REAL NUMBERS C L A S S E S

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1 TOPICS PAGES. Rel Numers - 5. Liner Equtions in Two Vriles-I 6 -. Liner Equtions in Two Vriles-II Polnomils Qudrtic Equtions Arithmetic Progression Co-ordinte Geometr Tringles 6-76 C L A S S E S... the support REAL NUMBERS. DIVISIBILITY : A non-zero integer is sid to divide n integer if there eists n integer c such tht c. The integer is clled dividend, integer is known s the divisor nd integer c is known s the quotient. For emple, 5 divides 5 ecuse there is n integer 7 such tht If non-zero integer divides n integer, then it is written s nd red s divides, / is written to indicte tht is not divisile.. EUCLID S DIVISION LEMMA : E. E. Let nd e n two positive integers. Then, there eists unique integers q nd r such tht r, where 0 r. If, thn r 0. Show tht n positive odd integer is of the form 6q or, 6q or, 6q 5, where q is some integer. Let e n positive integer nd 6. Then, Euclid s division lemm there eists integers nd r such tht 6q r, where 0 r < 6. 6q or, 6q or, 6q or, 6 or, 6q 4 or, 6q 5. [ 0 r < 6 r 0,,,,4,5] 6q or, 6q or, 6q 5. [ is n odd integer, 6q, 6q, 6q 4] Hence, n odd integer is of the form 6q or, 6q or, 6q 5. Use Euclid s Division Lemm to show tht the cue of n positive integer is of the form 9m, 9m or 9 m 8, for some integer q. Sol, Let e n positive integer. Then, it is of the form q or, q or,. Cse - I When q (q) 7q 9(q ) 9m, where m 9q Cse - II when q (q ) q 7q 9q IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

2 9q (q q ) 9m, where m q (q q ). Cse -III when q (q ) 7q 54q 6q 8 9q(q 6q 4) 8 9m 8, where m q 6q 4) Hence, is either of the form 9m of 9m or 9m 8. E. Prove tht the squre of n positive integer of the form 5q is of the sme form. Let e n positive s integer of the form 5q. When 5q 5q 0q 5(5q ) Let m q (5q ). 5m. Hence, is of the sme form i.e. 5m.. EUCLID S DIVISION ALGORITHM : If nd re positive integers such tht q r, then ever common divisor of nd is common divisor of nd r nd vice-vers. E.4 Use Euclid s division lgorithm to find the H.C.F. of 96 nd 88. Appling Euclid s division lemm to 96 nd The reminder t the second stge is zero. So, the H.C.F. of 88 nd 96 is 98. E.5 If the H.C.F. of 657 nd 96 is epressile in the form (-5), find. Appling Euclid s division lemm on 657 nd So, the H.C.F. of 657 nd 96 is 9. Given : (-5) H.C.F. of 657 nd (-5) E.6 Wht is the lrgest numer tht divides 66, 7 nd 568 nd leves reminders of, nd respectivel. Clerl, the required numer is the H.C.F. of the numer 66-65, 7-5 nd Using Euclid s division lemm to find the H.C.F. of 65 nd Clerl, H.C.F. of 65 nd 5 is 65. Now, H.C.F. of 65 nd So, the H.C.F. of 65 nd 565 is 65. Hence, H.C.F. of 65, 5 nd 565 is 65. Hence, the required numer is 65. E.7 44 crtons of coke cns nd 90 crtons of Pepsi cns re to e stcked is cnteen. If ech stck is of sme height nd is to contins crtons of the sme drink, wht would e the gretest numer of crtons ech stck would hve? In order to rrnge the crtons of the sme drink is the sme stck, we hve to find the gretest numer tht divides 44 nd 90 ectl. Using Euclid s lgorithm, to find the H.C.F. of 44 nd IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

3 So, the H.C.F. of 44 nd 90 is 8. Numer of crtons in ech stck 8..4 FUNDAMENTAL THEOREM OF ARITHMETIC : Ever composite numer cn e epressed s product of primes, nd this fctoristion is unique, ecept for the order in which the prime fctors occurs. SOME IMPORTANT RESULTS : (i) Let p e prime numer nd e positive integer. If p divides, then p divides. (ii) Let e rtionl numer whose deciml epnsion termintes. Then, cn e epressed in the form p, where p nd q re co-primes, nd prime fctoristion of q is of the form m 5 n, where m, n re non- q negtive integers. p (iii) Let e rtionl numer, such tht the prime fctoristion of q is not of the form m 5 n where q m, n re non - negtive integers. Then, hs deciml epnsion which is non - terminting repeting. E.8 Determine the prime fctors of E.9 E Check whether 6 n cn end with the digit 0 for n nturl numer. An positive integer ending with the digit zero is divisile 5 nd so its prime fctoristions must contin the prime 5. 6 n ( ) n n n The prime in the fctoristion of 6 n is nd. 5 does not occur in the prime fctoristion of 6 n for n n. 6 n does not end with the digit zero for n nturl numer n. Find the LCM nd HCF of 84, 90 nd 0 ppling the prime fctoristion method. 84 7, 90 nd 0 5. Prime fctors Lest eponent HCF 6. Common prime fctors Gretest eponent 5 7 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

4 E. LCM In morning wlk three persons step off together, their steps mesure 80 cm, 85 cm nd 90 cm respectivel. Wht is the minimum distnce ech should wlk so tht the cn cover the distnce in complete steps? Required minimum distnce ech should wlk so, tht the cn cover the distnce in complete step is the L.C.M. of 80 cm, 85 cm nd 90 cm LCM LCM LCM 40 cm, m 40 cm. E. Prove tht is n irrtionl numer. Let ssume on the contrr tht is rtionl numer. Then, there eists positive integer nd such tht where, nd re co primes i.e. their HCF is. ( ) is multiple of is multiple of c for some integer c. 4c 4c c...(i) is multiple of is multiple of...(ii) From (i) nd (ii), nd hve t lest s common fctor. But this contrdicts the fct tht nd re coprime. This mens tht is n irrtionl numer. E. Prove tht 5 is n irrtionl numer. Let ssume tht on the contrr tht 5 is rtionl. Then, there eist co-prime positive integers nd such tht, is rtionl [,, re integer is rtionl numer] This contrdicts the fct tht 5 is irrtionl Hence, 5 is n irrtionl numer. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 4

5 E.4 Without ctull performing the long division, stte whether 5 hs terminting deciml epnsion or not. 5 This, shows tht the prime fctoristion of the denomintor is of the form 0 5 m 5 n. 5 Hence, it hs terminting deciml epnsion. E.5 Wht cn ou s out the prime fctoristions of the denomintors of the following rtionls : (i) (ii) (i) Since, hs terminting deciml, so prime fctoristions of the denomintor is of the form m 5 n, where m, n re non - negtive integers. (ii) Since, hs non-terminting repeting deciml epnsion. So, its denomintor hs fctors other thn or 5. DAILY PRACTICE ROBLEMS # SUBJECTIVE DPP.. Use Euclid s division lgorithm to find the HCF of : (i) 56 nd 84 (ii) 665 nd Find the HCF nd LCM of following using Fundmentl Theorem of Arithmetic method. (i) 46 nd 576 (ii) 65, 5 nd 5. Prove tht is n irrtionl numer. [CBSE - 008] 4. Prove tht 5 is irrtionl numer. [CBSE - 008] 5. Prove tht 5 is irrtionl. 6. Prove tht is irrtionl. 7. Cn we hve n n N, where 7 n ends with the digit zero. 8. Without ctull performing the long division, stte whether the following rtionl numer will hve terminting deciml epnsion or non - terminting deciml epnsion : 77 (i) 0 5 (ii) An rm contingent of 66 memers is to mrch ehind nd rm nd of memers in prde. The two groups re to mrch in the sme numer of columns. Wht is the mimum numer of columns in which the cn mrch? 0. There is circulr pth round sports field. Soni tkes 8 minutes to drive one round of the field, while Rvi tkes minutes for the sme. Suppose the oth strt t the sme point nd t the sme time, nd go in the sme direction. After how mn minutes will the meet gin t the strting point?. Write rtionl numer etween nd. [CBSE - 008]. Use Euclid s Division Lemm to show tht the squre of n positive integer is either of the form m of m for some integer m. [CBSE - 008] ANSWERS (Sujective DPP.). (i) (ii) 79. (i) 6,40896 (ii) 5, No 8. (i) Non-terminting (ii) Terminting 9. 8 columns 0. 6 minutes. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 5

6 C L A S S E S... the support LINEAR EQUATIONS IN TWO VARIABLES. LINEAR EQUATIONS IN TWO VARIABLES : An eqution of the form A B C 0 is clled liner eqution. Where A is clled coefficient of, B is clled coefficient of nd C is the constnt term (free form & ) A, B, C, R [ elongs, to R Rel No.] But A nd B c not e simultneousl zero. If A 0, B 0 eqution will e of the form A C 0. If A 0, B 0, eqution will e of the form B C 0. If A 0, B 0, C 0 eqution will e of the form A B 0. If A 0, B C, C 0 eqution will e of the form A B C 0. [Line to Y-is] [Line to X-is] [Line pssing through origin] It is clled liner eqution in two vrile ecuse the two unknown ( & ) occurs onl in the first power, nd the product of two unknown equlities does not occur. Since it involves two vrile therefore single eqution will hve infinite set of solution i.e. indeterminte solution. So we require pir of eqution i.e. simultneous equtions. Stndrd form of liner eqution : (Stndrd form refers to ll positive coefficient) c 0 c 0 For solving such equtions we hve three methods. (i) Elimintion sustitution (ii) Elimintion equting the coefficients (iii) Elimintion cross multipliction.. Elimintion B Sustitution : E. Solve (i) (ii)...(i)...(ii) From eqution (i) (iii) Sustitute the vlue of in eqution (ii) 7 (4-4) Now sustitute vlue of in eqution (iii) 7 - () () So, solution is nd IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 6

7 7 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL. () Elimintion Equting the Coefficients : E. Solve (i) (ii) Multipl eqution (i) 7 nd eqution (ii) 4, we get Add Sustitute 4 in eqution (i) 9 (4) So, solution is 4 nd 7.. (c) Elimintion Cross Multipliction : c 0 c 0 [Write the coefficient in this mnner] c c c c c c c c Also, c c c c E. Solve (i) (ii) Here,, c 5, c ; (i) 0, 5 X 5, - 0 So, solution is 5 nd - 0. c c

8 . CONDITIONS FOR SOLVABILITY (OR CONSISTENCY) OF SYSTEM OF EQUATIONS:. () Unique Solution : Two lines c 0 nd c 0, if the denomintor - 0 then the given sstem of equtions hve unique solution (i.e. onl one solution) nd solutions re sid to e consistent () No Solution : Two lines c 0 nd c 0, if the denomintor - 0 then the given sstem of equtions hve no solution nd solutions re sid to e consistent. 0. (c) Mn Solution (Infinite Solutions) E.4 Two lines c 0 nd c 0, if mn solution nd solutions re sid to e consistent. then sstem of equtions hs Find the vlue of P for which the given sstem of equtions hs onl one solution (i.e. unique solution). P (i)...(ii) P, -, c - E.5 6 -, c - Conditions for unique solution is P 6 P cn hve ll rel vlues ecept. 6 P P Find the vlue of k for which the sstem of liner eqution k 4 k k k hs infinite solution. k, 4, c -(k - 4) 6, k, c - k Here condition is k 6 4 (k 4) k (k) c c k 4 4 k 4 lso 6 k k k k 64 4k k - 4k k ± 8 k(k-8) 0 k 0 or k 8 ut k 0 is not possile other wise eqution will e one vrile. k 8 is correct vlue for infinite solution. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 8

9 E.6 Determine the vlue of k so tht the following liner equtions hs no solution. ( ) - 0 (k ) (k - ) Here k, nd c - k, k - nd c - 5 For no solution, condition is Now, k k k 5 k nd k k k k k k (k ) (k - ) (k ) 5 c c k - 5k - k -5k - - 5k 5 k - Clerl, k 5 for k -. Hence, the given sstem of equtions will hve no solution for k -. OBJECTIVE DPP -. DAILY PRACTIVE PROVBLEMS #. The equtions - 5 0, nd hve : (A) No solution (C) Two solutions (B) A single solution (D) An infinite numer of solution. If p q nd the ordered pir (p, q) stisf then is lso stisfies : (A) 4 5 (B) (C) (D) None of these.. If, - 4 nd 6 then the vlue of z is : (A) (B) (C) (D) 4 4. The sstem of liner eqution 0, c d 0 hs no solution if : (A) d - c > 0 (B) d - c < 0 (C) d c 0 (D) d - c 0 5. The vlue of k for which the sstem k 7 nd - 5 hs no solution is : (A) 7 & k (B) 4 & k (C) If 9 7 0, then : 4 & k (D) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL & k (A), (B), (C), (D), On solving, 5 we get : (A) 8, 6 (B) 4, 6 (C) 6, 4 (D) None of these 8. If the sstem - 5 0, 4 k hs n infinite numer of solutions then : (A) k (B) k (C) k 6 (D) k 6 4 9

10 9. The eqution 4 nd 5 0. If (A) Are consistent nd hve unique solution (C) re inconsistent then z will e : z (A) - (B) - (C) SUBJECTIVE DPP. (B) Are consistent nd hve infinitel mn solution (B) Are homogeneous liner equtions Solve ech of the following pir of simultneous equtions.. 7 nd 6 (D) , ; nd Prove tht the positive squre root of the reciprocl of the solutions of the equtions 9 nd ( 0, 0) stisf oth the eqution ( 4) (4 5) 5 nd 7 8 5(7 5) For wht vlue of nd, the following sstem of equtions hve n infinite no. of solutions. 7; 7. Solve : (-) ( ) - (i) 8. Solve: 9. Solve : ; (ii) ; ; ; 0. Solve - z z 5 - z. Solve, p q r nd q r. Find the vlue of k for which the given sstem of equtions (A) hs Unique solution. 5 (B) ecomes consistent. (i) 5 (ii) k - 49 l -. Find the vlue of k for which the following sstem of liner eqution ecomes infinitel mn solution. or represent the coincident lines. (i) 6 k - (ii) 7 0 k 6 6 k 4 0 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 0

11 4. Find the vlue of k or C for which the following sstems of equtions e in consistent or no solution. (i) k k 0 (ii) C k 8 k 0 C 6 5. Solve for nd : ( - ) ( ) - - ( ) ( ) [CBSE - 008] 6. Solve for nd : [CBSE - 008] C L A S S E S... the support LINEAR EQUATIONS IN TWO VARIABLES. GRAPHICAL SOLUTION OF LINEAR EQUATIONS IN TWO VARIABLES : Grphs of the tpe (i) E. Drw the grph of following : (i), (ii) (iii) 4 0 (iv) 0 (i) (ii) (iii) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

12 (iv) 0 Grphs of the tpe (ii). E. Drw the grph of following : (i) 0, (ii) - 0, (iii) 4 0 (i) 0 (ii) - 0 (iii) Grphs of the tpe (iii) 0 (Pssing through origin) E. Drw the grph of following : (i) (ii) - (i) (ii) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

13 E.4 Grphs of the Tpe (iv) c 0. (Mking Interception - is, -is) Solve the following sstem of liner equtions grphicll : -, 8. Shde the re ounded these two lines nd -is. Also, determine this re. (i) (ii) 8 (ii) X 0 Y Solution is nd Are of is nd Are of ABC BC AD 9.5 Sq. unit.. NATURE OF GRAPHICAL SOLUTION : Let equtions of two lines re c 0 nd c 0. (i) Lines re consistent (unique solution) i.e. the meet t one point condition is (ii) Lines re inconsistent (no solution) i.e. the do not meet t one point condition is c c IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

14 (iii) Lines re coincident (infinite solution) i.e. overlpping lines (or the re on one nother) condition is c c. WORD PROBLES : E.5 For solving dil - life prolems with the help of simultneous liner eqution in two vriles or equtions reducile to them proceed s :- (i) Represent the unknown quntities sme vrile nd, which re to e determined. (ii) Find the conditions given in the prolem nd trnslte the verl conditions into pir of simultneous liner eqution. (iii) Solve these equtions & otin the required quntities with pproprite units. Tpe of Prolems : (i) Determining two numers when the reltion etween them is given, (ii) Prolems regrding frctions, digits of numer ges of persons. (iii) Prolems regrding current of river, regrding time & distnce. (iv) Prolems regrding menstrution nd geometr. (v) Prolems regrding time & work (vi) Prolems regrding mitures, cots of rticles, porting & loss, discount et. Find two numers such tht the sum of twice the first nd thrice the second is 89 nd four times the first eceeds five times the second. Let the two numers e nd. E.6 Then, eqution formed re 89...(i) On solving eq. (i) & (ii) we get 5 Hence required numers re & (ii) The numertor of frction is 4 less thn the denomintor If the numertor is decresed nd the denomintor is incresed, then the denomintor is eight time the numertor, find the rection. Let the numertor nd denomintor of frction e nd Then, eqution formed re (i) On solving eq. (i) & (ii) we get nd 7 Hence, frctions is 7. 8 ( - )...(ii) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 4

15 E.7 A numer consists of two digits, the sum of the digits eing. If 8 is sutrcted from the numer, the digits re reversed. Find the numer Let the two digits numer e Then, equtions formed re (i) nd...(ii) On solving eq. (i) & (ii) we get 5 nd 7 Hence numer is 75. E.8 The sum of two - digit numer nd the numer otined reversing the order of its digits is 65. If the digits differ, find the numer Let unit digit e ten s digit e no. will e 0. Acc. to prolem (0 ) (0 ) (i) nd -...(ii) or -( - )...(iii) On solving eq. (i) nd (ii) we gets 9 nd 6 The numer will e 69. Ans. On solving eq. (i) nd (iii) we gets 6 nd 9 The numer will e 96. Ans. E.9 Si ers hence men s ge will e three times the ge of his son nd three ers go he ws nine times s old s his son. Find their present ges Let mn s present ge e rs & son s present ge e rs. According to prolem 6 ( 6) [After 6 rs] nd - 9 ( - ) [Before rs.] On solving eqution (i) & (ii) we gets 0, 6. So, the present ge of mn 0 ers, present ge of son 6 ers. E.0 A ot goes km upstrem nd 40 km downstrem in 8 hrs. It cn go 6 km. upstrem nd km downstrem in the sme time. Find the speed of the ot it still wter nd the speed of the strem. Let the speed of the ot in still wter e km/hr nd the speed of the strem e km/hr then speed of ot in downstrem is ( ) km/hr nd the speed of ot upstrem is ( - ) km/hr. Time tken to cover km upstrem hrs. 40 Time tken to cover 40 km downstrem hrs. But, totl time tken 8 hr (i) 6 Time tken to cover 6 km upstrem hrs. Time tken to cover km downstrem hrs. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 5

16 Totl time tken 8 hr (ii) Solving eqution (i) & (ii) we gets 6 nd. Hence, speed of ot in still wter 6 km/hr nd speed of strem km/hr. E. Rmesh trvels 760 km to his home prtl trin nd prtl cr. He tken 8 hr, if he trvels 60 km trin nd the rest cr. He tkes minutes more, if he trvels 40 km trin nd the rest cr. Find the speed of trin nd the cr. Let the speed of trin e km/hr & cr e km/hr respectivel Acc. to prolem 8...(i) (ii) 5 Solving eqution (i) & (ii) we gets 80 nd 00. Hence, speed ot trin 80 km/hr nd speed of cr 00km/hr. E. Points A nd B re 90 km prt from ech other on highw. A cr strts from A nd nother from B t the sme time. If the go in the sme direction, the meet in 9 hrs nd if the go in opposite direction, the meet in 7 9 hrs. Find their speeds. Let the speeds of the crs strting from A nd B e km/hr nd km/hr respectivel. Acc to prolem (i) 9 9 & 90..(ii) 7 7 Solving (i) & (ii) we gets 40 & 0. Hence, speed of cr strting from point A 40 km/hr & speed of cr strting from point B 0 km/hr. E. In cclic qudrilterl ABCD, A ( ) 0, B ( ) 0, C ( 6) 0 nd D (5-5) 0, find the ngles of the qudrilterl. Acc. to prolem ( ) 0 ( 6) Solving we get nd ( ) 0 (5-5) & 49 A 75 0, B 45 0, C 05 0, D 5 0 E.5 A vessel contins miture of 4 l milk nd 6 l wter nd second vessel contins miture of 5 l milk & 0 l wter. How much miture of milk nd wter should e tken from the first nd the second vessel seprtel nd kept in third vessel so tht the third vessel m contin miture of 5 l milk nd 0 l wter? Let l of miture e tken from Ist vessel & l of the miture e tken from nd vessel nd kept in rd vessel so tht ( ) l of the miture in third vessel m contin 5 l of milk & A miture of l from st vessel contins l of wter. 4 l of milk & l of wter nd miture of 5 5 l from IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 6

17 E.5 nd vessel contins l of milk & l of wter (i) (ii) Solving (i) & (ii) 0 litres nd 5 litres. A ld hs 5 p nd 50 p coins in her purse. If in ll she hs 40 coins totling Rs..50, find the numer of coins of ech tpe she hs. Let the ld hs coins of 5 p nd coins of 50 p. E.6 E.7 Then cc. to prolem 40...(i) nd (ii Solving for & we get 0 (5 p coins) & 0 (50 P coins). Students of clss re mde to stnd in rows. If one student is etr in row, there would e rows less. If one students is less in row, there would e rows more. Find the totl numer of students in the clss. Let e the originl no. of rows & e the originl no. of student s in ech row. Totl no. of students. Acc. to prolem ( ) ( - )...(i) nd ( - ) ( )...(ii) Solving (i) & (ii) to get & 5 Totl no. of students 60 A mn strted his jo with certin monthl slr nd erned fied increment ever er. If his slr ws Rs fter 5 ers. of service nd Rs fter ers of service, wht ws his strting slr nd wht his nnul increment. Let his initil monthl slr e Rs nd nnul increment e Rs. E.8 Then, Acc. to prolem (i)...(ii) Solving these two equtions, we get Rs. 750 Rs 50. A deler sold A VCR nd TV for Rs mking profit of % on CVR nd 5% on TV. B selling them for Rs. 860, he would hve relised profit of 5% on CVR nd % on TV. Find the cost price of ech. Let C.P. of CVR e Rs & C.P. of T.V. e Rs. 5 Acc. to prolem (i) nd (ii) Solving for & we get Rs & Rs DAILY PRACTIVE PROBLEMS # OBJECTIVE DPP.. The grphs of - 6 0, , nd intersects in : (A) Four points (B) one point (C) two point (D) infinite numer of points. The sum of two numers is 0, their product is 40. The sum of their reciprocl is : (A) (B) (C) 4 (D) 0 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 7

18 . If Rs. 50 is distriuted mong 50 children giving 50 p to ech o nd 5 p to ech girl. Then the numer of os is : (A) 5 (B) 40 (C) 6 (D) In covering distnce of 0 km. Amit tkes hrs. more thn suresh. If Amit doules his speed, he would tke one hour less thn suresh. Amits speed is : (A) 5 km/hr. (B) 7.5 km/hr. (C) 6 km/hr. (D) 6. km/hr. 5. If in frction less from two times of numertor & dd in denomintor then new frction will e : (A) (B) SUBJECTIVE DPP. ( ) (C) (D). The denomintor of frction is greter thn its numertor 7. If 4 is dded to oth its numertor nd denomintor, then it ecomes. Find the frction.. In certin numer is divided the sum of its two digits, the quotient is 6 nd reminder is. If the digits re interchnged nd the resulting numer is divided the sum of the digits, then the quotient is 4 nd the reminder is 9. Find the numer.. men nd os together cn do piece of work is 8 ds. The sme work si done in 6 ds men nd os together. How long would o lone or mn lone tke to complete the work 4. The um of two no s is 8. the sum of their reciprocl is 4. Find the numers. 5. In cclic qudrilterl ABCD, A ( 4) 0, B ( ) 0, C ( 0) 0 nd D (4-5) 0 then find out the ngles of qudrilterl. 6. Solve grphicll nd find the pints where the given liens meets the - is : - 0, Use single grph pper & drw the grph of the following equtions. Otin the vertices of the tringles so otined : - 8, 5-4 & Drw the grph of - 0 ; - 0. Clculte, the re ounded these lines nd - is. 9. A mn sold chir nd tle together for Rs. 50 there mking profit of 5% on chir nd 0% on tle. B selling them together for Rs. 55 he would hve mde profit of 0% on the chir nd 5% on the tle. Find cost price of ech. 0. A mn went to the Reserve Bnk of Indi with note or Rs He sked the cshier to give him Rs. 5 nd Rs. 0 notes in return. The cshier gve him 70 notes in ll. Find how mn notes of Rs. 5 nd Rs. 0 did the mn receive.. Solve grphicll: ; Also find the vertices of the tringle formed the ove two lines nd -is.. The sum of the digits of two-digit numer is. The numer otined interchnging the two digits eceeds the given numer 8. Find the numer.. Drw the grphs of the following equtions nd solve grphicll: 6 0 ; 8-0 Also determine the co-ordintes of the vertices of the tringle formed these lines nd the - is. 4. A frmer wishes to purchse numer of sheep found the if the cost him Rs 4 hed, he would not hve mone enough Rs 5; But if the cost him Rs 40 hed, he would them hve Rs 40 more thn he required; find the numer of sheeps, nd the mone which he hd. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 8

19 ANSWERS (Ojective DPP.) Que Ans. D A B D D A C D A D (Sujective DPP.). 9, 6. 0., , , 7. (i) -, - (ii) -, , 9., 0., -, 5. q r(p q) r(q p) p,. () k is n rel numer () k 4 p q p q. () k 6 () k 4 4. () k - 4 () C , - 6., (Ojective DPP.) Que. 4 5 Ans. B A D A D (Sujective DPP.). / One o cn do in 0 ds nd one mn cn do in 0 ds. 4. No. s re nd 6 5. A 70 0, B 5 0, C 0 0, D ,, Point of contct with - is (0, ), (0, -) 7. (-4, ), (, ), (,5) Squre units. 9. Chir Rs. 600, Tles Rs rupees notes 40 & 0 rupees notes 0. (0,5) vertices (0,5) (-6,0), (4, 0) ,, Lines meets -is t (-, 0) & (4, 0) 4. 4 sheep, Rs 400 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 9

20 C L A S S E S... the support POLYNOMIALS 4. POLYNOMIALS : An lgeric epression f() of the form (f) 0... n n, where 0,,... n re rel numers nd ll the inde of re non-negtive integers is clled polnomils in nd the highest Inde n in clled the degree of the polnomil, if n () Zero Degree Polnomil : An non-zero numer is regrded s polnomil of degree zero or zero degree polnomil. For emple, f(), where 0 is zero degree polnomil, since we cn write f() s f() () Constnt Polnomil : A polnomil of degree zero is clled constnt polnomil. For emple, f() (c) Liner Polnomil : A polnomil of degree is clled liner polnomil. For emple : p() 4 - nd f(t) t 5 re liner polnomils. 4. (d) Qudrtic Polnomil : A polnomil of degree is clled qudrtic polnomil. For emple : f() 5-5 nd g() - 5 re qudrtic polnomils with rel coefficients. IMPORTANT FORMULAE : ( ) ( - ) - - ( ) ( - ) ( ) ( - ) ( ) - ( ) - ( - ) ( ) ( - ) ( - ) ( c) c c c ( ) ( ) ( - ) - - ( - ) c - c ( c) ( c - - c - c) Specil Cse : If c 0 then c c. 4. GRAPH OF POLYNOMIALS : In lgeric or in set theoretic lnguge the grph of polnomil f() is the collection (or set) of ll points (, ), where f(). In geometricl or in grphicl lnguge the grph of polnomil f() is smooth free hnd curve pssing through points, ), (, ), (, ),... etc. where,,,... re the vlues of the polnomil f() t,,,... respectivel. In order to drw the grph of polnomil f(), follow the following lgorithm. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 0

21 ALGORITHM : Step (i) Find the vlues,,... n of polnomil f() on different points,,... n nd prepre tle tht gives vlues of or f() for vrious vlues of. : n n. f() f( ) f( ). Y nf( n) n f( n).. Step (ii) Plot tht points (, ), (, ), (, ),...( n, n)... on rectngulr co-ordinte sstem. In plotting these points use different scles on the X nd Y es. Step (iii) Drw free hnd smooth curve pssing through points plotted in step to get the grph of the polnomil f(). 4. () Grph of Liner Polnomil : Consider liner polnomil f(), 0 Grph of is stright line. Tht in wh f() ) is clled liner polnomil. Since two points determine stright line, so onl two points need to plotted to drw the line. The line represented crosses the X-is t ectl one point, nmel, 0. E. Drw the grph of the polnomil f() - 5. Also, find the coordintes of the point where it crosses X- is. Let - 5. The following tle list the vlues of corresponding to different vlues of. 4 - The points A (, - ) nd B (4, ) re plotted on the grph pper on suitle scle. A line is drwn pssing through these points to otin the grphs of the given polnomil. 4. () Grph of Qudrtic Polnomil : Let,,c e rel numers nd 0. Then f() c is known s qudrtic polnomil in. Grph of the qudrtic polnomil i.e. he curve whose eqution is c, 0 Grph of qudrtic polnomil is lws prol. Let c, where c c 4 ( ) - ( - 4c) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

22 4 ( - 4c) ( ) 4 ( - 4c) 4 ( /) 4c D (i) 4 where D - 4c is the discriminte of the qudrtic eqution. REMARKS : Shifting the origin t, D 4 Sustituting these vlues in (i), we otin Y X (D), we hve X - nd Y (ii) which is the stndrd eqution of prol Clerl, this is the eqution of prol hving its verte t, The prol opens upwrds or downwrds ccording s > 0 or < SIGN OF QUADRTIV EXPRESSIONS : D. 4 Let α e rel root of c 0. Then α α c 0 Point (α,0) lies on c. Thus, ever rel root of c 0 represents point of intersecting of the prol with the X-is. Conversel, if the prol c intersects the X-is t point (α,0) then (α,0) stisfies the eqution c α α c 0 [ α is rel root of c 0] Thus, the intersection of the prol c with X-is gives ll the rel roots of c 0. Following conclusions m e drwn :- (i) If D>0, the prol will intersect the -is in two distinct points nd vice-vers. D The prol meets -is t α nd β D Roots re rel & distinct (ii) If D 0, the prol will just touch the -is t one point nd vice-vers. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

23 Roots re equl (iii) If D<0, the prol will not intersect -is t ll nd vice-vers. Roots re imginr REMARKS R, > 0 onl if > 0 & D 4c < 0 R, < 0 onl if < 0 & D 4c < 0 E. Drw the grph of the polnomil f() Let The following tle gives the vlues of or f() for vrious vlues of Let us plot the points (-4, 6), (-, 7), (-, 0), (-, -5), (0, - 8), (, - 9), (, - 8), (, - 5), (4, 0), (5, 7) nd (6, 6) on grphs pper nd drw smooth free hnd curve pssing through these points. The curve thus otined represents the grphs of the polnomil f() This is clled prol. The lowest point P, clled minimum points, is the verte of the prol. Verticl line pssing through P is clled the is of the prol. Prol is smmetric out the is. So, it is lso clled the line of smmetr. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

24 Oservtions : Fro the grphs of the polnomil f() - - 8, following oservtions cn e drwn : (i) The coefficient of in f() is ( positive rel numer) nd so the prol opens upwrds. (ii) D - 4c 4 6 > 0. So, the prol cuts X-is t two distinct points. (iii) On compring the polnomil with c, we get, - nd c - 8. D The verte of the prol hs coordintes (, -9) i.e.,, where D - 4c. 4 (iv) The polnomil f() ( - 4) ( ) is fctorizle into two distinct liner fctors ( - 4) nd ( ). So, the prol cuts X-is t two distinct points (4, 0) nd (-, 0). the -coordintes of these points re zeros of f(). E. Drw the grphs of the qudrtic polnomil f() - -. Let f() or, - -. Let us list few vlues of - - corresponding to few vlues of s follows : Thus, the following points lie on the grph of the polnomil - - : (-5, -), (-4, -5), (-, 0), (-, 4), (-, 4), (0, ), (, 0), (, - 5), (, -) nd (4, - ). Let plot these points on grph pper nd drw smooth free hnd curve pssing through these points to otin the grphs of - -. The curve thus otined represents prol, s shown in figure. The highest point P(-, 4), is clled mimum points, is the verte of the prol. Verticl line through P is the is of the prol. Clerl, prol is smmetric out the is. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 4

25 Oservtions : Following oservtions from the grph of the polnomil f() - - is s follows : (i) The coefficient of in f() - - is - i.e. negtive rel numer nd so the prol opens downwrds. (ii) D > 0. So, the prol cuts -is two distinct points. (iii) On compring the polnomil - - with c c, we hve -, - nd c. The verte D of the prol is t the point (-, 4) i.e. t,, where D - 4c. 4 (iv) The polnomil f() - - ( - ) ( ) is fctorizle into two distinct liner fctors ( - ) nd ( ). So, the prol cuts X-is t two distinct points (, 0) nd (-, 0). The co-ordintes of these points re zeros of f(). 4.4 GRAPH OF A CUBIC POLYNOMIAL : Grphs of cuic polnomil does not hve fied stndrd shpe. Cuic polnomil grphs will lws cross X-is t lest once nd t most thrice. E.4 Drw the grphs of the polnomil f() - 4. Let f() or, - 4. The vlues of for vrile vlue of re listed in the following tle : Thus, the curve - 4 psses through the points (-, -5), (-, 0), (-, ), (0,0), (, -), (, 0), (, 5), (4, 48) etc.plotting these points on grph pper nd drwing free hnd smooth curve through these points, we otin the grph of the given polnomil s shown figure. Oservtions : For the grphs of the polnomil f() - 4, following oservtions re s follows :- (i) The polnomil f() - 4 ( - 4) ( - ) ( ) is fctorizle into three distinct liner fctors. The curve f() lso cuts X-is t three distinct points. (ii) We hve, f() ( - ) ( ) Therefore 0, nd - re three zeros of f(). The curve f() cuts X-is t three points O (0, 0), P(, 0) nd Q (-, 0). IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 5

26 4.5 RELATIONSHIP BETWEEN ZEROS AND COEFFICIENTS OF A QUADRATIC POLYNOMIAL : E.5 Let α nd β e the zeros of qudrtic polnomil f() c. B fcto r theorem ( nd ( β) re the fctors of f(). f() k ( α)( β) re the fctors of f() c k{ - ( α β) αβ } c k - k ( α β) kαβ Compring the coefficients of, nd constnt terms on oth sides, we get k, - k ( α β) nd k αβ Hence, c α β nd α β Coefficient of Cons tn t term α β nd αβ Coefficient of Coefficient of Sum of the zeros Product of the zeros Coefficient of Coefficient of c Cons tn t term Coefficient of REMAKRS : If α nd β re the zeros of qudrtic polnomil f(). The, the polnomil f() is given f() k{ - ( α β) αβ } or f() k{ - (Sum of the zeros) Product of the zeros} α ) Find the zeros of the qudrtic polnomil f() nd verif nd the reltionship etween the zeros nd their coefficients. f() f() f() ( - 4) ( - 4)] f() ( - 4) ( ) Zeros of f() re given f() ( - 4) ( ) 0 4 or - So, α 4 nd β sum of zeros α β 4 - Coefficient of ( ) Also, sum of zeros Coefficient of Coefficient of So, sum of zeros α β Coefficient of Now, product of zeros αβ (4) (-) - 8 Cons tn t term 8 Also, product of zeros 8 Coefficient of Constn t term Product of zeros αβ. Coefficient of IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 6

27 E.6 Find qudrtic polnomil whose zeros re 5 nd 5 Given α 5, β 5 f() k{ - ( α β) αβ } Here, α β nd αβ ( 5 )(5 ) 5 - f() k { - 0 }, where, k is n non-zero rel numer. E.7 Sum of product of zeros of qudrtic polnomil re 5 nd 7 respectivel. Find the polnomil. Given : Sum of zeros 5 nd product of zeros 7 So, qudrtic polnomil is given f() k { - (sum of zeros) product of zeros} f() k{ - 5 7}, where, k is n non-zero rel numer, 4.6 RELATIONSHIP BETWEEN ZEROS AND COEFFICIENTS OF A CUBIC POLYNOMIAL : Let α, β, γ e the zeros of cuic polnomil f() c d, 0 Then, fctor theorem, α, β nd γ re fctors of f(). Also, f() eing cuic polnomil cnnot hve more thn three liner fctors. f() k( α)( β)( γ) c d k ( α)( β)( γ) c d k{ - ( α β γ) c d k - k ( α β γ) ( αβ βγ γα) αβγ } k( αβ βγ γα) kαβγ Compring the coefficients of,, nd constnt terms on oth sides, we get k, - k ( α β γ), c ( αβ βγ γα)ndd k( αβγ) And, α β γ c αβ βγ γα d αβγ Sum of the zeros Coefficient of Coefficient of Sum of the products of the zeros tken two t time Product of the zeros REMARKS : Cuic polnomil hving d f() k ( α)( β)( γ) f() k { - ( α β γ) Cons tn t term Coefficient of α, β nd γ s its zeros is given c Coefficient of Coefficient of ( αβ βγ γα) αβγ } where k is n non-zero rel numer. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 7

28 E.8 Verif tht, re zeros of cuic polnomil - 5. Also verif the reltionship etween, the zeros nd their coefficients. f() - 5 E.9 f f() () () 5() f(-) (-) (-) - 5(-) Let α, β nd γ Now, Sum of zeros Also, sum of zeros So, sum of zeros α β γ (Coefficient of Coefficient of α β γ Sum of product of zeros tken two t time Also, Coefficient of 5 ββ βγ γα Coefficient of IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL ) (Coefficient of Coefficient of So, sum of product of zeros tken two t time Now, Product of zeros αβγ Also, product of zeros ()( ) Cons tn t term Coefficient of Cons tn t term Product zeros αβγ Coefficient of ) αβ βγ γα 5 ( ) ( ) αβ βγ γα Coefficient of Coefficient of Find polnomil with the sum, sum of the product of its zeros tken two t time, nd product its zeros s, - nd - respectivel. 8

29 Given α β γ, αβ βγ γα nd αβγ So, polnomil f() k { - ( α β γ) ( αβ βγ γα) αβγ } f() k { - - }, where k is n non-zero rel numer. 4.7 VALUE OF A POLYNOMIAL : The vlue of polnomil f() t α is otined sustituting denoted f( α ). For emple : If f() - 7 then its vlue t is f() () - () 7() ZEROS OF ROOTS OF A POLYNOMIAL : α in the given polnomil nd is A rel numer is zero of polnomil f(), if f() 0, Here is clled root of the eqution f() 0. E.0 Show tht is root of p() E. Then, p() () () - 7() Hence is root of p(). If 4 is root of the polnomil f() 6 - k - 0 then find the vlue of k. f() 6 - k f k k k k k k 8 k 9. E. If & 0 re roots of the polnomils (f) - 5, then find the vlues of nd / f() () - 5() () (i) f(0) (0) - 5(0) (0) 0 0 4, FACTOR THEOREM : Let p() e polnomil of degree greter thn or equl to nd e rel numer such tht p() 0. then ( - ) is fctor of p(). Conversel, if ( - ) is fctor of p(), then p() 0. E. Show tht nd - re fctors of - 9. To prove tht ( ) nd ( - ) re fctors of p() - 9 it is sufficient to show tht p(-) nd p oth re equl to zero. p(-) (-) - 9(-) (-) And p IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 9

30 E.4 Find α nd β if nd re fctors of p() - α β. nd re the fctor of p(). Then, p(-) 0 & p(-) 0 Therefore, p(-) (-) (-) - α ( ) β 0 α β 0 β α...(i) p( ) ( ) ( ) α( ) β 0-8 4α β 0 β 4α 4...(ii) From eqution () nd () α 4α 4 α α Put α - eqution () β - (-) Hence α, β 0 E.5 Wht must e dded to - 9 so tht the result is ectl divisile 7-6. Let p() - 9 nd q() 7-6 We know if p() is divided q() which is qudrtic polnomil then the reminder e r() nd degree of r() is less thn q() or Divisor. B long division method Let we dded (liner polnomil) in p(), so tht p() is ectl divisile 7-6. Hence, p() s() ( - ) (9 ). 7 6 ± ( ) 9 6 ( ) 9 ( 6 ) ± ( ) ( ) 0 Hence, ( - ) & - 0 nd Hence if in p() we dded then it is ectl divisile 7-6. E.6 Wht must e sutrcted from so tht the result is ectl divisile -. Let e sutrcted from p() so tht it is ectl divisile -. s() ( ) (5 ) (80 - ) Dividend Divisor quotient reminder But reminder will e zero. Dividend Divisor quotient s() ( - ) quotient s() (5 ) (80 - ) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 0

31 6 ± 7 (5 ) (5 ) 80 ( ) ± 84 (4 ) ( 4 ) 0 Hence, (4 - ) (-4 -) & (-4 - ) 0 4 nd - 4 Hence, if in p() we sutrct 4-4 then it is ectl divisile -. E.7 Using fctor theorem, fctorize : p() ±, ±, ± 5, ± 9, ± 5, ± 45 If we put in p() p() () 4-7() - () 6() - 45 p() or - is fctor of p(). Similrl if we put in p() p() () 4-7() - () 6() - 45 p() Hence, or ( - ) 0 is the fctor of p(). p() p() ( - ) - 5 ( - ) - 8( - ) 45( - ) p() ( - ) ( ) p() ( - ) ( ) p() ( - ) [ -( - ) ( - ) - 5( - )] p() ( - ) ( - ) ( - 5) p() ( - ) ( - ) ( ) p() ( - ) ( - )[( ) - 5( )] p() ( - ) ( - ) ( ) ( - 5). 4.0 REMAINDER THEOREM : Let p() e n polnomil of degree greter thn or equl to one nd e n rel numer. If p() is divided - ), then the reminder is equl to p(). Let q() e the quotient nd r(x) e the reminder when p() is divided ( - ), then Dividend Divisor Quotient Reminder E.8 Find the reminder when f() is divided g() f E.9 Appl division lgorithm to find the quotient q() nd reminder r() on dividing f() () -. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

32 ± ± ± 9 5 So, quotient q() 5-8 nd reminder r() Now, dividend Quotient Divisor Reminder E.0 (5-8) ( - ) (-9 5) Hence, the division lgorithm is verified. Find ll the zeros of the polnomil f() , if two of its zeros re nd. Since nd re zeros of f(). Therefore, or - is fctor of f() ( - ) ( - - ) ( - ) ( - ) ( ) ( )( ) So, the zeros re,,, OBJECTIVE DPP - 4. DAILY PRACTICE PROBLESM # 4. If is divided - then reminder will e (A) 57 8 (B) 59 (C) (D) 55 8 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

33 . The polnomils - nd - 5 when divided ( - 4) leves reminders R & R respectivel then vlue of if R - R 0. (A) 8 8 (B) (C) 7 (D). A qudrtic polnomil is ectl divisile ( ) & ( ) nd leves the reminder 4 fter division ( ) then tht polnomil is (A) 6 4 (B) 6 4 (C) 6-4 (D) The vlues of & so tht the polnomil - - is divisile ( - ) & ( ) re (A) 5, (B), 5 (C) c -, 5 (D), Grph of qudrtic eqution is lws - (A) stright line (B) circle (C) prol (D) Hperol 6. If the sign of is positive in qudrtic eqution then its grph should e (A) prol open upwrds (C) prol open leftwrds (B) prol open downwrds (D) cn t e determined 7. The grph of polnomil - is lws pssing through the point - (A) (0, 0) (B) (, ) (C) (, -) (D) ll of these 8. How mn time, grph of the polnomil f() - will intersect X-is - (A) 0 (B) (C) (D) 4 9. Which of the following curve touches X-is - (A) - 4 (B) - 6 (C) (D) In the digrm given elow shows the grphs of the polnomil f() c, then (A) < 0, < 0 nd c > 0 (B) < 0, < 0 nd c < 0 (C) < 0, > 0 nd c > 0 (D) < 0, > 0 nd c < SUBJECTIVE DPP 4.. Drw the grph of following polnomils.. f() -. f() - 4 c. f() d. f() - 9 e. f() f. f() ( - ) g. f() - h. f(). Find the zeros of qudrtic polnomil p() nd verif the reltionship etween the zeros nd their coefficients.. Find qudrtic polnomil whose zeros re 5 nd Sum nd product of zeros of qudrtic polnomil re nd 5 respectivel. Find the qudrtic polnomil. 5. Find qudrtic polnomil whose zeros re 5 nd Verif tht 5,, re zeros of cuic polnomil Also verif the reltionship etween 4 the zeros nd the coefficients. 7. Divide If α, β re zeros of 5 5, find the vlue of α β. 9. Appl the division lgorithm to find the quotient nd reminder on dividing p() g() -. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL

34 0. On dividing - polnomil g(), the quotient reminder were nd - 4, respectivel. Find g().. α, β, γ re zeros of cuic polnomil - 44 c. If α, β, γ re in A.P., find the vlue of c.. Otin ll the zeros of , if two of its zeros re 5 5 nd.. Wht must e dded to so tht the result is ectl divisile - 6? 4. Wht must e sutrcted from so tht the result is ectl divisile - 4? 5. If α, β re zeros of qudrtic polnomil k 4 4, find the vlue of k such tht ( α β) αβ Find the qudrtic polnomil sum of whose zeros is 8 nd their product is. Hence find f the zeros of the polnomil. [CBSE - 008] 7. Is - 4 solution of the equtions 5-0 > [CBSE 008] 8. Write the numer of zeros of the polnomil f() whose grph is given figure [CBSE - 008] 9. If the product of zeros of the polnomil is 4, find the vlue of. [CBSE - 008] ANSWERS (Ojective DPP 4.) Que Ans. B B B B C A A B D A (Sujective DPP 4.). -, -. k{ - 5} 4. k{ - 5 } 5. k{ - 6 4} Quotient -, Reminder c ,,- nd k, 6. k{ - 8 } nd zeros re 6 &. 7. Yes 8. No. of zeros 9. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 4

35 C L A S S E S... the support 5. QUADRATIC EQUATION : QUADRATIC EQUATIONS If P() is qudrtic epression in vrile, then P() 0 is known s qudrtic eqution. 5. () Generl form of Qudrtic Eqution : The generl form of qudrtic eqution is c 0, where,,c re rel numers nd 0 Since 0, qudrtic equtions, in generl re of the following tpes :- (i) 0, c 0 i.e., of he tpe c0. (ii) 0, c 0, i.e. of the tpe 0. (iii) 0, c 0, i.e. of the tpe 0. (iv) 0, c 0, i.e., of the tpe c ROOTS OF A QUADRATIC EQUATION : The vlue of which stisfies the given qudrtic eqution is known s its root. The roots of the given eqution re known s its solution. Generl form of qudrtic eqution is : c 0 or 4 4 4c - 4c [Multipling 4] or 4 4-4c [B dding oth sides] or or 4 4c - 4c ( ) - 4c Tking squre root of oth the sides ± 4c or ± 4c Hence, roots of the qudrtic eqution c 0 re 4c nd 4c IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 5

36 REMARK : A qudrtic eqution is stisfied ectl two vlues of which m e rel or imginr. The eqution, c 0 is : A qudrtic eqution if 0 Two roots A liner eqution if 0, 0 One root A contrdiction if 0, c 0 No root An identif if c 0 Infinite roots A qudrtic eqution cnnot hve more thn two roots. If follows from the ove sttement tht if qudrtic eqution is stisfied more thn two vlues of, then it is stisfied ever vlue of nd so it is n identit. 5. NATURE OF ROOTS : Consider the qudrtic eqution, c 0 hving of roots of qudrtic eqution. It is denoted D or. Roots of the given qudrtic eqution m e (i) Rel nd unequl (ii) Rel nd equl (iii) Imginr nd unequl. α β s its roots nd - 4c is clled discriminte Let the roots of the qudrtic eqution c 0 (where 0,, c R ) e α nd β then 4c α...(i) nd 4c β...(ii) The nture of roots depends upon the vlue of epression - 4c with in the squre root sign. This is known s discriminte of the given qudrtic eqution. Consider the Following Cses : Cse- When - 4c > 0, (D > 0) In this cse roots of the given eqution re rel nd distinct nd re s follows α 4c nd β 4c (i) When ( 0),, c Q nd - 4c is perfect squre In this cse oth the roots re rtionl nd distinct. (ii) When ( 0),, c Q nd - 4c is not perfect squre In this cse oth the roots re irrtionl nd distinct. Cse- When - 4c 0, (D 0) In this cse oth the roots re rel nd equl to. [See remrks lso] Cse- When - 4c < 0, (D < 0) In this cse - 4c < 0, then 4c - > 0 or α i 4c (4c ) nd β (4c α [ i] i 4c nd β i.e. in this cse oth the root re imginr nd distinct. ) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 6

37 REMARKS : If,,c Q nd - 4c is positive (D > 0) ut not perfect squre, then the roots re irrtionl nd the lws occur in conjugte pirs like nd. However, if,,c re irrtionl numer nd - 4c is positive ut not perfect squre, then the roots m not occur in conjugte pirs. If - 4c is negtive (D > 0), then the roots re comple conjugte of ech other. In fct, comple roots of n eqution with rel coefficients lws occur in conjugte pirs like i nd - i. However, this m not e true in cse of equtions with comple coefficients. For emple, - i - 0 hs oth roots equl to i. If nd c re of the sme sign nd hs sign opposite to tht of s well s c, then oth the roots re positive, the sum s well s the product of roots is positive ( D 0). If,, re of the sme sign then oth the roots re negtive, the sum of the roots is negtive ut the product of roots is positive ( D 0). 5.4 METHODS OF SOLVING QUADRATIC EQUATION : 5.4 () B Fctoristion : ALGORITHM : Step (i) Fctorise the constnt term of the given qudrtic eqution. Step (ii) Epress the coefficient of middle term s the sum or difference of the fctors otined in step. Clerl, the product of these two fctors will e equl to the product of the coefficient of nd constnt term. Step (iii) Split the middle term in two prts otined in step. Step (iv) Fctorise the qudrtic eqution otined in step. ILLUSTRATIONS : E. Solve the following qudrtic eqution fctoristion method: Here, Fctors of constnt term ( - ) re ( - ) nd ( ). Also, Coefficient of the middle term - - [( - ) ( )] {( - ) ( )} ( - )( ) 0 - ( - ) - ( ) ( - ) ( ) { - ( - )}- ( ) { - ( - )} 0 { - ( - )} { - ( )} 0 - ( - ) 0 or, - ( ) 0 - or E. Solve We hve or (8) - (5) 0 or (8 5) (8-5) 0 i.e o This gives or Thus,, re solutions of the given equtions. 8 8 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 7

38 E. Solve the qudrtic eqution The given eqution m e written s 8( - ) 0 This gives 0 or. 0,, re the required solutions. E.4 Solve : is equivlent to (5) - (5) () 0 or (5 - ) This gives, or simpl 5 5 s the required solution. 5 E.5 Find the solutions of the qudrtic eqution The qudrtic polnomil 6 5 cn e fcorised s follows : ( 5) ( 5) ( 5) ( ) Therefore the given qudrtic eqution ecomes ( 5) ( ) This gives - 5 or - Therefore, - re the required solutions of the given eqution. 9 E.6 Solve : 0. ( )( ) Oviousl, the given eqution is vlid if - 0 nd 0. Multipling throughout ( - ) ( - ), we get ( ) ( - ) 9 0 or or 5 0 or ( ) ( ) 0 But 0, so we get 0. This gives - s the onl solution of the given eqution. 5.4 () B the Method of Completion of Squre : ALGORITHM : Step-(i) Otin the qudrtic eqution. Let the qudrtic eqution e c 0, 0. Step-(ii) Mke the coefficient of c unit, if it is not unit. i.e., otined 0. Step-(iii) Shift the constnt term c on R.H.S. to get Step-(iv) Add squre of hlf of the coefficient of i.e. on oth sides to otin Step-(v) Write L.H.S. s the perfect squre of inomil epression nd simplif R.H.S. to get 4c. 4 c c Step-(vi) Tke squre root of oth sides to get ± 4c 4 Step (vii) Otin the vlues of shifting the constnt term on RHS. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 8

39 9 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL E.7 Solve :- 0 We hve 0 Add nd sutrct ( coefficient of ) in L.H.S. nd get ± This gives ( ) 5 or 5 Therefore 5, 5 re the solutions of the given eqution. E.8 B using the method of completing the squre, show tht the eqution hs no rel roots. We hve, Clerl, RHS is negtive But, 8 cnnot e negtive for n rel vlue of. Hence, the given eqution hs no rel roots. 5.4 (c) B Using Qudrtic Formul : Solve the qudrtic eqution in generl form viz. c 0. We hve, c 0 Step (i) B comprison with generl qudrtic eqution, find the vlue of, nd c. Step (ii) Find the discriminte of the qudrtic eqution. D - 4c Step (iii) Now find the roots of the eqution given eqution D, D REMARK : If - 4c < 0 i.e. negtive, then c 4 is not rel nd therefore, the eqution does not hve n rel roots.

40 E.9 Solve the qudrtic eqution Compring the given eqution with c 0, we find tht, - 7 nd c -5. E.0 Therefore, D (-7) - 4 (-5) > 0 Since D is positive, the eqution hs two roots given , re the required solutions , For wht vlue of k, (4 - k) (k 4) (8k ) is perfect squre. The given eqution is perfect squre, if its discriminte is zero i.e. (k 4) - 4(4 - k) (8k ) 0 E. 4(k ) - 4(4 - k) (8k ) 0 4[4(k ) - (4 - k) (8k )] 0 [(k 4k 4) - (-8k k 4)] 0 9k - 7k 0 9k (k - ) 0 k 0 or k Hence, the given eqution is perfect squre, if k 0 or k. If the roots of the eqution ( - c) (c - ) c( - ) 0 re equl, show tht. c Since the roots of the given equtions re equl, so discriminnt will e equl to zero. (c - ) - 4( - c). c( - ) 0 (c - c) - 4c( - c - c) 0, c 4 c c - 4c c - 4c 0 ( c - c) 0 c - c 0 c c c. c Hence Proved. E. If the roots of the eqution ( - c) (c - ) ( - ) 0 re equl, then prove tht c. If the roots of the given eqution re equl, then discriminnt is zero i.e. (c - ) - 4( - c) ( - ) 0 c - c 4-4 4c - 4c 0 c 4 c - 4-4c 0(c - ) 0c Hence Proved. E. If the roots of the eqution re rel nd distinct, then find ll possile vlues of. Since the roots of the given eqution re rel nd distinct, we must hve D > ( - 6) > 0 4[6-6] > 0-4( - 6-6) > < 0 ( - 8) ( ) < 0 - < < 8 Hence, the roots of the given eqution re rel if lies etween - nd APPLICATIONS OF QUADRATIC EQUATIONS : ALGORITHM : The method of prolem solving consist of the following three steps : Step (i) Trnslting the word prolem into smolic lnguge (mthemticl sttement) which mens identifing reltionship eisting in the prolem nd then forming the qudrtic eqution. Step (ii) Solving the qudrtic eqution thus formed. Step (iii) Interpreting the solution of the eqution, which mens trnslting the result of mthemticl sttement into verl lnguge. REMARKS : Two consecutive odd nturl numers e -, where N Two consecutive even nturl numers e, where N Two consecutive even positive integers e, where Consecutive multiples of 5 e 5, 5 5, Z IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 40

41 E.4 The sum of the squres of two consecutive positive integers is 545. Find the integers. Let e one of the positive integers. Then the other integer is, Z Since the sum of the squres of the integers is 545, we get E.5 ( ) 545 or or or ( 7) - 6( 7) 0 or ( - 6) ( 7) 0 Here, 6 or -7 But, is positive integer. Therefore, reject - 7 nd tke 6. Hence, two consecutive positive integers re 6 nd (6 ), i.e., 6 nd 7. The length of hll is 5 m more thn its reth. If the re of the floor of the hll is 84 m, wht re the length nd the redth of the hll? Let the redth of the hll e metres. Then the length of the ll is ( 5) metres. The re of the floor ( 5) m Therefore, ( 5) 84 or or ( ) ( - 7) 0 This given 7 or -. Since, the redth of the hll cnnot e negtive, we reject - nd tke - onl. Thus, redth of the hll 7 metres, nd length of the hll (7 5), i.e., metres. E.6 Out of group of swns 7 times the squre root of the totl numer re pling on the shore of tnk. The two remining ones re pling, in deep wter. Wht is the totl numer of swns? Let us denote the numer of swns. 7 Then, the numer of swns pling on the shore of the tnk. There re two remining swns. 7 Therefore, 7 or or or ( ) 7 4( - 4 4) 49 or or or 4( - 6) - ( - 6) 0 or ( - 6) (4 - ) 0 This gives 6 or 4 We reject nd tke 6. 4 Hence, the totl numer of swns is 6. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 4

42 E.7 The hpotenuse of right tringle is 5 cm. The difference etween the lengths of the other two sides of the tringle is 5 cm. Find the lengths of these sides. Let the length of the shorter side cm. Then, the length of the longer side ( 5) cm. Since the tringle is right-ngled, the sum of the squres of the sides must e equl to the squre of the hpotenuse (Pthgors Theorem). ( 5) 5 or or or or ( 0) ( - 5) 0 This gives 5 or - 0 We reject - 0 nd tke 5. Thus, length of shorter side 5 cm. Length of longer side (5 5) cm, i.e., 0 cm. E.8 Swti cn row her ot t speed of 5 km/h in still wter. If it tkes her hour more to row the ot 5.5 E.9 Sol km upstrem thn to return downstrem, find the speed of the strem. Let the speed of the strem e km/h Speed of the ot in upstrem (5 - )km/h Speed of the ot in downstrem (5 )km/h Time, s t (in hours), for going 5.5 km upstrem Time, s t (in hours), for returning 5.5 km downstrem 5 Oviousl t > t Therefore, ccording to the given condition of the prolem, t t i.e., 5 5 or or 4 5 or or or or ( 5) ( - ) 0 5 This gives, since we reject. Thus, the speed of the strem is km/h. The sum of the squre of two positive integers is 08. If the squre of the lrger numer is 8 times the smller numer, find the numers. [CBSE - 007] Let e the smller numer. Then, squre of the lrger numer will e 8. Therefore, 8 08 or IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 4

43 E.0 or ( - 8) ( 6) 0 This gives 8 or - 6 Since the numers re positive integers, we reject - 6 nd tke 8. Therefore, squre of lrger numer So, lrger numer 44 Hence, the lrger numer is nd the smller is 8. n(n ) The sum S of first n nturl numer is given the reltion S. Find n, if the sum is 76. We hve n(n ) S 76 or n n This gives or n n 09 08,, or n, or n, We reject n - 4, since -4 is not nturl numer. Therefore, n. DAILY PRACTIVE PROBLEMS # 5 OBJECTIVE DPP If one root of 5 k 0 is reciprocl of the other then k (A) 0 (B) 5 (C) 6 (D) 6. The roots of the eqution re (A) Imginr (B) Rtionl (C) Irrtionl (D) None of these. The difference etween two numers is 5 different in their squres is 65. The lrger numer is (A) 9 (B) 0 (C) (D) 4. The sum of ges of fther nd son is 45 ers. Five ers go, the product of their ges ws 4 times the ge of the fther t tht time. The present ge of the fther is (A) 0 rs (B) rs (C) 6 rs (D) 4 rs 5. If one of the roots of the qudrtic eqution is then find the qudrtic eqution. (A) - ( ) 0 (B) ( ) 0 (C) (D) 4-0 SUBJECTIVE DPP If - nd re solutions of the equtions k λ 0. Find the vlue of k nd λ. 5. Find the vlue of k for which qudrtic eqution (k - ) (k - ) 5k hs equl roots.. The sum of the squres of two consecutive positive integers is 545. Find the integers. 4. A mn is five times s old s his son nd the sum of the squres of their ges is 06. Find their ges. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 4

44 5. The sides (in cm) of right tringle contining the right ngles re 5 nd -. If the re of the tringle is 60 cm. Find its perimeter. 6. The lengths of the sides of right tringle re 5, 5 nd -. If > 0 find the length of ech sides. 7. A two digit numer is four times the sum nd three times the product of its digits, find the numer [CBSE 000] 8. The numer of frction is less thn its denomintor. If is dded to ech of the numertor nd denomintor, the frction is incresed. Find the frction 8 [CBSE - 007] Solve the qudrtic eqution An eroplne left 0 minutes lter then its scheduled time nd in order to rech its destintion 500 km w in time. it hs to increse its speed 50 km/h from its usul speed. Determine its usul speed. [CBSE-005]. A motor ot whose speed is 8 km/h in still wter tkes hours more to go 4 km upstrem thn to return downstrem to the sme spot. Find the speed of the strem. [CBSE-008]. Two wter tps together cn fill tnk in 9 hours. The tp of lrger dimeter tkes 0 hours less tht the 8 smller one to fill the tnk seprtel. Find the time in which ech tp cn seprtel fill the tnk. [CBSE-008] ANSWERS (Ojective DPP # 5.) Que. 4 5 Ans. B C A C C (Sujective DPP # 5.). k 9 λ -. k or. 6, ers & ers cm 6. 7, 5, km/h. 6 km/hr. Smller tp hr, Lrger tp 5 hr IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 44

45 C L A S S E S... the support ARITHMATIC PROGRESSIONS 6. PROGRESSIONS : Those sequence whose terms follow certin ptterns re clled progression. Generll there re three tpes of progression. (i) Arithmetic Progression (A.P.) (ii) Geometric Progression (G.P.) 6. ARTHMETIC PROGRESSION : IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL (iii) Hrmonic Progression (H.P.) A sequence is clled n A.P., if the difference of term nd the previous term is lws sme. i.e. d t n t n Constnt for ll n N. The constnt difference, generll denoted d is clled the common difference. E. Find the common difference of the following A.P. :,4,7,0,, (constnt). Common difference (d). 6. GENERAL FORM OF AN A.P. : If we denote the strting numer i.e. the st numer nd fied numer to the dded is d then, d, d, d, 4d,... forms n A.P. E. Find the A.P. whose st term is 0 & common difference is 5. Given : First term () 0 & Common difference (d) 5. A.P. is 0, 5, 0, 5, 0, n th TERM OF AN A.P. : Let A.P. e, d, d, d,... Then, First term ( ) 0.d Second term ( ).d Third term ( ).d n th term ( n) (n - ) d n (n - ) d is clled the n th term. E. Determine the A.P. whose their term is 6 nd the difference of 5 th term from 7 th term is. Given : ( - ) d d 6...(i) 7-5 ( 6d) - ( 4d) 6d - - 4d d d 6 Put d 6 in eqution (i) 6-4 A.P. is 4, 0, 6,, 8,......(ii) 45

46 E.4 Which term of the sequence 7, 70, 68, 66,... is 40? Here st term 7 nd common difference d For finding the vlue of n n (n - )d 40 7 (n - ) (-) n - - n -4 - n n 7 7 th term is 40. E.5 Is 84, term of the sequence,7,,...? Here st term () nd common difference (d) 7-4 n th term ( n) (n - ) d 84 (n - ) 4 8 4n n 85 n 4 Since, n is not nturl numer. 84 is not term of the given sequence. E.6 Which term of the sequence 0, 9,8,7 is the st negtive term. 4 Here st term () 0, common difference (d) Let n th term of the given A.P. e st negtive term n < 0 i.e. (n - ) d < 0 8 n 0 (n - ) < 0 < n > 8 n > n > 7 Since, 8 is the nturl numer just greter then 7. st negtive term is 8 th. E.7 If p th, q th nd r th term of n A.P. re,,c respectivel, then show thn (q - r) ( - p) c(p - q) 0. p A (p - ) D...() E.8 q A (q - ) D...() r c A (r ) D c...() Now, L.H.S. (q - r) (r - p) c (p - q) {A (p - )D} (q - r) {A (q - )D} (r - p) {A (r - )D} (p - q) 0. R.H.S If m times the m th term of n A.P. is equl to n times its n th term. Show tht the (m n) th term of the A.P. Let A the st term nd D e the common difference of the given A.P. Then, m m n n m[a (m - )D] n[a (n - )D] A(m - ) D[m n (m - n) - (m - n)] 0 A (m n - )D 0 mn 0 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 46

47 E.9 If the p th term of n A.P. is q nd the q th term is p, prove tht its n th term is (p q - n). p q A (p - ) D q...(i) & q p A (q - ) D p Solve (i) & (ii) to get D - & A p q - n A (n - ) D n (p q - ) (n - ) (-) n p q - n. E.0 If the m th term of n A.P. n nd n th term e m then show tht its (mn) term is. & m A (m )D n n...(i) m A (n )D m m...(ii) B solving (i) & (ii) D mn & A mn mn A (mn - ) D. 6.5 m th TERM OF AN A.P. FROM THE END : Let e the st term nd d e the common difference of n A.P. hving n terms. Then m th term from the end is (n - m ) th term from eginning or {n - (m - )} th term from eginning. E. Find 0 th term from the end of n A.P.,7, (n - )4 n 0 0 th term from end m 0 0-(0-) from the eginning. 8 (8 ) SELECTION OF TERMS IN AN A.P. : E. Sometimes we require certin numer of terms in A.P. The following ws of selecting terms re generll ver convenient. No. of Terms Terms Common Difference For terms d,, d d For 4 terms d, d, d, d d For 5 terms d, d,, d, d d For 6 terms 5d, d, d, d, d, 5d d The sum of three numer in A.P. is - nd their product is 8. Find the numers. Three no. s in A.P. e - d,, d - d d & ( - d) ( d) 8 ( - d ) 8 (-) ( - d ) 8 - d - 8 d 9 d ± If 8 & d numers re -4, -,. If 8 & d - numers re, -, -4. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 47

48 6.7 SUM OF n TERMS OF AN A.P. : Let A.P. e, d, d, d,... (n - )d Then, S n ( d)... { (n - ) d} { (n - ) d}...(i) lso, S n { (n - )d} { (n - )d}... ( d)...(ii) Add (i) & (ii) S n (n - )d (n - )d... (n - )d S n n [ (n - ) d] n S n [ (n )d] S n n n [ (n )d] [ l] n S n [ l] where l is the lst term. E. Find the sum of 0 terms of the A.P.,4,7,0... n 0, d S n [ (n )d] S 0 [() (0 )] E.4 Find the sum of ll three digit nturl numers. Which re divisile 7. st no. is 05 nd lst no. is 994. Find n n (n )7 8 Sum, S 8 [05 994] 6.8 PROPERTIES OF A.P. : (A) (B) (C) (D) (E) For n rel numers nd, the sequence whose n th term is n n is lws n A.P. with common difference (i.e. coefficient of term contining n) If n n th term of sequence is liner epression in n then the given sequence is n A.P. If constnt term is dded to or sutrcted from ech term of n A.P. then the resulting sequence is lso n A.P. with the sme common difference. If ech term of given A.P. is multiplied or divided non-zero constnt K, then the resulting sequence is lso n A.P. with common difference Kd or the common difference of the given A.P. respectivel. Where d is In finite A.P. the sum of the terms equidistnt from the eginning nd end is lws sme nd is equl to the sum of st nd lst term. (F) If three numers,,c re in A.P., then c. E.5 Check whether n n is n A.p. or not. n n Then n (n ) n - n (n n ) - n - n 4n - n - 4n, which is not constnt The ove sequence is not n A.P. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 48

49 DAILY PRACTIVE PTOBLEMS # 6 OBJECTIVE DPP p th term of the series... will e n n n p p (A) (B) (C) n n. 8 th term of the series 0... will e n (D) p n p (A) 5 (B) 5 (C) 0 (D) 0. If 9 th term of n A.P. e zero then the rtio of its 9 th nd 9 th term is (A) : (B) : (C) : (D) : 4. Which term of the sequence,8,,8,... is 498 (A) 95 th (B) 00 th (C) 0 th (D) 0 th 5. Which of the following sequence is n A.P. (A) f(n) n N (B) f(n) kr n, n N (C) f(n) (n )kr n, n N (D) f(n), n N n n 6. If the n th term of n A.P. e (n - ) then the sum of its firs n terms will e (A) n - (B) (n - ) (C) n (D) n 7. The interior ngles of polgon re in A.P. if the smllest ngles e 0 0 nd the common difference e 5, then the numer of sides is (A) 8 (B) 0 (C) 9 (D) 6 8. In the first, second nd lst terms of n A.P. e,, respectivel, then its sum will (A) SUBJECTIVE DPP - 6. (B) ( ). Is 5 term of the A.P. 5, 8,, 4,...? (C) ( ) (D) 4( ). Find the common difference of n A.P. whose first term is 00 nd the sum of whose first si terms is five times the sum of the net si terms.. Find three numer in A.P. whose sum is nd their product is A student purchsed pen for Rs. 00. At the end of 8 ers, it ws vlued t Rs. 0. Assuming the erl deprecition is constnt mount, find the nnul deprecition./ 5. The fourth term of n A.P. is equl to three times the first term nd the seventh term eceeds twice the third one. Find the first term nd the common difference. 6. Which term of the sequence 7,6,5,4... is the first negtive term If S n n p nd S m m p ( m n) in n A.P. Prove tht S p p. 8. Find the sum of ll the three digit numers which leve reminder when divided Find the sum of ll two digit odd positive numers 0. Find the 0 th term from end of the A.P. 4,9,4,..., logs re stcked in the following mnner: 0 logs in the ottom row, 9 in the net row, 8 in the row IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 49

50 net to it nd so on. In how mn rows the 00 logs re plced nd how mn logs re in the top row?. The sum of the first n term of n A.P. is given S n n 4n. Determine the A.P. nd its th term. [CBSE - 004]. Find the sum of the first 5 terms of n A.P. whose n th term is given t n -n [CBSE - 004] 4. Find the numer of terms of A.P. 54, 54, so tht their sum is 5. [CBSE - 005] 5. In n A.P., the sum of first n terms is n 5n Find its 5 th term. [CBSE - 006] 6. Which term of the rithmetic progression 8, 4 0, 6,... will e 7 more thn its 4 st term?[cbse - 006] 7. The first term, common difference nd lst term of n A.P. re, 6 nd 5 respectivel. Find the sum of ll terms of this A.P. [CBSE - 007] 8. Write the net term of the 8, 8,,... [CBSE - 008] 9. The sum of the 4 th nd 8 th terms of n A.P. is 4 nd the sum of the 6 th nd 0 th terms is 44. Find the first three terms of the A.P. [CBSE - 008] ANSWERS (Ojects DPP # 6.) Que Ans. B A B B A C C C (Sujective DPP # 6.). No ,7, , 6. rd rows, 5 logs. -,5,,... & , rd , -8, - IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 50

51 C L A S S E S... the support CO-ORDINATE GEOMETRY 7. RECTANGULAR CO-ORDINATES : Tke two perpendiculr lines X OX nd Y OY intersecting t the point O. X OX nd Y OY re clled the coordinte es. X O is clled the X-is, Y OY is clled the Y-is nd O is clled the origin. Lines X OX nd Y OY re sometimes lso clled rectngulr es. 7. () Co-ordintes of Point : Let P e n point s shown in figure. Drw PL nd PM perpendiculrs on Y-is nd X-is, respectivel. The length LP (or OM) is clled the - coordinte of the sciss of point P nd MP i clled the - coordinte or the ordinte of point P. A point whose sciss is nd ordinte is nmed s the point (,) or P(,). IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 5

52 The two liens X OX nd Y OY divide the plne into four prts clled qudrnts. XOY, YOX X OY nd Y OX re, respectivel, clled the first, second third nd fourth qudrnts. The following tle shows the signs of the coordintes of pins situted in different qudrnts : Qudrnt X-coodrinte Y-coordinte Point First qudrnt (, ) Second qudrnt - (-, ) Third qudrnt - - (-, -) Fourth qudrnt - (, -) REMAKS (i) (ii) (iii) (iv) Asciss is the perpendiculr distnce of point from -is (i.e., positive to the right of -is nd negtive to the left of - is) Ordinte is positive ove - is nd negtive elow -is. Asciss of n point on -is is zero. Ordinte of n point of -is is zero. (v) Co-ordintes of the origin re (0,0) 7. DISTACE BETWEEN TWO POINTS : Let two points e P (, ) nd Q(, ) Tke two mutull perpendiculr lines s the coordinte is with O s origin. Mrk the points P(, ) nd Q (, ). Drw lines PA, QB perpendiculr to X-is from the points P nd Q, which meet the X-is in points A nd B, respectivel. Drw lines PC nd QD perpendiculr to Y-is, which meet the Y-is in C nd D, respectivel. Produce CP to meet BQ in R. Now OA sciss of P Similrl, OB, OC nd OD Therefore, we hve PR AB OB - OA - Similrl, QR QB - RB QB - PA - Now, using Pthgors Theorem, in right ngled tringle PRQ, we hve PQ Pr RQ or PQ ( - ) ( - ) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 5

53 Since the distnce or length of the line-segment PQ is lws non-negtive, on tking the positive squre root, we get the distnce s PQ ( ) ( ) This result is known s distnce formul. Corollr : The distnce of point P(, ) from the origin (0,0) is given OP Some useful points :.In questions relting to geometricl figures, tke the given vertices in the given order nd proceed s indicted. (i) For n isosceles tringle - We hve to prove tht t lest two sides re equl. (ii) For n equilterl tringle - We hve to prove tht three sides re equl. (iii) For right -ngled tringle - We hve to prove tht the sum of the squres of two sides is equl to the squre of the third side. (iv) for squre - We hve to prove tht the four sides re equl, two digonls re equl. (v) For rhomus - We hve to prove tht four sides re equl (nd there is no need to estlish tht two digonls re unequl s the squre is lso rhomus). (vi) For rectngle - We hve to prove tht the opposite sides re equl nd two digonls re equl. (vii) For Prllelogrm - We hve to prove tht the opposite sides re equl (nd there is no need to estlish tht two digonls re unequl st the rectngle is lso prllelogrm).. for three points to e colliner - We hve to prove tht the sum of the distnces etween two pirs of points is equl to the third pir of points. E. Find the distnce etween the points (8, -) nd (, -6). Let the points (8, -) nd (, -6) e denoted P nd Q, respectivel. Then, distnce formul, we otin the distnce PQ s PQ ( 8) ( 6 ) ( 5) ( 4) 4 unit E. Prove tht the points (, ),, nd (, ) re the vertices of n isosceles tringle. Let the point (, -),, nd (, ) e denoted P, Q nd R, respectivel. Now PQ 8 4 QR 8 4 E. PR ( ) ( ) 9 From the ove, we see tht PQ QR The tringle is isosceles. Using distnce formul, show tht the points (-, ), (, -) nd (9, -0) re colliner. Let the given points (-, ), (, -) nd (9, -0) e denoted A, B nd C, respectivel. Points A, B nd C will e colliner, if the sum of the lengths of two line-segments is equl to the third. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 5

54 Now, AB ( ) ( ) BC (9 ) ( 0 ) AC (9 ) ( 0 ) Since, AB BC 4 8 AC, the, points A, B nd C re colliner. E.4 Find point on the X-is which is equidistnt from the points (5, 4) nd (-, ). Since the required point (s P) is on the X-is, its ordinte will e zero. Let the sciss of the point e. Therefore, coordintes of the point P re (, 0). Let A nd B denote the points (5, 4) nd (-, ), respectivel. Since we re given tht AP BP, we hve AP BP i.e., ( - 5) (0-4) ( ) (0 - ) or or -4-8 or Thus, the required point is (, 0). E.5 The vertices of tringle re (-, 0), (, ) nd (, -). Is the tringle equilterl, isosceles or sclene? Let the points (-, 0), (, ) nd (, -) e denoted A, B nd C respectivel. Then, AB ( ) ( 0) 5 BC ( ) ( ) 7 E.6 nd AC ( ) ( 0 0) Clerl, AB BC AC. Therefore, ABC is sclene tringle. The length of line-segments is 0. If one end is t (, -) nd the sciss of the second end is 0, show tht its ordinte is either or -9. Let (, -) e the point A. let the ordinte of the second end B e. Then its coordintes will e (0, ). AB (0 ) ( ) 0 (Given) or or or or ( 9) ( - ) 0 Therefore, 9 or. E.7 Show tht the points (-, 5), (, -4) nd (7, 0) re the vertices of right tringle. Let the three points e A(-, 5), B(, - 4) nd C(7, 0). Then AB ( ) (-4-5) 06 BC (7 - ) (0 4) AC (7 ) (0-5) 06 We see tht BC AB AC A 90 0 Thus, ABC is right tringle, right ngled t A. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 54

55 E.8 If the distnce of P (, ) from A (5, ) nd B(-, 5) re equl, prove tht. P(, ), A (5, ) nd B (-, 5) re the given points. AP BP (Given) AP BP or AP - BP 0 or {( - 5) ( - )} - {( ) ( - 5) } 0 or or or. 7. SECTION FORMULAE : 7. () Formul for Internl Division : The coordintes of the pint which divided the line segment joining the pints (, ) nd, ) internll m n m m in the rtio m : n re given, m n m n Proof : Let O e the origin nd let OX nd OY e the X-is nd Y-is respectivel. Let A(, ) nd B(, ) et the given points. Let (, ) e the coordintes of the point p which divides AB internll in the rtio m : n Drw AL OX, BM OX, PN O. Also, drw AH nd PK perpendiculr from A nd P on PN nd BM respectivel. Then OL, ON, OM, AL, PN nd BM. AH LN ON - OL -, PH PH - HN PN - AL -, PK NM OM - ON - nd BK BM - MK BM - PN -. Clerl, AHP nd PKB re similr. AP AH PH BP PK BK m n Now, m n m - m n - n m n m n m n nd m n m n m - m n - n m n m n m n m n m n m n Thus, the coordintes of P re, m n m n REMARKS If P is the mid-point of AB, then it divides AB in the rtio :, so its coordintes re, 7. () Formul for Eternl Division : The coordintes of the points which divides the line segment joining the points (, ) nd (, ) eternll in the rtio m : n re given m n m n, m n m n IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 55

56 E.9 Find the coordintes of the point which divides the line segment joining the points (6, ) nd (-4, 5) in the rtio : (i) internll (ii) eternll. Let P(, ) e the required point. (i) For internl division, we hve nd 0 nd 5 So the coordintes of P re (0, /5) (ii) For eternl division, we hve n - 4 nd 9 So the coordintes of P re (-4, 9). E.0 In which rtio does the point (-, -) divides the line segment joining the pints (4, 4) nd (7, 7)? Suppose the point C(-, -) divides the line joining the points A(4, 4) nd B(7, 7) in the rtio k : Then, the 7k 4 7k 4 coordintes of C re, k k But, we re given tht the coordintes of the points C re (-, -). Thus, C divides AB eternll in the rtio 5 : 8. 7k 4 k k E. In wht rtio does the X-is divide the line segment joining the points (, -) nd (5, 6)? 5λ 6λ Let the required rtio e k :. Then the coordintes of the point of division re,. But, it is k k E. point on X-is on which -coordinte of ever point is zero. 6λ 0 k k Thus, the required rtio is : or :. A (, ) nd B(, -) re two points nd D is point on AB produced such tht AD AB. Find the coordintes of D. We hve, AD AB. Therefore, BD AB. Thus D divides AB eternll in the rtio AD : BD : Hence, the coordintes of D re (4, -)., E. Determine the rtio in which the line divides the segment joining the pints (, ) nd (, 7). Suppose the line divides the line segment joining A(, ) nd B(, 7) in the rtio k : t point k 7k C. The, the coordintes of C re, But, C lies on - 9 0, therefore k k k 7k 9 0 k k So, the required rtio is : 4 internll. A(, ) B(,-) D 6k 7k - 9k k IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 56

57 57 IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 7.4 CENTROID OF A TRIANGLE : Prove tht the coordintes of the tringle whose vertices re (, ), (, ) nd (, ) re,. Also, deduce tht the medins of tringle re concurrent. Proof : Let A(,, B(, ) nd C(, ) e the vertices of ABC whose medins re AD, BE nd CF respectivel. So. D,E nd F re respectivel the mid-points of BC, CA nd AB. Coordintes of D re,. Coordintes of point dividing AD in the rtio : re ), (,, ), ( ), (,.,. The coordintes of E re,. The coordintes of point dividing BE in the rtio : re, ) (., ) (. Similrl the coordintes of point dividing CF in the rtio : re, Thus, the point hving coordintes, is common to AD, BE nd CF nd divides them in the rtio :. Hence, medins of tringle re concurrent nd the coordintes of the centroid re,.,

58 7.5 AREA OF A TRIANGLE : Let ABC e n tringle whose vertices re A(, ) B(, ). Drw BL, AM nd CN perpendiculr from B,A nd C respectivel, to the X-is. ABLM, AMNC nd BLNC re ll trpeziums. Are of ABC Are of trpezium ABLM Are of trpezium AMNC - Are of trpezium BLNC We know tht, Are of trpezium Therefore Are of ABC Are of ABC (Sum of prllel sides) (distnce /w them) (BL AM) (LM) (AM CN) MN- (BL CN) (LN) ( ) - ) ( ) ( - ) - ( ) ( - ) Are of ABC [ ( ) ( ) ( )] 7.5 () Condition for collinerit : Three points A (, ) B(, ) nd C(, ) re colliner if Are of ABC AREA OF QUADRILATERAL : Let the vertices of Qudrilterl ABCD re A(, ), B(,, C(, ) nd D( 4, 4 ) So, Are of qudrilterl ABCD Are of ABC Are of ACD E.4 The vertices of ABC re (-, ), (5, 4) nd (, -) respectivel. Find the re of tringle. A(-, ), B(-, ) nd C(, -) e the vertices of tringle. So, -, ; 5, 4; - Are of ABC [ ( ) ( ) ( )] IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 58

59 [( )(4 ) (5)( ) ( 4) ] [ 4 ( 0) ( 6) ] E Sq. unit. The re of tringle is 5. Two of its vertices re (, ) nd (, -). The third verte lies on. Find the third verte. Let the third verte e (, ) re of tringle [ ( ) ( ) ( )] As ;, - ; Are of 5 sq. unit 5 ( ) ( ) ( ) ± 0 Tking positive sign (i) Tking negtive sing (ii) Given tht (, ) lies on So, -...(iii) Solving eq. (i) & (iii) 7 Solving eq. (ii) & (iii),, 7 So the third verte re, or, E.6 Find the re of qudrilterl whose vertices, tken in order, re (-, ), B(5, 4), (7, -6) nd D(-5, -4). Are of qudrilterl Are of ABC Are of ACD So, Are of ABC ( )(4 6) 5( 6 ) 7( 4) Sq. units IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 59

60 Are of ACD ( 6 4) 7( 4 ) ( 5)( 6) So, Sq. units Are of qudrilterl ABCD Sq. units. DAILY PRACTIVELY PROBLEMS # 7 OBJECTIVE DPP The points (-, -), (0, 0), (, ) nd (, ) re (A) Colliner (B) Vertices of prllelogrm (C) Vertices of rectngle (D) None of these. If the points (5, ), (, p) & (4, ) re colliner then the vlue of p will e (A) (B) 5 (C) (D) -. Length of the medin from B on AC where A(-, ), B(, -), (5, ) is (A) 8 (B) 0 (C) (D) 4 4. The points (0, -), (-, ), (6, 7) nd (8, ) re - (A) Colliner (C) Verticls of rectngle, which is not squre (D) None of these (B) Vertices of prllelogrm which is not rectngle 5. If (, -4) nd (-6, 5) re the etremities of the digonl of prllelogrm nd (-, ) is third verte, then its fourth verte is - (A) (-, 0) (B) (0, -) (C) (-, ) (D) None of these 6. The re of tringle whose vertices re (, c ), (, c) nd (-, c - ) re (A) (B) (C) c (D) c 7. The re of the qudrilterl s the coordintes of whose verticls re (, -,) (6, ), (5, ) nd (, 4) re (A) 9 (B) 5 SUBJECTIVE DPP Find the distnce etween the points : (i) P (-6, 7) nd Q(-, -5). (ii) A(t, t ) nd B(t, t ). (C) (D). If the point (, ) is equidistnt from the points (, - ) nd ( -, ), prove tht.. Find the vlue of, if the distnce etween the points (, -) nd (, ) is Show tht the points (, ), (-, -) nd, ) re the vertices of n equilterl tringle. 5. Show tht the points (, ), (-, 7) nd (, -) re colliner. 6. Prove tht (, -), (-, ) nd (5, ) re the vertices of right ngled tringle. Find the re of the tringle nd the length of the hpotenuse. 7. If A(-, ), B(, -) nd C(5, ) re the vertices of tringle ABC, find the length of the medin pssing through the verte A. 8. Show tht the points A(,), B(5, 4), C(, 8) nd D(-, 6) re the vertices of squre. 9. The sciss of point is twice its ordinte nd the sum of the sciss nd the ordinte is -6. Wht re the coordintes of the point? 0. If two vertices of tringle re (, 7) n (-, 5) nd its centroid is (, ), find the coordintes of the third verte. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 60

61 . If the mid point of the line-segment joining the points (-7, 4) nd (K, 4) is (, ), where 5, find the vlue of K.. Prove ht the points (, 0), (0, ) nd (, ) re colliner if.. The co-ordintes of two points A & B re (, 4) nd (5, -) respectivel. Find the co-ordinte of point P if PA PB, the re of APB Four points A(6, ), B(-, 5) C(4, -) nd D(, ) re given in such w tht Are ( DBC) Are ( ABC) find. 5. Show tht the points A(, -), B(4, 0), C(, ) nd D(-, ) re the vertices of rectngle. [CBSE-004] 6. Determine the rtio in which the point (-6, ) divides the join of A(-, -) nd B(-8, 9). Also find the vlue of. [CBSE 004] 7. Find pint on X-is which is equidistnt from the points (7, 6) nd (-, 4). [CBSE - 005] 8. The line segment joining the points (, -4) nd (, ) is trisected t the pints P nd Q. if the coordintes of P nd Q re (p, -) nd (5/, ) respectivel. Finds the vlue of p nd q. [CBSE 005] 9. If A(-, -), B(, 0), C(4, ) nd D(, ) re the verities of prllelogrm, find the vlues of nd.[ -006] 0. The coordintes of one end point of dimeter of circle re (4, -) nd the coordintes of the centre of the circle re (, -). Find the coordintes of the other end of the dimeter. [CBSE-007]. The pint R divides the line segment AB, where A(-4, 0) nd B(0, 6) re such tht AR 4 AB. Find the coordintes or R. (CBSE - 008). For wht vlue of k re the pints (, ), (, k) nd (-, 4) colliner?[cbse - 008]. Find the re of the ABC with vertices A(-5, 7), B (-4, -5) nd C(4, 5).[CBSE - 008] 4. If the point P(,) is equidistnt from the points A(,6) nd B(-,4) prove tht [CBSE - 008] 5. If A(4-8), B(,6) nd C(5,- 4) re the vertices of ABC, D is the mid-point of BC nd is P is point on AD AP joined such tht find the coordintes of P. [CBSE - 008] PD ANSWERS (Ojective DPP # 7.) Que Ans. A B B C A A C (Sujective DPP # 7.). (i) (ii) (t t ) 5 (t t ) 4. 7 or - 6. sq. units, units 9. (-4, -) 0. (, -). K -5. (7, ) or (, 0) 4., :, 5 7. (, 0) 8. p 7/, q 0 9., 0. (-, -5). 9 (, ). k -. 5 sq. units 5. (4, -) IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 6

62 C L A S S E S... the support TRIANGLES 8. CONGRUENT AND SIMILAR FIGURES: Two geometric figures hving the sme shpe nd size re known s congruent figures. Geometric figures hving the sme shpe ut different sizes re known s similr figures. 8. SIMILAR TRIANGLES: Two tringles ABC nd DEF re sid to e similr if their (i) Corresponding ngles re equl. i.e. A D, B E, C F And, (ii) Corresponding sides re proportionl i.e. 8. () Chrcteristic Properties of Similr Tringles : AB DE BC EF AC DF (i) (AAA Similrit) If two tringles re equingulr, then the re similr. (ii) (SSS Similrit) If the corresponding sides of two tringles re proportionl, then the re similr. (iii) (SAS Similrit) If in two tringle s one pir of corresponding sides re proportionl nd the included ngles re equl then the two tringles re similr. 8. () Results Bsed Upon Chrcteristic Properties of Similr Tringles : (i) If two tringles re equingulr, then the rtio of the corresponding sides is the sme s the rtio of the corresponding medins. (ii) If two tringles re equingulr, then the rtio of the corresponding sides is sme t the rtio of the corresponding ngle isector segments. (iii) if two tringles re equingulr then the rtio of the corresponding sides is sme t the rtio of the corresponding ltitudes. (vi) If one ngle of tringle is equl to one ngle of nother tringle nd the isectors of these equl ngles divide the opposite side in the sme rtio, then the tringles re similr. (v) If two sides nd medin isecting one of these sides of tringle re respectivel proportionl to the two sides nd the corresponding medin of nother tringle, then the tringles re similr. (vi) If two sides nd medin isecting the third side of tringle re respectivel proportionl to the corresponding sides nd the medin nother tringle, then two tringles re similr. 8. THALES THEOREM (BASIC PROPROTIONALITY THEOREM) : Sttement : If line is drwn prllel to one side of tringle to intersect the other sides in distinct points, then the other two sides re divided in the sme rtio. Given: To Prove : A tringle ABC in which line prllel to side BC intersects other two sides AB nd AC t D nd E respectivel. AD AE DB EC IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 6

63 Construction : Join BE nd CD nd drw DM AC nd EN AB. Proof : Are of ADE ( se height) AD EN. Are of ADE is denoted s re (ADE) So, r(ade) DB EN And r(bde) DB EN, Therefore, Similrl, AD EN r(ade) AD.(i) r(bde) DB DB EN r(ade AE DM nd r(dec EC DM. Corollr : And AE DM r(ade) AE...(ii) r(dec) EC DM EC Note tht BDE nd DEC re on the sme se DE nd etween the two prllel lines BC nd DE. So, r(bde) r(dec)...(iii) Therefore, from (i), (ii) nd (iii), we hve : AD AE Hence Proved. DB EC If in ABC, line DE BC, intersects AB in D nd AC in E, then (i) (iv) DB EC AB AC AD AE (ii) (ii) AD AE AD AE AB AC AB AC DB EC (v) DB EC AB AC 8. () Converse of Bsic Proportionlit Theorem : If line divides n two sides of tringle in the sme rtio, then the line must e prllel to the third side. 8. () Some Importnt Results nd Theorems : (i) The internl isector of n ngle of tringle divides the opposite side internll in the rtio of the sides contining the ngle. (ii) In tringle ABC, if D is point on BC such tht D divides BC in the rtio AB : AC, then AD is the isector of A. (iii) The eternl isector of n ngle of tringle divides the opposite sides eternll in the rtio of the sides contining the ngle. (iv) The line drwn from the mid-point of one side of tringle prllel to nother side isects the third side. (v) The line joining the mid-points of two sides of tringle is prllel to the third side. (vi) The digonls of trpezium divide ech other proportionll. (vii) If the digonls of qudrilterl divide ech other proportionll, then it is trpezium. (viii) An line prllel to the prllel sides of trpezium divides the non-prllel sides proportionll. (i) If three or more prllel lines re intersected two trnsversl, then the intercepts mde them on the trnsversl re proportionl. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 6

64 E. ILLUSTRATIONS : In ABC, D nd E re points on the sides AB nd AC respectivel such tht DE BC. If AD 4 -, AE 8-7, BD - nd CE 5 -, find the vlue of. [CBSE - 006] In ABC, we hve DE BC AD AE [B Bsic Proportionlit Theorem] DB EC ( ) ( - ) 0 or - So, the required vlue of is. [ - is neglected s length cn not e negtive]. E. E. D nd E re respectivel the points on the sides AB nd AC of ABC such tht AB cm, AD 8 cm, AE cm nd AC 8 cm, show tht DE BC. We hve, AB cm, AC 8 m, AD 8 cm nd AE cm. BD AB - AD ( - 8) cm 4 cm CE AC - AE (8 ) cm 6 cm Now, And, AD BC 8 4 AE CE 6 AD AE BD CE Thus, DE divides sides AB nd AC of ABC in the sme rtio. Therefore, the conserve of sic proportionlit theorem we hve DE BC. In trpezium ABCD AB DC nd DC AB. EF drwn prllel to AB cuts AD in F nd BC in E such BE tht. Digonl DB intersects EF t G. Prove tht 7FE 0AB. EC 4 In DFG nd DAB, FDG ADB [Common] [Corresponding s AB FG] DFG ~ DAB [B AA rule of similrit] DF FG...(i) DA AB Agin in trpezium ABCD IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 64

65 EF AB DC AF DF BE EC AF BE DF 4 (given) EC 4 AF DF 4 AF DF DF AD 7 DF DF 4.(ii) AD 7 From (i) nd (ii), we get FG 4 4 i.e. FG AB...(iii) AB 7 7 In BEG nd BCD, we hve BEG BCD GBE DBC [Corresponding ngle EG CD] [Common] BEG ~ BCD [B AA rule of similrit] BE EG BC CD E.4 EG BE EC 4 EC BE 4 BC 7 i.e.. 7 CD EG 7 BE BE BE EG CD ( AB) 7 7 [ CD AB (given)] 6 EG AB...(iv) 7 Adding (iii) nd (iv), we get FG EG AB AB AB EF AB i.e., 7EF 0AB. Hence proved. 7 In ABC, if AD is the isector of A, prove tht Are ( ABD) Are ( ACD) AB AC In ABC, AD is the isector of A. AB BD...(i) [B internl isector theorem] AC DC From A drw AL BC BD.AL Are ( ABD) Are ( ACD) DC.AL BD DC IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL AB AC [From (i)] Hence Proved. 65

66 E.5 E.6 BAC 90 0, AD is its isector. IF DE AC, prove tht DE (AB AB) AB AC. It is given tht AD is the isector of A of ABC. AB BD AC DC AB BD [Adding on oth sides] AC DC AB AC BD DC AC DC AB AC AC BC DC In s CDE nd CBA, we hve DCE BCA [Common] DEC BAC [Ech equl to 90 0 ] So, AA-criterion of similrit CDE ~ CBA CD DE CB BA...(i) AB BC...(ii) DE DC From (i) nd (ii), we hve AB AC AC AB DE DE (AB AC) AB AC. In the given figure, PA, QB nd RC re ech perpendiculr to AC. Prove tht In PAC, we hve BQ AP BQ CB [ CBQ ~ CAP] AP CA CB (i) CA In ACR, we hve BQ CR BQ AB [ ABQ ~ ACR] CR AC AB (ii) z AC Adding (i) nd (ii), we get z CB AC AB AC AB BC z AC z z AC AC Hence Proved. z z IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 66

67 E.7 In the given figure, AB CD. Find the vlue of. Since the digonls of trpezium divide ech other proportionll. AO BO OC OD ( - 8) ( - ) 0 8 or. 8.4 AREAS OF SIMILAR TRIANGLS : Sttement : The rtio of the res of two similr tringles is equl to the squre of the rtio of their corresponding sides. Given : Two tringles ABC nd PQR such tht ABC ~ PQR [Shown in the figure] To Prove : r(abc) AB r(pqr) PQ BC QR CA RP Construction : Drw ltitudes AM nd PN of the tringle ABC n PQR. Proof : r(abc) BC AM And r(pqt) QR PN BC AM r(abc) BC AM So,...(i) r(pqr) QR PN QR PN Now, in ABM nd PQN, And B Q [As ABC ~ PQR] M N [90 0 ech] So, ABM ~ PQN [AA similrit criterion] AM AB Therefore,...(ii) PN PQ Also, ABC ~ PQR [Given] So, AB BC CA...(iii) PQ QR RP IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 67

68 Therefore, r(abc) BC AB [From (i) nd (ii)] r(pqr) QR PQ AB AB [From (iii)] PQ PQ AB PQ Now using (iii), we get r( ABC) AB r( PQR) PQ BC QR 8.4 () Properties of Ares of Similr Tringles : E.8 CA RP (i) The res of two similr tringles re in the rtio of the squres of corresponding ltitudes. (ii) The res of two similr tringles re in the rtio of the squres of the corresponding medins. (iii) The re of two similr tringles re in the rtio of the squres of the corresponding ngle isector segments. Prove tht the re of the equilterl tringle descried on the side of squre is hlf the re of the equilterl tringle descried on this digonls. [CBSE - 00] Given : A squre ABCD. Equilterl tringles BCE nd ACF hve een descried on side BC nd digonls AC respectivel. To prove : Are ( BCE). Are ( ACF) Proof : Since BCE nd ACF re equilterl. Therefore, the re equingulr (ech ngle eing equl to 60 0 ) nd hence BCE ~ ACF. Are( BCE) BC Are( ACF) AC Are( BCE) Are( ACF) Are( BCE) Are( ACF) BC ( BC) ABCD is squre Digonl (side) AC BC Hence Proved. 8.5 PYTHAGOREOUS THEOREM : Sttement : In right tringle, the squre of the hpotenuse is equl to the sum of the squre of the other two sides. Given : A right tringle ABC, right ngled t B. To prove : AC AB BC Construction : BD AC Proof : ADB & ABC DAB CAB BDA CBA [Common] [90 0 ech] IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 68

69 So, ADB ~ ABC [B AA similrit] AD AB [Sides re proportionl] AB AC or, AD. AC AB...(i) Similrl So, CD BC BC AC BDC ~ ABC or CD. AC BC...(ii) Adding (i) nd (ii), AD. AC CD. AC AB BC or, AC (AD CD) AB BC or AC.AC AB BC or, AC AB BC Hence Proved. 8.5 () Converse of Pthgorens Theorem : Sttement : In tringle, if the squre of one side is equl to the sum of the squres of the other two sides, then the ngle opposite to the first side is right ngle. Given : A tringle ABC such tht AC AB BC Construction : Construct tringle DEF such tht DE AB, EF BC nd E 90 0 Proof : In order to prove tht B 90. 0, it is sufficient to show ABC ~ DEF. For this we proceed s follows Since DEF is right - ngled tringle with right ngle t E. Therefore, Pthgors theorem, we hve DF DE EF DF AB BC [ DE AB nd EF BC (B construction)] DF AC [ AB BC AC (Given)] DF AC...(i) Thus, in ABC nd DEF, we hve AB DE, BC EF [B construction] And AC DF [From eqution (i)] ABC DEF [B SSS criteri of congruenc] B E 90 0 Hence, ABC is right tringle, right ngled t B. 8.5 () Some Results Deduced From Pthgorens Theorem : (i) In the given figure ABC is n otuse tringle, otuse ngled t B. If AD CD, then AC AB BC BC. BC (ii) In the given figure, if B of ABC is n cute ngle nd AD BC, then AC AB BC - BC. BD IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 69

70 (iii) In n tringle, the sum of the squres of n two sides is equl to twice the squre of hlf of the third side together with twice the squre of the medin which isects the third side. (iv) Three times the sum of the squres of the sides of tringle is equl to four times the sum of the squres o the medins of the tringle. E.9 In ABC, AB BC CA nd AD BC. Prove tht [CBSE - 00] (i) AD (ii) re ( ABC) (i) Here, AD BC. Clerl, ABC is n equilterl tringle. Thus, in ABD nd ACD AD AD [Common] And Now, ADB ADC AB AC RHS congruenc condition ABD ACD BD DC ABD is right ngled tringle [90 0 ech] AD AD AB BD [Using Pthgorens Theorem] 4 or (ii) Are ( ABC) BC AD E.0 BL nd Cm re medins of ABC right ngled t A. Prove tht 4(BL CM ) 5 BC In BAL BL AL AB...(i) [Using Pthgorens theorem] nd In CAM CM AM AC...(ii) [Using Pthgorens theorem] Adding () nd () nd then multipling 4, we get 4(BL CM ) 4(AL AB AM AC ) 4{AL AM (AB AC )} [ ABC is right tringle] 4(AL AM BC ) 4(ML BC ) [ LAM is right tringle] 4ML 4 BC [A line joining mid-points of two sides is prllel to third side nd is equl to hlf of it, ML BC/] BC 4BC 5BC Hence proved. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 70 [CBSE-006]

71 E. In the given figure, BC AB, AE AB nd DE AC. Prove tht DE.BC AD.AB. In ABC nd EDA, We hve ABC ADE [Ech equl to 90 0 ] ACB EAD [Alternte ngles] B AA Similrit ABC ~ EDA BC AD AB DE DE.BC AD.AB. Hence Proved. E. O is n point inside rectngle ABCD (shown in the figure). Prove tht OB OD OA OC Through O, drw PQ BC so tht P lies on A nd Q lies on DC. [CBSE - 006] Now, PQ BC E. Therefore, PQ AB nd PQ DC [ B 90 0 nd C 90 0 ] So, BPQ 90 0 nd CQP 90 0 Therefore, BPQC nd APQD re oth rectngles. Now, from OPB, OB BP OP...(i) Similrl, from ODQ, OD OQ DQ...(ii) From OQC, we hve OC OQ CQ...(iii) And form OAP, we hve OA AP OP...(iv) Adding (i) nd (ii) OB OD BP OP OQ DQ CQ OP OQ AP [As BP CQ nd DQ AP] CQ OQ OP AP OC OA [From (iii) nd (iv)] Hence Proved. ABC is right tringle, right-ngled t C. Let BC, CA, AB c nd let p e the length of perpendiculr form C on AB, prove tht (i) cp (ii) p Let CD AB. Then CD p Also, Are of ABC (Bse height) (AB CD) cp Are of ABC (BC AC) cp CP AB. (ii) Since ABC is right tringle, right ngled t C. AB BC AC c.. p cp c p IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 7

72 p p p E.4 In n equilterl tringle ABC, the side B is trisected t D. Prove tht 9 AD 7AB. ABC e cn equilterl tringle nd D e point on BC such tht [CBSE - 005] BC BC (Given) Drw AE BC, Join AD. BE EC (Altitude drown from n verte of n equilterl tringle isects the opposite side) BC So, BE EC In ABC AB AE EB...(i) AD AE ED...(ii) From (i) nd (ii) AB AD - ED EB AB AD - BC 6 BC ( BD DE 4 BC BC BC AB AD ( EB ) 6 4 AB AB AB AD ( AB BC) 6 4 6AB AB 6 9AB AD 8AB 6 7 AB 9AD DAILY PRACTIVE PROBLEMS # 8 OBJECTIVE DPP - 8. BC AD BC BC BC DE DE ) 6. The perimeters of two similr tringles re 5 cm nd 5 cm respectivel. If one side of first tringle is 9 cm, then the corresponding side of the other tringle is (A) 6. cm (B).4 cm (C) 5.4 cm (D) 8.4 cm. In the following figure, AE BC, D is the mid point of BC, hen is equl to (A) (C) d c d h 4 (B) (D) h d QR. Two tringles ABC nd PQR re similr, if BC : CA : AB : :, then is PR d 4 c (A) (B) (C) (D) 4. In tringle ABC, if ngle B 90 0 nd D is the point in BC such tht BD DC, then (A) AC AD CD (B) AC AD 5 CD C) AC AD 7 CD (D) AC AB 5 BD IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 7

73 5. P nd Q re the mid points of the sides AB nd BC respectivel of the tringle ABC, right-ngled t B, then (A) AQ CP AC (B) AQ CP (C) AQ CP 5 AC 4 6. In ABC, AD is the isector of A, meeting side BC t D. If AB 0 cm, AC 6 cm, BC cm, find BD. (A). (B) 8 (D) AQ CP 4 AC 5 AC 5 (C) 7.5 (D). 7. In tringle ABC, stright line prllel to BC intersects AB nd AC t point D nd E respectivel. If the re of ADE is one-fifth of the re of ABC nd BC 0 cm, then DE equls (A) cm (B) 5 cm (C) 4 cm (D) 4 5 cm 8. ABC is right-ngle tringle, right ngled t A. A circle is inscried in it. The lengths of the two sides contining the right ngle re 6 cm nd 8 cm, then rdius of the circle is (A) cm (B) cm (C) 4 cm (D) 8 cm SUBJECTIVE DPP Given GHE DFE 90 0, DH 8, DF, DG - nd DE 4. Find the lengths of segments DG nd DE.. In the given figure, DE is prllel to the se BC of tringle ABC nd AD : DB 5 :. Find the rtio : - AD (i) AB [CBSE - 000] (ii) Are of DEF Are of CFB. In Figure, ABC is right-ngled tringle, where ACB The eternl isector BD of ABC meets AC produced t D. If AB 7 cm nd BC 8 cm, find the AC nd BD. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 7

74 4. In figure, QPS RPT nd PST PQR. Prove tht PST ~ PQR nd hence find the rtio ST : PT, if PR : R 4 : In the figure, PQRS is prllelogrm with PQ 6 cm nd QR 0 cm. L is point on PR such tht RL : LP :. QL produced meets RS t M nd PS produced t N. Find the lengths of PN nd RM. 6. In ABC, D nd E re points on AB nd AC respectivel such tht DE BC. If AD.4 cm, AE. cm, DE cm nd BC 5 cm, find BD nd CE. 7. In tringle PQR, L n DM re two points on the se QR, such tht :PQ QRP nd RPM RQP. Prove tht : (i) PQL ~ RPM (ii) QL RM PL PM (iii) PQ QR QL 8. In figure, BAC 90 0, AD BC. prove tht AB BD - CD. 9. In figure, ACB 90 0, CD AB prove tht CD BD.AD. IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 74

75 0. In right tringle, prove tht the squre on the hpotenuse is equl to sum of the squres on the other two sides. Using the ove result, prove the following: In figure PQR is right tringle, right ngled t Q. If QS SR, show tht PR 4PS - PQ.. In ABC, ABC 5 0. Prove tht AC AB BC 4r ( ABC).. In figure, ABC nd DBC re two right tringles with the common hpotenuse BC nd with their sides AC nd DB intersecting t P. Prove tht AP PC DP PB. [CBSE - 000]. An point O, inside ABC, in joined to its vertices. From point D on AO, DE is drwn so tht DE AB nd EF BC s shown in figure. Prove tht DF AC. [CBSE-00] 4. In figure, D nd E trisect BC. Prove tht 8AE AC 5AD [CBSE - 006] 5. The perpendiculr AD on the se BC of ABC meets BC t D so tht DB CD. Prove tht 5AB 5AC BC. [CBSE - 007] 6. Prove tht the rtio of the res of two similr tringles is equl to the rtio of the squres on their corresponding sides. Using the ove, do the following :The digonls of trpezium ABCD, with AB DC, intersect ech other point O. If AB CD, find the rtio of the re of ΑΟΒ to the re of COD [CBSE - 008] IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 75

76 7. r( DEF) D, E nd F re the mid-points of the sides AB, BC nd CA respectivel of ABC. Find. [ - 008] r( ABC) 8. D nd E re points on the sides CA nd CB respectivel of ABC right-ngled t C. Prove tht AE BD AB DE. 9. BE AC In figure, DB BC, DE AB nd AC BC. Prove tht DE BC [CBSE - 008] ANSWERS (Ojective DPP # 8.) Que Ans. C A B B C C B B (Sujective DPP # 8.). 0 unit & 0 unit. (i) 8 5 (ii) cm., 8 4 cm : 4 5. PN 5 cm, RM 0.67 cm. 6. DB.6 cm, CE 4.8 cm 6. 4 : 7. : 4 ALL THE BEST FROM..SUHAAAAAAAAAAAAAAAAAAAAAAAAAG IITJEE Mths SUHAG SIR, Bhopl, Ph.(0755) , Pge Free Remed Clsses for clss 0 th, R-, ZONE-, M.P.NAGAR, BHOPAL 76