C D C D cosc cosd sinb + lsinb l 8 th Std. Business Mthemtics

147 Emple. Epress the following s sum or difference (i) sinicos i (ii) cosicos i (iii) sinisin i A 5A (iv) cos7i cos 5i (v) cos cos (vi) cos9isin 6i (vii) cosasin 5A (i) sinicos i sin] i+ ig+ sin] i ig sini+ sin i (ii) cosicos i cos] i+ ig+ cos] i ig cosi+ cos i (iii) sinisin i cos] i ig cos] i+ ig cosi cos 6i (iv) cos7i cos 5i 6 cos] i+ 5 ig+ cos] i 5 ig@ cosi+ cos i 5? A 5A (v) cos cos A 5A A 5A cosb + l+ cosb l 8A A ; cos + cosb le cos A cos A 6 + ] g@ cos A+ cos A 5? (vi) cos9isin 6i 6 sin] 9 i+ 6 ig sin] 9 i 6 ig@ 6 sin] 5ig sin i@ (vii) cosasin 5 A sin] A+ 5Ag sin] A 5Ag sin8a+ sin A Emple.5 Show tht sin0csin 0csin 80c 8 T ri g onom et ry 9

148 Emple.6 Emple.7 0 th Std. Business Mthemtics LHS sin0csin 0csin 80c sin 0csin^60c 0chsin^60c+ 0ch sin 0c6 sin 60c sin 0c@ sin 0c8 sin 0cB sin 0 sin 0c: c D sin0c sin 0c sin 60c 8 Show tht sin0csin 0csin60csin 80c 6 L.H.S sin0csin 0csin60csin 80c sin60csin 0csin] 60c 0cgsin] 60c+ 0cg sin 0c^sin 60c sin 0ch sin 0 cb sin 0cl b lb l^sin0c sin 0ch. b lb lsin 60c b lb lb l 6 R.H.S. Prove tht cos A + cos ] A+ 0cg+ cos ] A 0cg cos A + cos ] A+ 0cg+ cos ] A 0cg cos A + cos ] A + 0cg + ^ sin ] A 0c gh (Since cos A sin A) cos A cos A ^ ] + cgh sin ] A 0cg cos A + + cos[ ^ A+ 0 c ) + ( A 0 c h] cos[ ] A+ 0 c g ] A 0 c g]

149 cos A + + cosa cos0 o (Since cos A sin B cos] A+ Bgcos] ABg ) cos A + + ^ cos A hcos] 80 c + 60 c g (since cos A + cos A) cos A + + ^ cos A h] cos 60 c g cos A + + ^cos A hb l cos A+ cos A+ Emple.8 Convert the following into the product of trigonometric functions. (i) sin9a+ sin 7A (ii) sin7i sin i (iii) cos8a+ cos A (iv) cos cos 8 (v) cos0c cos 0c (vi) cos75c+ cos 5c (vii) 9A+ 7A 9A7A (i) sin9a+ sin 7A sinb lcosb l 6A A sinb lcosb l sin8acos A 7i+ i 7ii (ii) sin7i sin i cosb lsinb l i i cosb lsinb l 8A+ A 8AA (iii) cos8a+ cos A cosb lcosb 0A A cosb lcosb l cos0a cos] Ag cos0acos A (iv) cos cos 8 sinb lsinb l sinb lsinb l sin6sin 0c+ 0c 0c0c (v) cos0c cos 0c sinb lsinb l sinb 50 c lsinb 0 c l sin5csin 5c l T ri g onom et ry

150 75c+ 5c 75c5c (vi) cos75c+ cos 5c cosb lcosb l 0c 0c cosb lcosb l cos60ccos 5c (vii) cos55c+ sin 55c cos55c + cos] 90c 55cg Emple.9 cos55c+ cos 5c 55c+ 5c 55c5c cosb lcosb l cos5ccos 0c d n$ cos 0c cos 0c If three ngles A, B nd C re in rithmetic progression, Prove tht sina cotb cos C sina cosc Emple.0 Prove tht sin C cos A sin C cos A A C A+ C sinb lcosb l A+ C AC sinb lsinb l A+ C cosb l A+ C sinb l A cotb + C l cotb (since A, B, C re in A.P., B A + C ) sin5 sin + sin cos5 cos tn sin5 sin + sin cos5 cos sin5+ sin sin cos5 cos sin cos sin cos5 cos sin ] cos g sin sin sin cos sin sin cos tn th Std. Business Mthemtics

151 Emple. Find sin05 o + cos05 o sin05 o + cos05 o Emple. sin(90 o + 5 o )+cos05 o cos5 o + cos05 o cos05 o + cos5 o 05c+ 5c 05c5c cosb lcosb l 0c 90c cosb lcosb l cos60 o cos5 o b Prove tht ^cos+ cos bh + ^sin+ sin bh cos b l + b b cos+ cos b cosb l$ cosb l.() + b b sin+ sin b sinb l$ cosb l.() Squring nd dding () nd () ^cos+ cos bh + ^sin+ sin bh cos + b cos b sin + b cos b b l b l+ b l b l b + b + b cos b l; cos b l+ sin b le cos b b l Eercise.. Epress ech of the following s the sum or difference of sine or cosine: A A (i) sin 8 sin 8 (ii) cos] 60c+ Agsin] 0c+ Ag 7A 5A (iii) cos sin (iv) cos7i sin i T ri g onom et ry

152 Epress ech of the following s the product of sine nd cosine (i) sina + sina (ii) cosa + cosa (iii) sin6i sin i (iv) cosi cos i. Prove tht (i) cos0ccos 0ccos 80c 8 (ii) tn0ctn 0ctn 80c. Prove tht b (i) ^cos cos bh + ^sin sin bh sin b l (ii) sinasin( 60c+ A) sin( 60c A) sin A 5. Prove tht (i) sin] A Bgsin C + sin] B Cgsin A + sin] C Agsin B 0 r 9r r 5r (ii) cos cos + cos + cos 0 6. Prove tht (i) cosa cos A A sina sin A tn (ii) cos7a cos 5A sin7a sin 5A cot A 7. Prove tht cos0ccos 0ccos60ccos 80c 6 8. Evlute (i) cos0 + cos 00 + cos 0 (ii) sin50 sin 70c + sin 0 9. If cosa+ cos B nd sina+ sin B, prove tht tnb A+ B l 0. If sin^y+ z h, sin^z+ yh, sin^+ yzh re in A.P, then prove tht tn, tn y nd tn z re in A.P A+ B. If coseca+ sec A cosecb+ sec B prove tht cotb l tn Atn B th Std. Business Mthemtics

153 Inverse Trigonometric Functions.. Inverse Trigonometric Functions The quntities such s sin, cos, tn functions. etc., re known s inverse trigonometric i.e., if sin i, then i sin. The domins nd rnges (Principl vlue brnches) of trigonometric functions re given below Function Domin Rnge y sin [, ] r r :, D cos [, ] [0, r ] tn r r R ^h b, l, cosec r r R (, ) :, D, y! 0 sec r r R (, ) :, D r, y! cot R (0, r ).. Properties of Inverse Trigonometric Functions Property () (i) sin sin (ii) cos ] cos g (iii) tn tn (iv) cot ] cot g (v) sec sec (vi) cosec ] cosec g Property () (i) sin b l cosec () (ii) cos b l sec () (iii) tn b l cot () (iv) cosec b l sin () (v) sec b l cos () (vi) cot b l tn () T ri g onom et ry 5

154 Property () (i) sin ( ) sin () (iii) tn ( ) tn () (v) sec ( ) r sec () Property () r (i) sin () + cos () r (iii) sec () + cosec () (ii) cos ( ) r cos () (iv) cot ( ) cot () (vi) cosec ( ) cosec () r (ii) tn () + cot () Property (5) If y<, then (i) y tn () + tn () y tn + d n y (ii) y tn () tn () y tn d n + y Property (6) sin () sin y + ^ h sin _ y + y i Emple. Find the principl vlue of (i) sin _ i ( ii) tn ^ h (i) Let sin _ i r r y where # y # ` sin y _ i sin b r 6 l 6 th Std. Business Mthemtics y 6 r The principl vlue of sin _ i r is 6 (ii) Let tn ^ h r r y where # y # ` tn y tnb r l The principl vlue of tn ^ h is y r r

155 Emple. 5 Evlute the following (i) cosbsin 5 (i) Let bsin l i... () sin i 5 cos i sin i 5 69 From () nd (), we get 5 Now cosbsin l cos i (ii) Let bcos 7 8 l i 8 cos i 7 i... () sin i cos () From () nd (), we get Now tnbcos 7 8 l sin i tn i cos i l (ii) tnbcos 7 8 l Emple.5 Prove tht (i) tn 7 tn b l+ b l tn b 9 l (ii) cos 5 tn 5 b l+ b l tn b 7 l (i) tn b 7 l+ tn b l tn 7 > + _ i_ i H 7 T ri g onom et ry 7

156 tn b 90 0 l tn b 9 l (ii) Let cos b 5 l i. Then cos i ` tn i ` cos b 5 l tn b l 5 & i tn b l & sin i 5 Now cos Emple.6 b 5 l+ tn b 5 l tn b l + tn 5 tn + e o # tn b 7 l Find the vlue of tn; r tn b 8 le Emple.7 5 b 5 l Let tn b 8 l i Then tn i 8 ` tn; r tn b 8 le tn b r il Solve tn r tn tn i r + tn tn i b l+ tn b + + l r 8 tn + tn b l b l tn > tn + H b l _ i_ + i Given tht tn b r l+ tn b + + l 8 th Std. Business Mthemtics

157 ` tn b l r r tn & 0 &! Emple.8 Simplify: sin ` j+ sin ` j We know tht sin ] g + sin () y sin 7 y + y A ` sin b l+ sin b l sin : D sin : D Emple.9 Solve tn sin 5 c m 9 + ] + g+ tn ] g tn b l & tn + h + ^ h < F tn ^+ h^ ` h j ^ tn c ^ h m tn ` j & 6 & 9 ` + Emple.0 If tn( + y) nd tn (), then find y T ri g onom et ry 9

158 tn(+y) + y tn () tn ()+y tn () y tn () tn () tn ; + ] # g E tn b 85 0 l y tn : 7 8 D Eercise.. Find the pricipl vlue of the following (i) sin b l (ii) tn ]g (iii) cosec ^h (iv) sec ^ h. Prove tht (i) tn ] g sin c + (ii) tn b l+ tn b 7 l. Show tht tn tn b l+ b l tn b l r r. Solve tn + tn 5. Solve tn + + tn ] g ] g tn b 7 l 6. Evlute (i) cos8tn ` jb (ii) sin8 cos ` 5 jb 7. Evlute: cos sin 5 ` ` j+ sin ` jj 8. m m n Prove tht tn b n l r tn b m+ n l 9. Show tht sin 5 sin 7 8 b l 8 b l cos Epress tn cos : r sin, < < r D, in the simplest form. m 50 th Std. Business Mthemtics

159 ICT Corner Epected finl outcomes Step Open the Browser nd type the URL given (or) Scn the QR Code. GeoGebr Work book clled th BUSINESS MATHS will pper. In this severl work sheets for Business Mths re given, Open the worksheet nmed Inverse Trigonometric Functions Step Inverse Trigonometric Functions pge will open. Click on the check boes on the Lefthnd side to see the respective Inverse trigonometric functions. You cn move nd observe the point on the curve to see the vlue nd yvlue long the es. Sine nd cosine inverse re up to the m limit. But Tngent inverse is hving customised limit. St e p St e p St e p B r o w s e i n t h e l i n k th Business Mths: (or) scn the QR Code T ri g onom et ry 5

160 Eercise.5 Choose the correct nswer. The degree mesure of r 8 is () 0c60l (b) c0l (c) c60l (d) 0c0l. The rdin mesure of 7c0l is 5r () r (b) (c) 7r. If tn i nd i lies in the first qudrnt then cos i is 5 () 6 (b) 6 (c) 5 6 9r (d) (d) 5 6. The vlue of sin 5 o is () + (b) (c) (d) 5. The vlue of sin] 0cg is () (b) (c) (d) 6. The vlue of cos] 80cgis () (b) (c) (d) 7. The vlue of sin8 o cos7 o + cos8 o sin7 o is () (b) (c) (d) 0 8. The vlue of sin5º cos5º is () (b) (c) (d) 9. The vlue of seca sin(70 o + A) is () (b) cos A (c) sec A (d) 0. If sina + cosa then sina is equl to () (b) (c) 0 (d) 5 th Std. Business Mthemtics

161 The vlue of cos 5º sin 5º is () (b) (c) 0 (d). The vlue of sin 5º is () (b) (c) (d) 0. The vlue of cos 0º cos0º is (). The vlue of () (b) tn 0c + tn 0c (b) is (c) (c) (d) (d) 5. If sin A then cos A cos A is () (b) 0 (c) 6. The vlue of () tn0c tn 0c tn 0c (b) 7. The vlue of cosec d n is r r () (b) 8. sec + cosec () r r (b) 9. If nd b be between 0 nd then sin is () If tn A nd tn B is (c) (c) r (c) r (d) (d) (d) r 6 (d) r r nd if cos^+ bh nd sin^ bh 5 56 (b) 0 (c) 65 then tn^ A+ B h is equl to 6 (d) 65 () (b) (c) (d). tn b r l is () b + tn tn l (b) tn tn b + l (c) tn (d) + tn T ri g onom et ry 5

162 sinbcos () 5 5 lis. The vlue of () (b) 5 cosec] 5cg is (b) (c) 5 (d) 5 (c) (d). If p sec50c tn 50c then p is () cos50c (b) sin50c (c) tn50c (d) sec50c cos 5. b cosec l sin cos is () cos sin (b) sin cos (c) (d) 0 Miscellneous Problems. Prove tht cos+ cos + cos sin+ sin + sin cot. Prove tht cosec0c sin 0c. Prove tht cot ] sin5+ sin g cot ] sin5 sin g.. If r tn, r < < then find the vlue of sin nd cos 5. Prove tht r r r sin + cos + sec 0 6. Find the vlue of (i) sin75c (ii) tn5c 7. If sina, sin B then find the vlue of sin] A+ Bg, where A nd B re cute ngles. 8. Show tht cos sin 5 56 b l+ b l sin b 65 l 9. If cos( + b) 5 nd sin( b) 5 where + b nd bre cute, then find tn 0. Epress tn cos sin b cos+ sin l, 0 < < r in the simplest form. 5 th Std. Business Mthemtics

163 Summry If in circle of rdius r, n rc of length l subtends n ngle of θ rdins, then l r θ sin + cos tn + sec cot + cosec sin(α + β) sin α cos β + cos α sin β sin(α β) sin α cos β cos α sin β cos(α + β) cos α cos β sin α sin β cos^ bh cos $ cosb+ sin sin b tn+ tn b tn^+ bh tntn b tn tn b tn^ bh + tntn b sin sin cos cos cos sin sin cos tn tn tn sin sin sin cos cos cos tn tn tn tn + y y sin + sin y sin cos + y y sin sin y cos sin + y y cos+ cos y cos cos + y y cos cos y sin sin sin cos y 6 sin^+ yh+ sin^yh@ cos sin y 6 sin^+ yhsin^yh@ coscos y 6 cos^+ yh+ cos^yh@ sin sin y 6 cos^yh cos^+ yh@ T ri g onom et ry 55

164 GLOSSARY Allied ngles Angle Componendo nd dividendo. Compound ngle Degree mesure Inverse function Length of the rc. Multiple ngle Qudrnts Rdin mesure Trnsformtion formule Trignometric identities Trignometry Rtios த மணக சக ணஙகள சக ணம க ட டல றற ம கழ ததல வ க த ம க டட க சக ணம ப மக அளமவ சநர ற ர ப வ ல ல ன ந ளம டஙக சக ணம க ல பக த கள சரட யன அளவ / ஆமரயன அளவ உர றற ச தத ரஙகள த சக ண த ம றவற ர ம கள த சக ண த வ க தஙகள 56 th Std. Business Mthemtics

165 Chpter 5 DIFFERENTIAL CALCULUS Lerning Objectives Introduction After studying this chpter, the students will be ble to understnd the concept of limit nd continuity formul of limit nd definition of continuity bsic concept of derivtive how to find derivtive of function by first principle derivtive of function by certin direct formul nd its respective ppliction how to find higher order derivtives Clculus is Ltin word which mens tht Pebble or smll stone used for clcultion. The word clcultion is lso derived from the sme Ltin word. Clculus is primry mthemticl tool for deling with chnge. The concept of derivtive is the bsic tool in science of clculus. Clculus is essentilly concerned with the rte of chnge of dependent vrible with respect to n independent vrible. Sir Issc Newton (6 77 CE) nd the Germn mthemticin G.W. Leibnitz (66 76 CE) invented nd developed the subject independently nd lmost simultneously. Isc Newton In this chpter we will study bout functions nd their grphs, limits, derivtives nd differentition technique. 5. Functions nd their grphs Some bsic concepts 5.. Quntity Anything which cn be performed on bsic mthemticl opertions like ddition, subtrction, multipliction nd division is clled quntity. G.W. Leibnitz i e enti cu us 57

166 5.. Constnt A quntity which retins the sme vlue throughout mthemticl investigtion is clled constnt. Bsiclly constnt quntities re of two types (i) Absolute constnts re those which do not chnge their vlues in ny mthemticl investigtion. In other words, they re fied for ever. Emples:,, π,... (ii) Arbitrry constnts re those which retin the sme vlue throughout problem, but we my ssign different vlues to get different solutions. The rbitrry constnts re usully denoted by the letters, b, c, Vrible Emple: In n eqution ym+, m is clled rbitrry constnt. A vrible is quntity which cn ssume different vlues in prticulr problem. Vribles re generlly denoted by the letters, y, z,... Emple: In n eqution of the stright line y +, b nd y re vribles becuse they ssumes the coordintes of moving point in stright line nd thus chnges its vlue from point to point. nd b re intercept vlues on the es which re rbitrry. There re two kinds of vribles (i) A vrible is sid to be n independent vrible when it cn hve ny rbitrry vlue. (ii) A vrible is sid to be dependent vrible when its vlue depend on the vlue ssumed by some other vrible. Emple: In the eqution y 5 +, is the independent vrible, y is the dependent vrible nd is the constnt. 5.. Intervls The rel numbers cn be represented geometriclly s points on number line clled rel line. The symbol R denotes either the rel number system or the rel line. A subset of the rel line is n intervl. It contins tlest two numbers nd ll the rel numbers lying between them. A Fig. 5. B b 58 th Std. Business Mthemtics

167 (i) Open intervl The set { : < < b} is n open intervl, denoted by (, b). In this intervl, the boundry points nd b re not included. For emple, in n open intervl (, 7), nd 7 re not elements of this intervl. A B ( ) Fig. 5. b But,.00 nd 6.99 re elements of this intervl. (ii) Closed intervl The set { : < < b} is closed intervl nd is denoted by [, b]. In the intervl [, b], the boundry points nd b re included. For emple, in n intervl [, 7], nd 7 re lso elements of this intervl. A B [ ] Fig. 5. b Also we cn mke mention bout semi closed or semi open intervls. (, b] { : < < b} is clled left open intervl nd [, b) { : < < b} is clled right open intervl. In ll the bove cses, b h is clled the length of the intervl Neighbourhood of point Let be ny rel number. Let ε > 0 be rbitrrily smll rel number. Then ( ε, + ε ) is clled n ε neighbourhood of the point nd denoted by N, ε For emples, 5 b5, 5 + l N, N, 9 & : < < 0, + 5 b l & : < < i e enti cu us 59

168 Function Let X nd Y be two nonempty sets of rel numbers. If there eists rule f which ssocites to every element! X, unique element y! Y, then such rule f is clled function (or mpping) from the set X to the set Y. We write f : X $ Y. The function nottion y f() ws first used by Leonhrd Euler in 7 75 The set X is clled the domin of f, Y is clled the codomin of f nd the rnge of f is defined s f(x) {f() /! X}. Clerly f(x) Y. A function of is generlly denoted by the symbol f(), nd red s function of or f of Clssifiction of functions Functions cn be clssified into two groups. (i) Algebric functions Algebric functions re lgebric epressions using finite number of terms, involving only the lgebric opertions ddition, subtrction, multipliction, division nd rising to frctionl power. Emples : y , y, + + functions. y + 6 re lgebric () y is polynomil or rtionl integrl function. + 7 (b) y is rtionl function. + + (c) y + 6 is n irrtionl function. (ii) Trnscendentl functions A function which is not lgebric is clled trnscendentl function. Emples: sin, sin, e, log re trnscendentl functions. () sin, tn,... re trigonometric functions. (b) sin, cos,... re inverse trigonometric functions. (c) e,,,... re eponentil functions. (d) log, log e (sin ),... re logrithmic functions. 60 th Std. Business Mthemtics

169 Even nd odd functions A function f() is sid to be n even function of, if f( ) f( ). A function f() is sid to be n odd function of, if f( ) f( ). Emples: f() nd f() cos re even functions. f() nd f() sin re odd functions. NOTE f() + 5 is neither even nor odd function 5..9 Eplicit nd implicit functions A function in which the dependent vrible is epressed eplicitly in terms of some independent vribles is known s eplicit function. Emples: y + nd y e + e re eplicit functions of. If two vribles nd y re connected by the reltion or function f(, y) 0 nd none of the vrible is directly epressed in terms of the other, then the function is clled n implicit function. Emple: + y y 0 is n implicit function Constnt function If k is fied rel number then the function f() given by f() k for ll! R is clled constnt function. Emples: y, f() 5 re constnt functions. Grph of constnt function y f() is stright line prllel to is 5.. Identity function A function tht ssocites ech rel number to itself is clled the identity function nd is denoted by I. i.e. A function defined on R by f() for ll R is n identity function. Emple: The set of ordered pirs {(, ), (, ), (, )} defined by f : A " A where A{,, } is n identity function. Grph of n identity function on R is stright line pssing through the origin nd mkes n ngle of 5cwith positive direction of is i e enti cu us 6

170 5.. Modulus function if $ 0 A function f() defined by f(), where ) is clled the modulus if 0 function. NOTE It is lso clled n bsolute vlue function. 5 5, 5 Remrk Domin is R nd rnge set is [0, ) ( 5) Signum function A function f() is defined by f() if! 0 * is clled the Signum function. 0 if 0 Remrk Domin is R nd rnge set is {, 0, } 5.. Step function (i) Gretest integer function The function whose vlue t ny rel number is the gretest integer less thn or equl to is clled the gretest integer function. It is denoted by. i.e. f : R$ R defined by f() is clled the gretest integer function. NOTE.5,., 0.7 0, 0.,. (ii) Lest integer function The function whose vlue t ny rel number is the smllest integer greter thn or equl to is clled the lest integer function. It is denoted by. i.e. f : R$ R defined by f() is clled the lest integer function NOTE.7 5, 7. 7, 5 5, Remrk Function s domin is R nd rnge is Z [set of integers] 5..5 Rtionl function p () A function f() is defined by the rule f (), q ( )! 0 is clled rtionl q () function. + Emple: f (),! is rtionl function. 6 th Std. Business Mthemtics

171 Polynomil function For the rel numbers 0,,,..., n ; 0! 0 nd n is nonnegtive integer, function f() given by f n n n ] g n is clled s polynomil function of degree n. Emple: f () is polynomil function of degree Liner function For the rel numbers nd b with! 0, function f() + b is clled liner function. Emple: y + is liner function Qudrtic function For the rel numbers, b nd c with! 0, function f() + b + c is clled qudrtic function. Emple: f () + 7 is qudrtic function Eponentil function A function f(),! nd > 0, for ll R is clled n eponentil function Remrk Domin is R nd rnge is (0, ) nd (0, ) is point on the grph Emples: e, e + nd re eponentil functions Logrithmic function For > 0, > 0 nd!, function f() defined by f() log is clled the logrithmic function. Remrk Domin is (0, ), rnge is R nd (, 0) is point on the grph Emple: f() log e (+), f() log e (sin) re logrithmic functions. 5.. Sum, difference, product nd quotient of two functions If f() nd g() re two functions hving sme domin nd codomin, then (i) (f + g) () f() + g() (ii) (fg)() f() g() i e enti cu us 6

172 f (iii) c m() g f (), g ( )! 0 g () (iv) (cf)() cf(), c is constnt. 5.. Grph of function The grph of function is the set of points _, f] gi where belongs to the domin of the function nd f() is the vlue of the function t. To drw the grph of function, we find sufficient number of ordered pirs _, f] gi belonging to the function nd join them by smooth curve. NOTE It is enough to drw digrm for grph of function in white sheet itself. Emple 5. Find the domin for which the functions f() nd g() re equl. Given tht f() g() Emple 5. & + 0 (+) ( ) 0 `, Domin is $,. f {(, ), (, )} be function described by the formul f() +b. Determine nd b? Given tht f() + b, f() nd f() ` + b nd + b & nd b 6 th Std. Business Mthemtics

173 Emple 5. Emple 5. If f() +, then show tht [f()] f( ) + f ` j f() +, f( ) + nd f` j f() Now, LHS [f()] 8 + B + + ` + j f ( ) + f () f ( ) + f` j RHS Let f be defined by f() 5 nd g be defined by 5 g() + 5 if! * 5 Find λ such tht f() g() for ll R m if 5 Given tht ` f() g() for ll R f( 5) g( 5) 5 5 λ ` λ 0 Emple 5.5 If f(), show tht f ^ h$ fy ^ h f ^ + yh f() (Given) ` f( + y) +y. y f ^ h$ fy ^ h i e enti cu us 65

174 Emple 5.6 If f( ), 0,,, then show tht + f( ) (Given) + f] g ` f 7f] ga f] g + Emple 5.7 If + f( ) log, f( ) log (Given) + ` f c m + log f[ f( )] show tht f f( ) log + ( + ) log ( ) + log f( ). Emple 5.8 If f() nd g(), then find (i) (f+g)() (ii) (f g)() (iii) (fg)() (i) (ii) (f + g) () f() + g() + + if $ 0 ) ) if < 0 0 (f g) () f() g() if $ 0 if < 0 if $ 0 0 ) ) ( ) if < 0 if $ 0 if < 0 66 th Std. Business Mthemtics

175 (iii) (f g) () f() g() ) if $ 0 if < 0 Emple 5.9 A group of students wish to chrter bus for n eductionl tour which cn ccommodte tmost 50 students. The bus compny requires t lest 5 students for tht trip. The bus compny chrges ` 00 per student up to the strength of 5. For more thn 5 students, compny chrges ech student ` 00 less 5 times the number more thn 5. Consider the number of students who prticiptes the tour s function, find the totl cost nd its domin. Let be the number of students who prticipte the tour. Then 5 < < 50 nd is positive integer. Formul : Totl cost (cost per student) (number of students) (i) If the number of students re between 5 nd 5, then the cost per student is ` 00 ` The totl cost is y 00. (ii) If the number of students re between 6 nd 50, then the cost per student is ` $ 00 ( 5) ` The totl cost is y ` 09 j ; 5 # # 5 ` y * 09 nd ; 6 # # 50 5 the domin is {5, 6, 7,..., 50}. Emple 5.0 Drw the grph of the function f() y f() ) if $ 0 if < 0 i e enti cu us 67

176 Choose suitble vlues for nd determine y. Thus we get the following tble. 0 y 0 Tble : 5. Plot the points (, ), (, ), (0, 0), (, ), (, ) nd drw smooth curve. y (, ) (, ) (, ) (, ) y, < 0 y, > 0 The grph is s shown in the figure Emple 5. l (0, 0) yl Fig. 5. Drw the grph of the function f() 5 Let y f() 5 Choose suitble vlues for nd determine y. Thus we get the following tble. 0 y 5 Tble : 5. y Plot the points (, ), (, ), (, ), (0, 5), (, ), (, ), (, ) nd drw smooth curve. (, ) (, ) y 5 The grph is s shown in the figure Emple 5. Drw the grph of f(),! nd > 0 l (, ) (, ) (0, 5) yl Fig 5.5 (, ) (, ) We know tht, domin set is R, rnge set is (0, ) nd the curve pssing through the point (0, ) 68 th Std. Business Mthemtics

177 Cse (i) w h e n > < if < 0 y f() * if 0 > if > 0 We noticed tht s increses, y is lso increses nd y > 0. ` The grph is s shown in the figure. y y, > l (0, l) o yl Fig 5.6 Cse (ii) w h e n 0 < < y f() > if < 0 * if 0 < if > 0 We noticed tht s increses, y decreses nd y > 0. ` The grph is s shown in the figure. y, 0<< y NOTE l (0, l) O yl Fig 5.7 (i) The vlue of e lies between nd. i.e. < e < (ii) The grph of f() e is identicl to tht of f() if > nd the grph of f() e is identicl to tht of f() if 0 < < Emple 5. Drw the grph of f() log ; > 0, > 0 nd!. i e enti cu us 69

178 We know tht, domin set is (0, ), rnge set is R nd the curve pssing the point (, 0) Cse (i) when > nd > 0 < 0 if 0 < < f () log * 0 if > 0 if > We noticed tht s increses, f() is lso increses. ` The grph is s shown in the figure. y f() log, > l O (, 0) yl Fig 5.8 Cse (ii) when 0 < < nd > 0 > 0 if 0 < < f () log * 0 if < 0 if > We noticed tht increses, the vlue of f() decreses. ` The grph is s shown in the figure. y f() log, 0<< l O (, 0) 70 th Std. Business Mthemtics yl Fig 5.9

179 ICT Corner Epected finl outcomes Step Open the Browser nd type the URL given (or) Scn the QR Code. GeoGebr Work book clled th BUSINESS MATHS will pper. In this severl work sheets for Business Mths re given, Open the worksheet nmed Types of Functions Step Types of Functions pge will open. Click on the check boes of the functions on the Righthnd side to see the respective functions. You cn move the slider nd chnge the function. Observe ech function. St e p St e p St e p B r o w s e i n t h e l i n k th Business Mths: (or) scn the QR Code i e enti cu us 7

180 Eercise 5.. Determine whether the following functions re odd or even? (i) f() (ii) f ] g log_ + + i + (iii) f() sin + cos (iv) f() (v) f() +. Let f be defined by f() k +, R. Find k, if f is n odd function.. If f (). If f () 5. For f () 6. If f (), then show tht f () + f` j 0 +, then prove tht f^f() h +, write the epressions of f ` j nd e nd g () log e, then find f () (i) (f+g)() (ii) (fg)() (iii) (f)() (iv) (5g)() 7. Drw the grph of the following functions : (i) f () (iv) f () 6 (ii) f () (iii) f () e (v) f () 5. Limits nd derivtives e (vi) f () In this section, we will discuss bout limits, continuity of function, differentibility nd differentition from first principle. Limits re used when we hve to find the vlue of function ner to some vlue. Definition 5. Let f be rel vlued function of. Let l nd be two fied numbers. If f() pproches the vlue l s pproches to, then we sy tht l is the limit of the function f() s tends to, nd written s lim f() l " Nottions: (i) (ii) L[f()] lim f() is known s left hnd limit of f() t. " R[f()] lim f() is known s right hnd limit of f() t. " + 7 th Std. Business Mthemtics

181 5.. Eistence of limit lim f() eist if nd only if both L[f()] " nd R[f()] eist nd re equl. i.e., lim f() eist if nd only if L[f()] " R[f()]. NOTE L[f()] nd R[f()] re shortly written s L[f()] nd R[f()] 5.. Algorithm of left hnd limit : L[f()] (i) Write lim f(). " (ii) Put h, h > 0 nd replce " by h"0. (iii) Obtin lim f( h). h " 0 (iv) The vlue obtined in step (iii) is clled s the vlue of left hnd limit of the function f() t. 5.. Algorithm of right hnd limit : R[f()] (i) Write lim f(). " + (ii) Put +h, h>0 nd replce " + by h"0. (iii) Obtin lim f(+h). h " 0 (iv) The vlue obtined in step (iii) is clled s the vlue of right hnd limit of the function f() t. Emple 5. Evlute the left hnd nd right hnd limits of the function if! f() * t. 0 if L[f()] lim f() " lim f( h), h h " 0 lim ( h) h " 0 ( h ) i e enti cu us 7

182 h lim h " 0 h lim h 0 h " h lim h " 0 R[f()] lim f() " + Emple 5.5 lim f( + h) h " 0 lim ( + h) h " 0 ( + h ) lim h h " 0 h lim h " 0 Verify the eistence of the function f() NOTE Here, L[f()]! R[f()] ` lim f() does not eist. " 5 if 0 # ) t. if L[f()] lim f() " lim f( h), h h " 0 lim [5( h) ] h " 0 lim ( 5h) h " 0 R[f()] lim f() " + lim f( + h), + h h " 0 lim [( + h) ( + h)] h " 0 () () Clerly, L[f()] R[f()] ` lim f() eists nd equl to. " 7 th Std. Business Mthemtics

183 NOTE Let be point nd f() be function, then the following stges my hppen (i) lim f() eists but f() does not eist. " (ii) The vlue of f() eists but lim f() does not eist. " (iii) Both lim f() nd f() eist but re unequl. " (iv) Both lim f() nd f() eist nd re equl. " 5.. Some results of limits If lim f() nd lim g() eists, then " " (i) lim (f+g)() lim f() + lim g() " " " (ii) lim (fg)() lim f() lim g() " " " f lim f () (iii) limc g m " ^h lim g (), whenever lim g ( )! 0 " " " (iv) lim k f() k lim f(), where k is constnt. " " 5..5 Indeterminte forms nd evlution of limits Let f() nd g() re two functions in which the limits eist. f ^ h If lim f() lim g() 0, then tkes " " g ^ h f_ i imply tht lim is meningless. " g_ i 0 0 form which is meningless but does not In mny cses, this limit eists nd hve finite vlue. Finding the solution of such limits re clled evlution of the indeterminte form. 0 The indeterminte forms re 0,, 0#,, 0 0, 0 nd. 0 Among ll these indeterminte forms, 0 is the fundmentl one Methods of evlution of lgebric limits (i) Direct substitution (ii) Fctoriztion (iii) Rtionlistion (iv) By using some stndrd limits i e enti cu us 75

184 Some stndrd limits (i) lim " (ii) lim sin i 76 th Std. Business Mthemtics n i " 0 i (iii) lim " 0 ( ) (iv) lim log + " 0 n nn if n! Q lim tn i i " 0 i log e if >0 (v) lim + " 0 e (vi) lim ` + j e " e (vii) lim Emple 5.6 " 0 Evlute: lim( + 5). " lim( + 5) " () + () 5 Emple Evlute: lim +. " " + 6 lim + Emple 5.8 lim^ " + 6h lim^ + h " ^h ^h sin cos Evlute : lim cos + sin. " r 5sin cos lim cos + sin " r, θ is in rdin. lim^5sin cos h r " lim^cos + sin h r "

185 Emple 5.9 Evlute: lim. " r 5 sin r cos r r cos + sin 5 ^ h 0 ^ h 0 ^ h + ^ h lim 0 is of the type " 0 lim ^ h^ + + h lim ^ h " " lim^ + + h " () Emple 5.0 Evlute: lim " 0 +. lim " 0 + ^ + h^ + + h lim " 0 ^ + + h + lim " 0 ^ + + h lim " 0 ^ + + h. + Emple 5. Evlute: lim " lim " lim. 5 5 " ^ h 5 ^h i e enti cu us 77

186 Emple 5. sin Evlute: lim sin 5. " 0 sin lim sin 5 " 0 Emple 5. sin # lim 0 sin 5 # 5 5 lim sin 5 " 0 lim sin 5 5 " * " 0 5 ` j 6 5 Evlute: lim " lim " " lim Emple 5. Evlute: lim tn` j. " Let y nd y " 0 s " lim tn` j lim tn y y " y " 0. Emple 5.5 / n Evlute: lim n n. " n lim n n 78 th Std. Business Mthemtics nn ^ + h^n + h " / lim n " 6n

187 lim n n n 6 $ + + ` n j` n j` n j. n " 6 lim $ ` + nj` + nj. n " 6 ^ h^ h. Emple 5.6 log^ + h Show tht lim 0 sin. " log^ + h lim 0 sin " lim log ^ + h ' # " 0 sin lim log ^ + h # " 0 lim` sin j # " 0 Eercise 5.. Evlute the following (i) lim + + " (iv) lim " (ii) lim " (v) lim " (iii) lim n n " / (vi) lim sin " If lim + lim^ + 6h, find the vlues of. " ". If lim " n 7 n 8, then find the lest positive integer n. 8. If f^h 5, then find lim f^h " + b 5. Let f^h +, If lim f^h nd lim f^h, then show tht f( ) 0. " 0 " i e enti cu us 79

188 Derivtive Before going into the topic, let us first discuss bout the continuity of function. A function f() is continues t if its grph hs no brek t If there is ny brek t the point then we sy tht the function is not continuous t the point. If function is continuous t ll the point in n intervl, then it is sid to be continuous in tht intervl Continuous function A function f() is continuous t if (i) f() eists (ii) lim f() eists " (iii) lim f() f() " Observtion In the bove sttement, if tlest ny one condition is not stisfied t point by the function f(), then it is sid to be discontinuous function t 5..9 Some properties of continuous functions Let f() nd g() re two rel vlued continuous functions t, then (i) f() + g() is continuous t. (ii) f() g() is continuous t. (iii) kf() is continuous t, k be rel number. (iv) (v) (vi) f^h is continuous t, if f() ] 0. f^h g^h is continuous t, if g() ] 0. f^h is continuous t. 80 th Std. Business Mthemtics

189 Emple 5.7 Show tht f() 5, if 0 # ) is continuous t, if < L7f] ga lim f] g " lim f] hg, h h " 0 lim 65] h " 0 5() R7f] ga lim f] g " + lim f] + hg, + h h " 0 lim 7] + hg ] hga h " 0 () () Now, f() 5() 5 ` lim f] g lim f] g f(), " " + f() is continuous t. Emple 5.8 Verify the continuity of the function f() given by if< f] g ) t. + if$ L7f] ga lim f] g " lim f] hg, h h " 0 lim " ] hg, h $ 0 lim h 0 h " 0 i e enti cu us 8

190 R7f] ga lim f] g + + lim f] + hg, + h h " 0 lim" + ] + hg, h " 0 lim] + hg h " 0 f] g Now, f() +. Here lim f] g! lim $ $ Hence f() is not continuous t. Z () if < ] If f() [ k if then ] b() if > \ L[f()] lim f () lim () " " R[f()] lim f () lim () b + " " It is pplicble only when the function hs different definition on both sides of the given point. Observtions (i) A constnt function is everywhere continuous. (ii) The identity function is everywhere continuous. (iii) A polynomil function is everywhere continuous. (iv) The modulus function is everywhere continuous. (v) The eponentil function, >0 is everywhere continuous. (vi) The logrithmic function is continuous in its domin. (vii) Every rtionl function is continuous t every point in its domin. Eercise 5.. Emine the following functions for continuity t indicted points () f] g *, if! t 0, if (b) f] g * 9, if! t 6, if. Show tht f() is continuous t Differentibility t point Let f() be rel vlued function defined on n open intervl (, b) nd let 8 th Std. Business Mthemtics

191 f] g f] cg c! (, b). f() is sid to be differentible or derivble t c if nd only if lim c eists finitely. " c This limit is clled the derivtive (or) differentil coefficient of the function f() t c nd is denoted by (or) D f(c) (or) : d d _ f] gid. Thus. 5.. Left hnd derivtive nd right hnd derivtive f] g f] cg f] c hg f] cg (i) lim c " c or lim h h " 0 is clled the left hnd derivtive of f() t c nd it denoted by f l^c h or L[ f l (c)]. f] g f] cg f] c+ hg f] cg (ii) lim c " c or lim h is clled the right hnd derivtive of + h " 0 f() t c nd it denoted by f l^c + h or R[ f l (c)]. Result f() is differentible t c, L7fl] cga R7fl] cga c Remrks: (i) If L7fl] cga! R7fl] cga, then we sy tht f() is not differentible t c. (ii) f() is differentible t c f() is continuous t c. A function my be continuous t point but my not be differentible t tht point. Emple 5.9 Show tht the function f() is not differentible t 0. f() ) if if $ 0 < 0 L[ f l (0)] f] g f] 0g lim, " 0 0 f] 0 hg f] 0g lim h " 0 0h 0, 0 h i e enti cu us 8

192 lim f] hg f] 0g h h " 0 lim h 0 h h " 0 h lim h h " 0 h lim h h " 0 lim] g h " 0 f] g f] 0g R[ f l (0)] lim 0 " 0 + f] 0 + hg f] 0g lim h " 0 0+ h 0, 0+h lim h " 0 lim h " 0 f] hg f] 0g h h 0 h h lim h " 0 h h lim h " 0 h lim h " 0 Here, L[ f l (0)]! R[ f l (0)] ` f() is not differentible t 0 Emple 5.0 Show tht f() differentible t nd find f l (). f() 8 th Std. Business Mthemtics

193 f] g f] g L[ f l ()] lim " f] hg f] g lim h " 0 h ] hg lim h h " 0 + h h lim h h " 0 lim] h + g h " 0 f] g f] g R[ f l ()] lim " + f] + hg f] g lim h " 0 + h ] + hg lim h h " 0 + h + h lim h h " 0 lim] h + g h " 0 Here, L[ f l ()] R[ f l ()] ` f() is differentible t nd f l () 5.. Differentition from first principle The process of finding the derivtive of function by using the definition tht f] + hg f] g f l] g lim h $ 0 h is clled the differentition from first principle. It is dy convenient to write f l() s d. i e enti cu us 85

194 5.. Derivtives of some stndrd functions using first principle d. For! R, ^ d h n Proof: n n Let then f() n f(+h) (+h) n d d _ f] gi lim h " 0 f] + hg f] g h n ] + hg ] g lim h h " 0 n ] + hg ] g lim h " 0 ] + h g n n z lim z z" n n, where z + h nd z " s h " 0 n n n < lim n F " n n i.e., d d ^ h n n n. d d ^e h e proof: Let f] g e then f(+h) e + h d d _ f] gi lim h " 0 f] + hg f] g h e lim h " 0 + h e h e e e lim h h " 0 lim e h " 0 h h ; e E h 86 th Std. Business Mthemtics

195 e e lim; E h h " 0 h e # < e lim; E h F e h " 0 h i.e., d d ^ h e. e. log d d ^ h Proof: e Let Then, f() log e f(+h) log e (+h) d d _ f] gi lim h " 0 f] + hg f] g h loge] + hg loge lim h h " 0 log lim h " 0 log lim h " 0 e e b + h b h log lim h " 0 e + b h h h + l l h l # [ lim log e ] + g ] " 0 Emple 5. d Find d ^ h from first principle. i e enti cu us 87

196 Let f() then f] + hg ] + hg d d _ f] gi lim h " 0 f] + hg f] g h ] + hg lim h h " 0 h+ h + h lim h h " 0 h " 0 lim^ + h+ h h + () 0 + () 0 Emple 5. d d ^ h d Find d Let f() e ^ e h from first principle. then f( + h) e (+h) d d _ f] gi lim h " 0 f] + hg f] g h e lim h " 0 ] + hg e h e $ e e lim h h " 0 h e ^e h lim h h " 0 h 88 th Std. Business Mthemtics

197 e e lim h " 0 h h e # e [ lim ] e " 0 Eercise 5.. Find the derivtive of the following functions from first principle. (i) (ii) e (iii) log(+) 5. Differentition techniques In this section we will discuss bout different techniques to obtin the derivtives of the given functions. 5.. Some stndrd results [formule] d d ^ h n n n d d () d d (k) 0, k is constnt d d (k) k, k is constnt d d b l d d ^ h d d (e ) e d d (e ) e d d (e +b ) e +b d d 0. _ e i e. d d ^ h loge i e enti cu us 89

198 d. ^log d h e d. log d e ] + g + d. _ loge ] + bgi d d 5. ^logh d log 6. sin d d ] g cos e ] + bg d d (cos) sin d d (tn) sec d d (cot) cosec d d (sec) sec tn d d (cosec ) cosec cot 5.. Generl rules for differentition (i) Addition rule d d 7f] g+ g] ga d d 7f] ga+ d d 7g] ga (ii) Subtrction rule (or) Difference rule d d 7f] g g] ga d d 7f] ga d d 7g] ga (iii) Product rule d d 7f] g$ g] ga d d f g f d 7 ] ga ] g+ ] g 7 g ] ga d d (or) If u f() nd v g() then, ] uv g d u d v d v d ] g+ ] u g d (iv) Quotient rule d d f g d f d f d ] g ] g 7 ] ga ] g 7 g ] ga G d g ] g 7g] ga d u (or) If u f() nd v g() then, b d v l v d ] u g d u d ] v g d v 90 th Std. Business Mthemtics

199 (v) Sclr product d d 7 cf ] ga c d 7 f ] ga, d where c is constnt. (vi) Chin rule: d d 7 f _ g ] gia f l_ g ] gi $ g l] g (or) If y f(t) nd t g() then, Here we discuss bout the differentition for the eplicit functions using stndrd results nd generl rules for differentition. Emple 5. dy d dy dt Differentite the following functions with respect to. (i) (ii) 7e (iii) + (iv) sin (v) sin (vi) + + $ d dt (i) d d ` j (ii) d d ^7e h 7 d d (iii) Differentiting y dy d ^e h 7e + with respect to ] + g d ] g d d ] g ] + g d + ] g ] + g] g ] g] g + ] g (iv) Differentiting y sin with respect to dy d d ] sin g d d + sin ^ d h cos+ sin ] cos+ sin g 6 ] + g i e enti cu us 9

200 (v) y sin (or) (sin ) Let u sin then y u Differentiting u sin with respect to We get, du d cos Differentiting y u with respect to u dy We get, du u sin dy dy du d du $ d sin $ cos (vi) y + + Let u + + then y u Differentiting u + + with respect to We get, du d + Differentiting y u with respect to u. dy We get, du u dy dy d du + + du $ d $ ] + g sin log sin log log d d d d ^ ^ hh ^ ^ hh d log d ^ h ^ h d d 5 ^ + + h d^ d ^ + + h + + h d d ^ + + h 5 d d ^e h d ^ h e d d d ^ h ^ + ^ + h h h ^ th Std. Business Mthemtics

201 Emple 5. If f() n nd f l] g 5, then find the vlue of n. f() n ` f l] g n n Emple 5.5 f l] g n() n f l] g n f l] g 5 (given) ( n 5 If y nd u u dy 9, find d. y u u dy du u dy du ^ 9h ( u 9) u 9 du d Now, dy dy d du du $ d ^ 9h $ ^ 9h. i e enti cu us 9

202 ICT Corner Epected finl outcomes Step Open the Browser nd type the URL given (or) Scn the QR Code. GeoGebr Work book clled th BUSINESS MATHS will pper. In this severl work sheets for Business Mths re given, Open the worksheet nmed Elementry Differentition Step Elementry Differentition nd Functions pge will open. Click on the function check boes on the Righthnd side to see the respective grph. You cn move the sliders, b, c nd d to chnge the coefficient nd work out the differentition. Then click the Derivtives check bo to see the nswer nd respective grphs St e p St e p St e p B r o w s e i n t h e l i n k th Business Mths: (or) scn the QR Code 9 th Std. Business Mthemtics