anihv ]T\ klmbn lbà sk - dn KWnXw PnÃm ] m-b v, ImkÀtKmUv 0-þ4
anihv 04 Advisory Committee Chairman : Adv. P.P. Syamala Devi President, District Panchayath, Kasaragod. Vice Chairman : Smt. K. Sujatha Standing Committee Chairman for Education & Health, District Panchayath, Kasaragod. Convenors : Sri. Srikrishna Kayarthaya, DDE, Kasaragod Dr. P.V. Krishna Kumar, Principal, DIET, Kasaragod. Smt. Vijayalakshmi H.S, DPO, SSA, Kasaragod. Members : Sri. I. Sathyanarayana Bhat, DEO Kasaragod. Sri. Mohanan, DEO, Kanhangad. Sri. K. Kamalakshan, Sr. Lecturer, DIET, Kasaragod. Academic Support : DIET Kasaragod
ADV. P.P. SYAMALA DEVI President District Panchayath Kasaragod Ph: 04994-5677 Fa: 5677 apj-samgn Athira Nelkkala Road Vidyanagar Kasaragod - 67 Ph: 04994-096 04994-079 Mob: 94956580 Date: 8.0.0 KpW-\n-e-hm-c-apÅ hnzym-`ymkw Hmtcm Ip«n-bp-sSbpw Ah-Im-i-am-Wv. AXv Dd- ph-cp- pi F Xv \½psS IÀ -hy-am-wv. hnzymà n-i-fpss ]T-\-\n-e-hmcw ani-hp-ä-xm- p-hm-\på bxv\- Ä FÃm kviqfp-i-fnepw \S- n-em n hcp- Xv Gsd kt m-j-i -c-am-wv. lbà sk - dn hnzymà n-i-fpss ]T-\-\n-e-hmcw DbÀ p-hm³ Imk-d-tKmUv PnÃm ] m-b- nsâ t\xr-xz- nâ \S- n-em- n-h-cp anihv ]²Xn PnÃ-bnse lbà sk - dn hnp-b-i-x-am\w KWy-amb coxn-bnâ DbÀ p-hm\pw PnÃ-bpsS hnzym-`ym-k -]n-t m- m-hø Hcp ]cn-[n-hsc ]cn-l-cn- p-hm\pw klm-bn- n-«p- v. lbà sk - dn ]co- -bv på aps m-cp {]hà -\- Ä FÃm kviqfp-i-fnepw \S- p-h-cp Cu kµà`- nâ Ip«n-I-fn Bi-b- -Ä IqSp-X ZrUo-I-cn- m³ DX-Ip-, A²ym-]-IÀ v klmb-i-amb Hcp ssi]p-kvxiw "anihv 04'Imk-d-tKmUv PnÃm ] m-b v Cu hàjhpw ]pd- n-d- p-i-bm-wv. tnmzy- Ä, NÀ -IÄ, sndp {Kq v {]hà -\- Ä XpS- nbhbneqss CXv km[n- p-sa p R Ä hniz-kn- p- p. Xmgv \ne-hm-c-apå Ip«n-IÄ pw DbÀ \ne-hm-c-apå Ip«n-IÄ pw Hcp t]mse KpW-I-c-am-Ip hyxykvx \ne-hm-c- nepå ]T\ {]hà -\- Ä CXn DÄs -Sp- m³ R Ä {i²n- n-«p- v. Cu hàjw PnÃbnse apgp-h³ Ip«n-I-sfbpw C+ \p api-fnepå t{kup-i-fn-te v DbÀ pi F \½psS e yw km m-xv -cn- m³ CXv klm-b-i-am-ip-sa v R Ä {]Xym-in- p- p. AXn-\mbn PnÃ-bnse apgph³ {][m\ A²ym-]-I-cp-sSbpw, A[ym-]-I-cp-sSbpw, Ip«n-I-fp-sSbpw, c n- Xm- -fp-ssbpw \m«p-im-cp-ssbpw BßmÀ -amb kl-i-c-whpw ]n p-w-bpw {]Xo- n- p-sim- v, {]kn- - U- v PnÃm ] m-b v, Imk-d-tKmUv.
DÅS w. KWnXw 5 þ 5 Resource Team :. Jayan T.V. GHSS Kakkat. A. Vidya GHSS Udma. V. Pavithran GHSS Udinoor 4
HIGHER SECONDARY MATHEMATICS 5
. RELATIONS AND FUNCTIONS Previous Knowledge Cartesian Product. A, B Ch iq\y-k-w- -fãmsb nâ A bnse Hmtcm AwKhpw BZy-kw-Jy-bmbpw B bnse Hmtcm AwKhpw c m-as kwjy-bmbpw hcp {Ia-tPm-Sn-I-fpsS KW-amWv A t{imêv B. CXns\ AXB F v kqnn- n- p- p. eample: A = (,) B = (,4) F nâ AXB ={(,), (,4), (,), (,4)}.. Relation (_Ôw) A F KW- nâ \n v B F KW- n-te- på relation FÃm-bvt mgpw AXB bpss D]- KWw (subset) Bbn-cn- pw. A bnâ \n v B bnte- på relation \nse AwK- Ä FÃmbvt mgpw {Ia-tPm-Sn-I-fm-bn-cn- pw.. Domain A bnâ \n v B bnte- på Hcp relation se {Ia-tPm-Sn-I-fpsS BZyAwK- -fpss KWs dnte-jsâ awvuew (domain) F p-]-d-bp- p. {Ia-tPm-Sn-bnse c m-as AwK- -fpss KWs Range of a relation F pw hnfn- p- p. 4. Functions A bnâ \n v B bnte- på dnte-j³ function BI-W-sa- nâ A bnse FÃm AwK- Ä pw Htc Hcp image ({]-Xn-_nw-_w) B bnâ D m-bn-cn- -Ww. hyxykvx coxn-bn-epå _Ô- Ä Refleive relations A relation R on A is a relfeive If ara, a A or (a,a), a A i.e, A bnse FÃm AwK- Ä pw AtX AwKw CtaPv Bbn-h-cp {Ia-tPm-Sn-IÄ R - Â D m-bn-cn- -Ww. 6
Symmetric Relation Ans: A relation R on A is said to be symmetric if (a,b) (b,a) AXm-bXv X n-«på R F _Ô- nâ (a,b) F {Ia-tPmSn R Â Ds - nâ (b,a) F {Ia-tPmSn RÂ \nà_- Ô-ambpw D m-bmâ am{xta R Hcp symmetric relation BsW v ]d-bp-i-bp-åq. DZm:-þ Let A = {,,, 4) Define R = {(, ), (, ), (, ), (, 4), (4, 4), (4, ), (, )} Here R is a refleive relation and symmetric relation. Because (a,a), a and (,4), (4, ) ; (, ) ; (, ) Eg:- A = {,,, 4}, A relation on R is defined as R = {(,), (,), (4,), (4,4), (,),,4)} X n-«på _Ôw Hcp refleive relation AÃ ImcWw (,), ]s R Symmetric relation BWv. ImcWw (,), (,), (4,), (,4). Transitive Relation Ans: A relation is transitive if (a,b), (b,c) (a,c) Hcp _Ôw R Transitive relation BsW v ]d-b-w-sa- nâ (a,b), (b,c) F nh R se AwK- -fm-bmâ (a,c) R-se Hcp AwK-am-bn-cn- -Ww. DZm: Z (]qà -kw-jym-k-ww) F KW- nse Hcp _Ô-amWv Xmsg simsp- n-«p-å-xv. R = {(a,b)/a-b sâ KpWn-X-am-Wv, a, b, Z} ChnsS Hcp {Ia-tPmSn R-se AwK-am-I-W-sa- nâ (b-a) sâ KpWn-X-am-bn-cn- -Ww. (a,b), (b,c), R b-a, c-b Ch sâ KpWn-X-am-Wv. (a b) m () ( sâ KpWn-Xw) b-a+c-b sâ KpWn-X-am-Wv. (b a) M() thus R is symmetric. c-a, sâ KpWn-X-am-Wv. (a,c) R, a,b,c R AXp-sIm v R - Hcp transitive relation BWv. Equivalence Relation A F KW- nâ \nàh-nn- -s -«n-«på Hcp _Ôw R (ie, R:AA) Equivalence relation BI-W-sa- nâ,. R - Refleive Bbn-cn- -Ww. ie, (a,a) R, a A. R - Symmetric Bbn-cn- -Ww. ie, (a,b) R (b,a) R. ie, (a,b) (b,c) R (a,c) R (transitive Bbn-cn- -Ww) 7
DZm-l-cWw Z F KW- nâ \nàh-nn- -s -«n-«på Hcp _Ô-amWv (relation) R = {a, b) : b-a, sâ Hcp KpWn-X-am-Wv a, bz} ChnsS R Hcp equialence relation BWv. F p-sim-s - mâ. R - Refleive BWv. Let ar, a-a = 0 a a 0 a0 a a =0 0, sâ Hcp KpWn-X-amWv (a,a) R, a R. R - Symmetric BWv. Imc-Ww, (a,b) R sâ KpWn-X-amWv b-a sâ KpWn-X-amWv a-b (b,a) R. R ~Hcp transitive relation BWv. For, (a,b), (b,c) Ch R se AwK- -fm-sw- n-cn- -s«. b-a, c-b Ch sâ KpWn-X- -fm-wv. b-a+c-b Ch sâ Hcp KpWn-X-am-Wv. AXm-bXv c-a sâ Hcp KpWn-X-am-Wv. (a,c) R hyxy-kvx-co-xn-bn-epå GI-Z- Ä (Types of Functions). One-one or injective function A F KW- nâ (sk-äv) \n v B F KW- n-te- på Hcp GIZw (function) one one BI-W-sa- nâ A bnse FÃm hyxykvx AwK- Ä pw B bnâ hyxykvx {]Xn-_nw_w (C-ta-Pv) D m-bn-cn- -Ww. AXm-bXv A bnse c v AwK- Ä v Htc CtaPv BWp-ÅsX- nâ B c v AwK- fpw XpÃy-am-bm am{xsa GIZw Hcp one one BIp-I-bp-Åq. AXymXv f( ) = f( ) =,, A DZm:þ A f B 4 One one 8 9 0 8 P a b c g Q d e g is not one one
Onto Function (Surjective function) A bnâ \n v B bnte- på 'f' F GIZw Hcp onto function BI-W-sa- nâ Bbnse FÃm AwK- Ä pw A bnâ {]nþ-c-tapv D m-bn-cn- -Ww. ie, f : A Note: f : A DZm-l-cWw: B Hcp onto function Bbm Bbnse GXv AwKw ]cn-k-wn- mepw B AwK- Äs Ãmw A bnâ Hcp AwKw D m-bn-cn- -Ww. ie, y B, we can find an element A such that f() = y B Hcp onto function BsW- nâ f() = y, (ie R(A)=B Bbn-cn- pw. A g B f y b a a b c d e g is onto f is not onto Bijective Function A F KW- nâ\n v B F KW- n-te- på f F function bijective BI-W-sa- nâ f one one Dw onto Dw Bbn-cn- -Ww. A one one and onto function is bijective. DZm:þ....4 bijective.a.b.c.d A F KW- nâ \n v B F KW- n-te- på Hcp function bijective Bbm n(a)=n(b) Bbn-cn- pw. 9
Composition Function Let f : A Band g : B C be two functions. Then the composition of 'f' and 'g' denoted by gof is defined as the function gof : A C given by (gof)() = g(f()), A f g A f B C A B g (f() g(f() gof C Eg:- Let A = {,, 4, 5}, B={, 4, 5, 9} Find gof and C = {7,,5}, F : A g(4) = 7 and g(5) = g(9) = Clearly from the figure gof = {(,7), (,7), (4,), (5,)} OR Domain of gof = A Range of gof = C (gof) () = g(f() = g() = 7 (gof) () = g(f() = g(4) = 7 (gof) (4) = g(f(4) = g(5) = (gof) (5) = g(f(5) = g(5) = gof B, g : B C defined as f()=, f()=4, f(4)=f(5)=5 and g() = 4 5 A f B g C gof 4 5 9 7 5 Identity Function An identity function I on A is a function from AA defined as I() =, A Inverse of a Function X F KW- nâ \n v Y F KWn- n-te- på Hcp function BWv 'f'. gof = I, fog - Iy F hyh-ø-iä ]men- - - -hn-[- nâ Y bnâ \n v X bnte v Hcp function I p]n-sn- m³ km[n- p- p-s - nâ g F function s\ f F function sâ inverse F p-]-d-bp- p. ChnsS I F Xv XX te på function BWv. IqSmsX I ()=, X AXymbXv I F -Xv X se Hcp sfu³ânän ^Mvj³ BWv. AXp-t]mse I y F Xv yy bnte- på ^Mvj³ IqSmsX I(y)=y, yy f F ^Mvj\v inverse Ds nâ f 0
invertible BsW v ]d-bp- p. f sâ inverse s\ f - Fs -gpxn kqnn- n- p- p. Note: f F ^Mvj³ invertible BsW- nâ f - bijective Bbn-cn- pw. f F ^Mvj³ bijective DZm-l-cWw (one - one and onto) BsW- nâ f \v \nà_-ô-ambpw inverse D m-bn-cn- pw. ie, f is invertible f is bijective. f : X Y F nâ f - :Y X Bbn-cn pw S = {,, } f F Xv S Â \n v S te på Hcp ^Mvj³ BWv. Xmsg-sIm-Sp- n-cn- p ^Mvj³kn\v inverse Dt m F v ]cn-tim-[n- p-i. Ds - nâ f - ImWp-I. a) f = {(, ), (, ), (, )} ChnsS f bijective function BWv. ImcWw f se hyxykvx AwK- Ä v hyxykvx CtaPv BWp-Å-Xv. IqSmsX S - se FÃm AwK- Ä pw pre-image D v. AXp-sIm f invertible BWv. f - = {(,), (,), (,)} Note: ChnsS 'f' Hcp identity function BWv. Identity function sâ inverse AtX function Xs Bbn-cn- pw. b) f = {(,), (,), (,)} ChnsS f() = = f(). AXp-sIm v f one one function AÃ. f \v inverse CÃ. c) f = {(,), (,), (,)} ChnsS 'f' one-one and onto BWv. (F- p-sim v?) f \v inverse D v. f - = {(,), (,), (,)} S f S Review Questions f - Xmsg-sIm-Sp- n-cn- p functions \v inverse Dt m F v ]cn-tim-[n- p-i. ImcWw Fgp-Xp-I.. f : {,,, 4} {0} f : {(,0), (,0), (,0), (4,0)}
. g : {5, 6, 7, 8} {,,, 4} g = {(5, 4), (6, ), (7, 4), (8, )}. h : {,, 4, 5} {7, 9,, } h = {, 7), (, 9), (4, ), (5, )} Inverse of a Function A function f : y has inverse 'g' if g : y X and (gof) ()=, X and ( fog)( y) y, y y. Then are write f- =g. 'f ' is inversible It is one one and onto... Eample : Let Y n n N. Consider f : N Y such that f(n)=n, show that f is inversible. Find the inverse of 'f'. Ans: Given Y n n N and f : N Y First we want to define a function from Y N Get y Y. Then y=n, n N, n y Define g : y N by g( y) y, y y ( gof )( n) g ( f ( n) g( n ) n n Let, y Y fog ( y) f g( y) f y y y g is the inverse of f. ie, f - =g State with reason whether the following function have inverse. ) f :,,, 4,0 f with (,0),(,0),(,0),(4,0) ) g : 5,6,7,8,,,4 g (5, 4)(6,),(7, 4),(8, ) ) h :,, 4,5 7,9,, h with (,7),(,9),(4,),(5,) with
Answers ) f is not one-one because different elements,,,, 4 have same image 0 f has no inverse ) The g : 5,6,7,8,,,4 Hence in the co-domain of g has no preimage, g is not onto Hence g has no inverse. ) h is one-one and onto h has inverse h (7, ),(9,),(, 4),(,5) Miscellaneous Eamples. Let A = N N and * be a binary operation on A defined by (a, b) * (c, d) = (a+c, b+d) a) Show that * is commutative and Associative. b) Find the identify element for * on A, if any. Answer : a) Given A = N N Any element in A is of the form (a, b), a, bn We want to S.T. '*' is commutaitve ie, *y = y*,, y A Let, y A (Then ( a, b), y ( c, d) = * y ( a, b)*( c, d ) = (a+c, b+d) by definition = (c+a, d+b), Since + is commutaitve on N. = (c,d) + (a,b), by definition of * = y* * is commutative on A. Now to Prove that * is associative. Let, y, z A. Then ( a, b), y ( c, d ) z = (u, v) ie, to Show that ( * y)* z *( y * z) * y ( a, b)*( c, d ) ( a c, b d ) ( * y)* z ( a c, b d)*( u, v)
= ( a c) u, ( b d) v by definition = a c u, b d v (since + is associative with N...() Now y * z ( c, d ) * ( u, v ) = ( c u, d v) *( y* z) ( a, b)*( c u, d v) a ( c n), b ( d v) a c n, b d v...() From () and () * is associative. b) Let e = (p,q) be the identity element in A. Then * e ( a, b)*( p, q) ( a, b) a p, b q ( a, b) a p a and b q b p 0 and q 0 But (p,q) = (0,0) N X N A has no identify element for *.. Consider the binary operation *: RXR R and 0: RXR R defind by a * b a b and aob a, a, b a) Show that * is commutative b) Is '*' associative. Justify your answer. c) Show that '0' is assocoative but not commutative. d) Show that a*(boc) = (a*b8) 0 (a*c), a, b, c Solution a) Let a, b a*b = ab = b*a '*' is commutative b) we have,, * * * * = * * * * 4
= 0 * * *(*) * is not associative. c) To Show that 0 is associative ie, to Show that (aob) oc=ao (boc) If a, b, c Then aob = a (aob) 0c = aoc = a...() we have (boc) = b ao (boc) = aob = a...() Then () and () (aob)oc = ao(boc) '*' is associative. To Show that '0' is not commutative. We have,. Then 0 = by definition. 0 = 0 0 0 is not commutative. Binary Operation GXv c v F Âkw-Jy-I-fpsS XpI Hcp F  kwjy Bbn-cn- pw. AXm-bXv a+bn, a, bn. GXv c v ]qà -kw-jy-i-fp-ssbpw KpW-^ew Hcp ]qà -kwjy Bbn-cn- pw. AXm-bXv a bz, for every a, b, Z ChnsS "+" F Hm -td-js\ * F v amän Fgp-Xnbm a*bn, a, bn ChnsS * F Hm -td-js\ N F KW- nse binary operation F p-]-d-bp- p. Definition * F operation A KW- nse binary operation BI-W-sa- nâ * A A \n v A bnte- på Hcp ^Mvj³ Bbn-cn- -Ww. DZm: KpW\w (multiplication) Z F KW- nse Hcp binary operation BWv. a bz, a, bz Subtraction N-se Hcp Binary operation AÃ. ImcWw a b Hcp natural number Bbn-cn- Ww F n-ã. 5
Eample a [if a, b, Z ] b A = {-, 0, } km[m-cw KpW\w A bnâ binary operation BWv. ChnsS composition table se FÃm AwK- fpw A F KW- nse AwK- -fm-wv. - 0-0 - 0 0 0-0 Usual addition A bnse Binary operation AÃ. ChnsS - + - = - A + - 0 - - - 0 0-0 0 AXp-sIm v '+' A bnse Hcp binary operation AÃ. Commutative property A F KW- nse * F binary operation commutative BI-W-sa- nâ a*b = b*a, b A Bbn-cn- pw. Usual addition "+" is commutative on Z ie, a+b = b+a, a, b Z a, Eample: Define an operation * on R defined by a*b = ab 4, a, b a) Is * a binary opration on R. Justify? b) Prove that * is commutative. Ans: a) a*b = ab 4 R, a, b * is a Binary operation on R b) a*b = ab 4, = b*a = ba 4 (multiplication is commutative on R) a*b = b*a, ie, * is commutative. R 6
Eample Define an operation * on Z as a*b = a+b - ab ) Is * a Binary operation on Z ) Prove that * is commutative. Ans: ) Given a*b = a+b - b Z, a, bz because sum of product of two integers is an integer and hence their difference is again an integer. ) a*b = a+b - ab = b+a-ba (beause additional multiplication and commutative on Z ) = b*a Associative Property Eample: Eample: Answer A F KW- nse "*' F ss_\dn Hm -td-j³ A bnâ associative BI-W-sa- nâ (a*b)*c = a*(b*c), for every element a, b, c in A Bbn-cn- -Ww. k -e\w (addition), KpW\w F o Hm -td-j³kv R F KW- nâ associative BWv. R F KW- nse Hcp ss_\dn Hm -td-j³ BWv Xmsg-sIm-Sp- n-cn- p- -Xv. ab a *b, a, br * associative BtWm F v ]cn-tim-[n- p-i. ab a *b, a, br a, b, c R (a*b)*c = a*(b*c) BtWm F v ]cn-tim-[n- mw. (a*b)*c = d*c, (d=a*b R ) = dc (\nàh-n-\-{]-im-cw) = = (a *b)c ab ( )c ab c abc ½...() 4 = (ImcWw km[m-cw KpWnXw R Â Atkm-kn-tb-äohv BWv.) a*(b*c) = a*d, d=b*c 7
= ad = = a.(b *c) bc a.( ) a(bc) = ½ = a(bc) 4 = abc...() 4, Ch-bn \n v a*(b*c) = a*(b*c) Fs -gp-xmw. Associative BWv. ) R F KW- nse Hcp Hm -td-j\mwv Xmsg-sIm-Sp- n-cn- p- -Xv. Ans: b a*b = a+b, a, R. '*' - ss_\dn Hm -td-j³ BtWm F v ]cn-tim-[n- p-i.. '*' Commutative BtWm F v ]cn-tim-[n- p-i. (*)* Dw *(*) Ch ImWp-I. Given a*b = a+b a, br a, br a+br a*br * -Hcp binary operation BWv.. a*b = a+b b*a = b+a * commutative AÃ.. \ap v ASp- -Xmbn (a*b)*c = a*(b*c) F v ]cn-tim-[n- -Ww. a, b, c Ch R se AwK- -fm-sw- n-cn- -s«. (a*b)*c = (d*c), d=a*b = d+c 8
= (a*b)+c = a+b+c...(a) a*(b*c) = a*p, p=b*c = a+p = a+(b*c) = a+(b+c) = a+b+4c...(b) (A), (B) F n-h-bnâ (a*b)*c a*(b*c) F v e`n- p- p. 4. (*)* = (d*), d=* = d+. = d+6 = (*)+6 = +*+6 = +4+6 = 5+6 = *(*) = *p, = * = +p = +(*) = +(+*) = +4+ = 7 Identity Element GsXmcp F Âkw-Jy-tbmSpw (H v) F F ÂkwJysIm v KpWn- mâ AsX kwjy e`n- p-sa- -dn-bmw. AXm-bXv a Hcp F Âkw-Jy-bm-bm a*=*a=a Fs -gp-xmw. C s\ e`n- p ''s\ multiplicative identity F p-hn-fn- p- p. GsXmcp integers t\mspw 0 Iq«n-bm AsX integer e`n- pw. AXmbXv a+0 = 0+a=a, a Z ChnsS '0' sb additional identity F p-hn-fn- p- p. *, A F KW- nse Hcp ss_\dn Hm -td-j-\mwv. A F KW- nse Hcw-K-amWv 'e'. e F AwK-t msv A bnse GsXmcp AwKhpw tnà v * F operation \S- n-bmâ AtX AwKw e`n- p- p-s - nâ e-sb Abnse identity element with respect to * F p-]- d-bp- p. 9
DZm-l-cWw ie, a*e = a, a A ie, a.e = a = e.a a A. N has no identity element with respect to addition because a+0=a, a N but 0 N. R F KW- nse identity element ImWp-I. R F KW- nse ss_\dn Hm -td-j³ Xmsg-sIm-Sp- n-cn- p- p. a*b = ab 4, a, br Solution R F KW- nse identity element e BsW- n-cn- -s«. a*e =, a R ie, ae a 4, by definition of '*' ae = 4a e = 4a,a 0 a = 4 R a=0 BsW- nâ 0*e = 0.e 0 4 4, R se identity element BWv. Eistence of inverse element Eample A F KW- nse Hcp ss_\dn Hmt -d-j-\mwv *, identity element e Dw BWv. A bnse 'a' F AwK- n\v inverse Ds v ]d-b-w-sa- nâ a*b = b*a = e F hn[- nâ A bnâ b F HcwKw D m-bn-cn- -Ww. ChnsS b sb a bpss inverse F v hnfn- p- p. CXns\ a - Fs -gpxn kqnn- n- p- p. AXm-bXv b=a - R F KW- nse identity element BWv ''. ChnsS binary operation km[m-cw KpW\w BWv. ChnsS sâ inverse ½ BWv. ImcWw *½==½*. ]qpy- n\v inverse CÃ. Eample R F KW- nse Hcp ss_\dn Hm -td-j³ BWv Xmsg-sIm-Sp- n-«p-å-xv. a*b = ab, a, br. Identity element ImWp-I. 0
Answer :. F AwK- nsâ inverse ImWp-I.. R se identity element 'e' BsW v hnnm-cn- p-i. Then ae = a, a R ae a (by definition of ) ae = a e =, a 0 When a=0, o.e = 0. = 0, R- - -se identity element BWv.. sâ inverse BWv b F n-cn- -s«. At mä, *b = e *b = b (by detention) b = 9 9 b R
Time : 45 mts Unit Test Ma Score : 0 ) f :,, 4,,5 and g,,5, f be given by (,),(,5),(4,) and g (,),(,),(5,). Find (gof) (). ) Let f : R R defined as f ( ). Choose the correct answer. ) f is one-one ) f is many one - onto ) f is one-one not onto 4) f is neither one-one nor onto. ) f : R R be given by f ( ) then ( )( ) 4) f : R R byf f() = then f ( )... fof 5,,, 5) Let f : R R with fof ( ), R. What is the inverse of 'f'. ( mark each) 6) ab Let '*' be a binary operation on R defind as a* b. 4 a) Prove that '*' is commutative and associative. () b) Find the identity element in R with respect to '*' () c) Find the inverse of 6 in R with respect to '*' () 7) Give an eample for not an equivalence relations which is symmetric and refleive on A,,, 4. () f, given by f ( ) a) Show that f is one-one () 8) Consider the function :, b) Does have preimage in,. Justify your answer. () c) Is 'f' onto? () 9) Let f : N Y be a function defined as f ( ) 4, where Solution Y Y N : y 4, N. a) Write one element in Y. () b) Show that f is inversible () c) Find the inverse of 'f' () ) gof () g f ( b) g(5) f () 5 ) f is one-one For, if f ( ) f ( )
) Given f is onto For, if y R with f ( ) y y y ie, for any y R we can find R with f ( ) f ( ) ( ) fof ( ) f f ( ) f ( ) Replace y by = = = = Answers 4) 5) f f ( ) 6) ab Given a * b, 4 a, b, a) * is commutative ab For, a * b 4 ba ( usual multiplication is commutative) 4 = b*a (by definition of *) * is associative For, (a*b)*c = p*c where p = a*b = 4 pc, by definition abc a b c abc = = 4 4 4 = * 4 = abc abc...() 4 4 6
= a *( b* c) a* q, q=b*c aq = by definition 4 = ( b* c) a 4 4 a = ( bc ) 4 4 abc...() 6 From () and () * is association. b) Let e be the identity element in w.r.t * The we've a*e = a, a ae a, by definition 4 ae = 4a dividing through by a, e=4, if a 0 When a=0, 0.4= 0 Hence 4 is the identity element in. c) Let 'b' is the inverse of 6. Then b*6=4 (If 'b' is the inverse of a then we have a*b = b*a = e) b6 4 4 6b 6 6 8 b 6 8 7)... is the inverse of 6. 8) Given f() =,, f is one-one a) For f ( ) f ( ) ( ) ( ) 4
b) Let has preimage [-,] Then [,] with f() = ie, 0 which is a absurd. Therefore has no preimage in. c) f is not onto, because has no preimage in [-,] 9) a) Any element in y is of the form y 4, N we have N Y 4 7 7 Y b) Given f : N Y For finding f, first our aim is to define a function g : Y Take y Y. Then y 4, N 4 y y N 4 y Define g( y) 4 clearly g : Y N Claim : ( gof )( ), N and ( fog)( g) y, g y Let N ( gof )( ) g f ( ) g (4 ) by definition of f. = 4 4 4 Let y Y f ( fog)( y) f g( y) y y 4 4 4 = y-+ = y ( fog)( y) y, y y Now g : Y N N such that ( fog)( y) y, y Y and ( gof )( ), Y y Hence g is the inverse of f. c) The inverse of f is g ie, f - =g 5
. INVERSE TRIGNOMETRIC FUNCTION Important Results. Function Domain Range Sine R (-, ) Cosine Tangent Cotangent Secant Cosecant R R (n ) n Z R { : n R { : (n ) } R { : n } (-, ) R R R (-, ) R (-, ). Function Y=Sin - Y=Cos - Y=Cosec - Y=Sec - Y=tan - Y=Cot - Domain [-, ] [-, ] R {,} R {,} R Range, [0,] [, ] {0} [0, ] { } p, p (0, ) 6
Values of Trignometric Functions in Particular angles Trignometric function Angles 0 6 4 Sine 0 0-0 Cosine 0-0 Tan 0 not defined 0 not defined 0 Cosec not defined not defined - not defined Sec not defined - not defined Cot not defined 0 not defined 0 not defined Important Trignometric Identities. Sin Cos. Cos Sin. Sin Cos 4. tan Sec 5. Cot Co sec 6. Sin SinCos 7. Cos Cos 8. Cos Sin 9. Cos Cos 0. Cos Sin 7
TanA TanB Tan A B Tan A.TanB.. TanA TanB Tan(A B) TanA.TanB tan. Sin tan 4. Cos tan tan tan 5. Tan tan Cos 6. Sin Sin 7. Cos Cot 8. Tan 9. Sec Co sec 0. Co sec Sec Angle Co-ratio Angle Same ratio ASTC Sine sâ co-ratio Cosine, Tan sâ co-ratio cot Dw, Secsâ co-ratio cosec Dw BWv. ASTC D]- tbm-kn v functional sign Xocp-am-\n- p-ibpw sn mw. DZm: Sin Cos ImcWw Sine sâ co-ratio Cosine Dw quadratâ Bb-XvsIm v -ve Dw BWv. 8 c m-as
For suitable value of Domain Y=Sin - = Siny Sin (Sin - ) =,, Sin - (Sin) =,, Sin - (/) = Cosec -, or Cos - (/) = Sec -, or Cos - (/) = Sec -, Tan - (/) = Cot -, >0 Cos - (-) = - Cos - (),, Cot - (-) = - Cot - (), Sec - (-) = - Sec - (), Sin - (-) = -Sin - (),, Tan - (-) = -Tan - (), Cosec - (-) = -Cosec - (), Sin - +Cos - = /,, Tan - +Cot - = /, Sec - +Cosec - = /, Tan - + Tan - y y = T an y, y < y y, y>- Tan Tan y Tan tan - = Sin -, tan - = Cos -, 0 tan tan, 9
Sin - ( )=Sin -, / / Sin - ( )=Cos -, / Complete the Table Function Sin - ( ) 6 Principal Value Cos - (- ) Tan - (-) 4 Sin - (- ) - 6 Cos - ( ) Tan - (- ) 4 Ley y=sin - ( ), then Siny= Principal value branch of Sin - is [, ]. Therefore = [, ] satisfies the condition 6 Sin = 6. Therefore is the principal value. 6 Let y = Cos - (- ). Then Cos y = - principal branch of Cos- is [0, ] Cos y = - shows that angle in the second quadrent. Principal Value is - =. Find the principal value of Cot X S T Y S C Y A X Let y Cot Coty Cot y 0
. Find the principal value of C ot Solution Let y Cot Then Cot y...() we've cot Also we've Cot Cot i.e., Cot i.e., Cot From () Cot y = Cot y 0, Problems. Find the principal value of Cos. Find the principal value of tan. Find the principal value of Cos Problems. Write principal value of Cos - function. Evaluate Tan - (-) + Cos - Sin
Answers a. 0, b. We've Tan - (-) = -Tan - = Cos 4 Cos Sin Cos ( ) Cos ( ) Sin 6 Sin Sin ( ) Sin ( ) Tan ( ) Cos ( ½) Sin ( ½) 4 6 8 8 5 4. Find the principal value of Sin Let y Sin Sin y Sin 4 y, 4 ie, Sin 4
Problem. Show that tan ½ tan tan 4 Ans: We've y Tan Tan (y) Tan y Tan ½ tan L.H.S = tan = tan 5 5 0 tan 5 tan 0 5 tan 0 tan R.H.S 4 Cos. Epress tan, Sin in the simplest form. Ans: We've tan - (tan ) = Hence write Cos Sin in tan function. Cos Sin = Sin Cos,
Sin Cos we've Cos Sin Sin Cos Sin Sin Sin Cos and Cos Sin ( ) Sin Cos 4 4 Sin 4 = Sin Cos 4 4 Sin.Sin 4 4 Cos 4 Sin 4 Cot 4 = Tan 4 Tan Cot Tan 4 Tan 4 4
Eample Solution Cos Tan Tan Tan Sin 4 4 Write Cot,, in the simplest form. we've Cot Cot write Sec Put = Sec = Sec - in the cot function. =, tan we've Sec tan tan cot tan Cot Cot Cot Cot, Sec Eample 4 Show that 8 84 Sin Sin Cos 5 7 85 5
Solution Let 8 Sin, y Sin 5 7 8 Then Sin and Sin y 5 7 A 4 5 B C adjacent side Cos hypotenu se AB +BC = AC AB = AC - BC = 5 - = 5-9 = 6 AB = 6 = 4 = 4 5 Also P Sin y = 8 7 y 7 PQ = 7-8 5 = 89-64 Q 8 R = 5 PQ = 5 5 Cosy 7 we've Cos(-y) = Cos Cosy+ Sin Siny 4 5 8 5 7 5 7 60 4 85 85 84 85 84 y Cos 85 8 84 Sin 5 Sin Cos 7 85 ie, 6
5. Show that 4 6 Sin Cos tan 5 6 Let 4 6 5 6 Sin ; y Cos ; tan z 4 6 Then Sin ;Cosy, tan z 5 6 tan tan y we've Tan(+y) = tan tan y Sin Op positesid e Tan Ad jacentsid e A y 5 B C = 5 4 Cos y 5 Tan y 4 Here Tan (+y) = Tan tan y tan.tan y AB 69 44 5 5 P 5 y Q 4 R 48 5 5 4 0 0 0 6. 5 4 0 0 6 0 6 0 6 6 Tan( y) tan z Tan( y) tan( z)ortan( y) tan( z) y z or y z Since and y are positive y z yz i.e, y z Hence the result. 7
Evaluate the following : Cos - (Cos ) Tan - (Tan 7 ) 6 6 Tan - - Cot - (- ) Sin - (Sin ) Tan - [Cos (Sin - ] Prove the following : Tan - ( )+Tan- ( 7 4 )= Tan an- ( ) Tan - ( ) - Tan- ( 7 ) = Tan- ( ) Tan - ( ) - Tan- ( 7 ) = Tan- ( 7 ) Sin - ( 5 ) = Tan- ( 4 7 ) Sin - ( 8 7 ) + Sin- ( 5 ) = Tan- ( 77 6 ) If 4Cos - + Sin - =, find the value of. Find the principal value of Sin - [Cos(Sin - )] Find the value of Tan[Cos - ( 4 5 ) + Sin- ( ) {Hint : Sin - = Tan - ( ), Cos - = Tan - ( b ) If Tan - (+) + Tan - (-) = Tan - ( 8 ). Find the value of, Prove that Cos - ( ) + Sin- ( 5 ) = Sin- ( 6 65 ) Show that Tan - ( 8 ) + Tan- ( 7 )+Tan- ( )+Tan- ( 8 ) = 4 Prove that Sin - = Tan - ( ) and Tan - - Cot - ( ) Hence find the value of Tan {Sin - ( 5 ) + Cot- ( )} If Tan - ( ) + Tan- ( ) =. Find then value of. 4 8
Miscellaneous Eamples ) Prove that tan, Cos, 0, tan we have tan R.H.S = q Cos q Cos take tan q, then tan q q tan tan Cos q tan Cos Cosq q = q = tan = L.H.S q Sin Sin 4,, ) Show that we have Sin Sin 4Sin R.H.S = Sin 4 Sin q = Sin = L.H.S ) If tan Sinq tan p 4 then find the value of. y We have tan tan y tan y = tan tan tan 9
Given tan ( ) tan tan tan tan 4 4 4 4 tan p 4 ie, tan 4 p 4 4 4 tan p 4 4, ) If Sin Sin Cos 5 then find the value of. 5 Given Sin Sin Cos Sin Cos Sin () 5 p But we have Sin Cos p Comparing 5 40
Chapter Marks : 0 ) Find the principal value of Sin UNIT TEST ) tan Sec ( )... p, p, p, p ) S i n t a n is equal to... Time : 45mts,,, if, 4) Sin Cos ( )..., 5) Find tan p tan 6 7 6) Find the value of tan () Cos Sin ( score each) () 7) a) Show that Cos( ) Sin () Sin Sin p b) Solve () c) Write in the simplest form tan, 0 () 5 6 tan 5 6 8) a) Show that Sin Cos b) Evaluate Cos Cos p 5 4 c) Prove that 7 tan tan tan () () () Answers ) We have Sinp Sin 6 Sin p 6 ) tan p Cos p tan p Sec p 4
Sec ( ) p Sec () = p p p ) Sin(tan ) 4) tan Sec ( ) p p p p Let tan y tan y Sin y Then tan p Sin Sin y 5) tan tan p 7 6 7 p 6 p p p tan tan tan tan 6 6 p, 6 p p 6) We have tan () p 4 Cos p Cos = p p = p Sin Sin p 6 Sin tan () Cos p p p 4 6 0p 5p 6 p 8p p 6 4
7. Prove that Cos Sin a) We have Cos Sin C o s S in L.H.S = Cos ( ) Take Sinq = Cos Sin q = Cos Cosq = q =.Sin R.H.S q Sin b) Given Sin ( ) Sin p Sin p Sin p q Sinp Sin = Cosq Sin CosSin = CosCos 0 ( ) 0 0 or 0 or When L.H.S = Sin Sin Sin Sin = Sin = p p 6 is not possible hence 0 c) Take tan q q tan Tan tan q tan tanq 4
= tan Sec tan = tan Sec tan = tan Cos Sin Cos Cos tan Sin tan Sin Cos = tan tan Cos Cos Sin Cos tan tan Sin tan Sin Cos 8) a) Let Sin and Cos 5 5 Then Sin Cos y 5 We want to Show that Sin ie, to Show that 5 y 44 5 6 Cos tan 5 6 6 y tan 6 ie, to Show that tan y Now tan y 5 4 tan( y) 5 4. 6 6 tan tan y tan.tan y 6 6 5 6 6 0 6 6 5 4 5 tan 5 Sin 5 5 Sin y 4 tan y
6 6 6 6 Here the result. b) Cos Cos 0, 5 5 Cos Cos Cos Cos 0, Cos Cos c) We have y tan tan y tan y 7 7 4 4 7 4 tan tan tan = tan 4 7 64 64 4 = 64 64 5 tan 50 tan 48 77 tan 50 *** 45
. MATRICES Definition A matri is a rectangular array of numbers or functions. These numbers or functions are called elements of the matri. Consider following matrices, A B 0 C 5 ½ D 4 0 0 0 0 0 0 0 0 E 4 5 6 F 4 5 G 0 0 H 0 0 7 6 0 0 0 0 Order of a matri: A matrice having m rows and n columns is called a matri of order mn. In the above eample. Order of the matrice A = Order of the matrice B = Order of the matrice C = Qn. Write the order of other matrices. General form of a matri A a a a... a a a,... a...... a a... a n n m m m n Which is also represented by (a ij ) mn Where a ij is the j th element of i th row. Qn: Construct a matrice whose elements are given by aij = i j a Ans: General form of a matrice is a a a a a here a= 46
a a a a a 5 hence the required matri is 5 Qn: Construct a matrice (aij) where aij = i j Qn: Construct a matrice (aij) where aij = Types of matrices Column matri - Matri having one column ( i j) Eg:- A B 0 4 Row matri - Matri having one row. Eg:- C 5 D 5 Square matri - No.of rows is equal to no.of columns. 4 Eg:- E F 4 5 6 0 4 Diagonal matri - A square matri in which non diagonal elements are zero. Eg:- 0 0 G 0 0 0 0 0 H 0 4 Scalar matri - A diagonal matri is said to be scalar matri if its diagonal elements are equal. 0 0 Eg:- J 0 0 0 0 0 K 0 Identity matri - In a square matri all its diagonal elements are and all its non diagonal elements are zeros then it is an identity matri which is usually denoted by I. 0 I 0 I 0 0 I 0 0 0 0 Zero matrice - matri with all its elements are zero. 0 0 L 0 0 0 0 0 M 0 0 0 47
Equality of Matrices Two matrices are said to be equal if they are of same order and corresponding elements are equal. Qn: 4 5 y Then find & y Ans: =4 and y=5 Qn: Find the value of a, b, c & d from the equation a b a c 5 a b c d = 0 Ans: Equate the corresponding elements and find the values. Addition of matrices: If A = (aij) B=(bij) then A+B = (aij+bij) Qn: Let P 4 5 6 Q 4 5 Then find P+Q Ans: P Q 4 4 5 5 6 0 8 8 5 Multiplication of a matri by a scalar: If A= (aij) mn is a matri and k is a scalar then KA = (ka ij ) Qn: A= andb 0 and then find A-B Ans: Qn: 4 6 A B 4 6 0 4 6 5 Then A-B = 4 6 0 5 6 0 Find X and Y if 7 0 0 X+Y = 5 X-Y 0 7 0 Ans: X+Y = 5 - () 0 X-Y = 0 - () 48
0 0 ()+() X = 8 5 0 X = 4 4 0 ()-() Y = 0 Y = Multiplication of Matrices The product of two matrices A&B defined if the number of columns of A is equal to no.of rows of B. If A=(a ij ) mp & B = (b ij )pn Then AB = (C ij )mn Where C ij = j p a ij b jk ie, ij th element of AB is product of i th row of A & j th column of B. Qn: Find AB Where A 4 5 6 7 and B = Ans: 6 6 0 8 AB 4 0 0 0 0 6 4 8 8 0 4 8 8 8 4 50 0 46 7 Qn: P 5 0 Find PQ where Q 4 5 0 8 6 4 0 0 9 PQ 5 0 4 5 0 0 0 0 6 4 0 5 5 9 5 6 0 Qn: ) Complete the indicated product. a b a b b a b a 49
) 4 ) 4) 4 4 5 4 5 6 5 0 4 0 5 5) 0 0 Qn: If F() = Cos Sin 0 0 Sin Cos 0 0 Show that F(), F(g) = F(+y) Cos Sin 0 0 Ans: = F( ) = Sin Cos F( y ) = 0 0 Cosy Siny 0 0 Siny Cosy 0 0 CosCosy SinSiny CosSiny SinCosy 0 F( ). F( y) 0 SinCosy CosSiny SinSiny CosCosy 0 0 Cos ( y) Sin ( y) 0 Sin( y) Cos ( y) 0 0 0 = F(+y) Qn: A 0 4 and I 0 Find k so that A = ka - I 9 8 6 4 Ans: A = A A = 4 4 = 8 8 4 = 4 4 KA - I = k 4k k k - 0 0 4 = k k 4 k k 50
k K A = ka - I 4k k = k- = k = k= Transpose of a Matri 4 4 If A=(a ij ), then the matri obtained by interchanging rows and columns of the given matri A is called transpose of A denoted by A T =(a ij ). Properties: (A T ) T = A (A+B) T = A T +B T (KA) T = KA T (AB) T = B T A T 4 5 Qn: If A= Find AT A T = 4 6 7 8 5 6 7 8 Symmetric and Skew symmetric matrices. Eg:- A matris is said to be symmetric if A T =A and is said to be skew symmetric if A T =-A A 4 B= 5 6 4 6 7 0 C 0 0 4 4 0 is symmetric is symmetric is skew symmetric D is skew symmetric Result: For any square matri A, A+A T is symmetric and A-A T is skew symmetric. Proof: (A+A T ) T = A T +(A T ) T = A T +A = A+A T A+A T is symmetric (A-A T ) T = A T -(A T ) T = A T -A = -(A-A T ) A-A T is skew symmetric Diagonal elements of a skew symmetric matrices are all zero. A skew symmetric matri is not a diagonal matri. 5
UNIT TEST Time : 40 mts Ma.Marks : 0 ) Is A 4 4 8 and B 0 0 then AB =... () ) 0 5 A 6 5 6 0 is a skew symmetric matri, then the value of =... () ) A 6 7 8, B 4 then order of AB =... () 4) Cos Sin If A Sin Cos and A+AT = I, then the value =... () 5) If A, B are symmetric matrices of same order, then AB - BA is... (A - Skew symmetric matri, B - Symmetric matri, C - Zero Matri, D-Identity matri () 6) Consider matri A=[a ij ] where a i j ij a) Write A () b) Find A+A () 7) Consider A 4 5 a) Using the Matri A, form Symmetric and Skew Symmetric Matrices. () b) Epress A as sum of symmetric and skew symmetric matrices () 8) a) If A is a square matri, such that A =A then (I+A) -7A is equal to... () b) Let (i) Find A and 5A () (ii) Show that A -5A+7I=0 () 9) Find the inverse of the matri A, using elementory row operation. 0) For the matrices A and B verify that (AB) = B.A. Where A 4, B = [- ] () 5
Answers ) We have A.I. = I, A (Ans: A) Here AB =A ) Since A is Skew Symmetric matri, diogonal elements are all zero. 0 ) Order of A =, order of B= Order of AB = 4) Cos A Sin Sin Cos Cos Sin A Sin Cos Given A+A = 0 A 0 First element in Ist row of A+A = Cos Cos = Cos = Cos = 5) Given A = A, B = B (AB - BA) T = (AB) T -(BA) T = B T A T - A T B T = BA - AB = -(AB-BA) AB - BA is Skew Symmetric. 6) a) Given A a ij a A a a a Given a i j ij a a 6 4 a 4 a 4 6 = Cos 4 A 5
b) A 4 4 A A 4 4 5 4 5 4 7) a) Given A 4 5 A = 4 5 let P = A+A 6 5 4 4 5 4 4 P 6 5 4 4 P 5 4 4 P is Symmetric Q A A 4 5 4 5 0 5 5 0 6 6 0 0 5 Hence Q 5 0 6 6 0 Q is Skew Symmetric. = 0 5 5 0 6 6 0 = -Q 54
b) We have A A A A P Q A = 6 5 6 5 P 4 4 4 4 5 4 4 5 4 4 5 5 = 5 0 Q 5 6 Q 0 6 0 5 5 0 5 A 0 5 0 8) a) (I+A) - 7A = I + I A+ IA +A - 7A = I + A + A + A - 7A = I + A + A + A - 7A = I + A + A + A - 7A = I b) i) A A 4 8 5 5 55
ii) 5A 5 = 5 5 5 5 = 5 5 5 0 0 7 70 7I 7 0 0 7 7 = 7 0 0 7 = = 8 5 5 5 7 0 A 5A 7I 5 5 0 0 7 7 0 7 0 0 7 0 7 0 0 0 0 9) Write A = IA 0 ie, 0 A 0 0 5 A R R R R R 5 0 A 0 5 5 R R 4 0 5 5 A 0 5 5 56
5 5 A = 5 5 5 5 A 5 5 0) 4 A, B AB 4 4 4 4 4 8 4 6 4 AB 8 6...() 4 B A 4 = 4 8 6...() 4 From () and () (AB) = B A *** 57
4. DETERMINANTS BapJw Ir-Xn-bn hn\y-kn- n-cn- p Hcp am{snivkpambn _Ô-s -Sp- p kwjy-bmwv UnäÀan-\â v. A F ka-n-xp-cm-ir-xn-bn-epå am{snivknsâ UnäÀan-\âns\ A simt m det(a) simt m kqnn- n- p- p. am{snivknsâ UnäÀan-\â v a A c b d F nâ A ad bc am{sn-ivknsâ UnäÀan-\â v a A d g b e h c f F nâ a A a e h f a - b d g f d a + C g e h kwjy-i-fpss {Inb-I-fp-ambn _Ô-s «Nne kqn-\-iä k -e\w +ve + +ve = +ve -ve + -ve = -ve hn]-cox NnÓ- fpå kwjy-iä Iq«p-t¼mÄ tih-e-hne hep-xnâ \n v tih-e-hne sndpxv Ipd v tih-e-hne hep-xnsâ NnÓw Fgp-Xp-I. DZm:þ þ8+ = -þ(8þ) - = -þ6 hyh-i-e\w kwjy-i-fpss hn\ym-k-amwv am{snivkv F v \n Ä a\-ên-em- n. F mâ ka-n-xp-cm- 9+- - þ4 = 9-þ4 = 5 Ipd-t kwjy-bpss NnÓw amän Iq«pI DZm:þ 0-þ- þ = 0+ = KpW\w (- -þ4) -þ(- þ 9) = þ4+9=9-þ4 = 5 (+ve)(+ve) = +ve (+ve)(-ve) =-ve 58
lcww (-ve)(+ve) =-ve (-ve)(-ve)=+ve ve ve ve ve ve ve ve ve ve ve ve ve `n -kw-jy-iä a b F cq]- n-epå kwjy-i-fmwv `n -kw-jy-iä. (a bpw b bpw ]qà -kw-jy-i-fm-bn-cn- -Ww) ]qà -kw-jybpw `n -kw-jybmwv. DZm:- -þ {Inb-IÄ 5 5 k -e\w: tozw kam-\-am-sw- nâ D -c- nsâ tozw s]mxp-hmb tozhpw Awiw Awi- -fpss XpI-bpw. DZm:þ 4 4 6 7 7 7 7 DZm:þ KpW\w lcww tozw hyxy-kvx- -fm-sw- nâ a c ad bc (t{imkv KpW-\w) b d bd 4 5 (4)() (5)(9) 44 45 89 9 (9)() 99 99 Awi- Ä X½nepw toz- Ä X½nepw KpWn- p-i. a c a. c b d b. d Awihpw tozhpw ]c-kv]cw amän KpWn- pi a c a d ad b d b c bc 59
Hcp {XntIm-W- nsâ hnkvxoà w Hcp {XntIm-W- nsâ ioàj- Ä (vertices) (,y ), (,y ), (,y ) F nâ hnkvxoà w ½ y y y ssa\dpw tim^m-ivsdpw (Minor and cofactor) a ij F ]Z- nsâ ssa\-dns\ M ij sim p kqnn- n- p- p. a ij F ]Z- nsâ tim^m-ivs-dns\ A ij sim p kqnn- n- p- p. M ij In«m³ a ij \nâ p row bpw column Dw Hgn-hm- n- n-«p am{sn-ivknsâ UnäÀan-\â v I mâ axn. A ij In«m³ A ij =(-) i+j M ij F formula D]-tbm-Kn- p-i. kqn\: (þ)sâ IrXn Hä-kw-Jy-bm-sW- nâ þdw Cc-«-kw-Jy-bm-sW- nâ Dw BWv. am{sn-ivknsâ AUvtPm-bnâ v A ij F Xv a ij F ]Z- nsâ tim^m-ivsà F nâ, a a A a a a a a a a F am{sn-ivknsâ AUvtPm-bnâ vv adj A A A A A A A A A A kq-n\: A bnse H m-as row bnse ]Z- -fpss cofactor I v column Bbn-«mWv adja In«m³ Fgp-tX- -Xv. am{sn-ivknsâ C³thgvkv (inverse) ImWm-\pÅ formula. A 0 F nâ A - = adja A System of linear equation \nà[m-cww sn p- -hn[w Xmsg-]-d-bp Sysytem of linear equation ]cn-k-wn- p-i. a +b y = c a +b y = c CXns\ am{snivkv cq]- nâ C s\ Fgp-Xmw. a a b b y = c c 60
a a b b s\ A F pw y sb X F pw c c sb B F pw hnfn- mâ, AX=B F v In«pw. A - I v A - s\ B sim v KpWn- p-i. A - B F am{snivkv y F cq]- nâ Bbn-cn- pw. At mä X y y F v In«pw. AXp-sIm v = F pw y=y F pw In«p- p. A 0 F nâ am{xta Cu coxn-bnâ \nà[m-cww sn m³ Ign-bp-I-bp-Åq. DZm:þ Evaluate the determinant Solution 4 5 4 5 = -- - 4 + 5 (Epansion along I st Row) () ( ) 4() ( ) 5() () 6 4 4 5 6 4 5 5 (7) 4(6) 5() 4 5 50 Find the area of the triangle with vertices at the point given in (,0), (6,0), (4,) Solution Area of the triangle with vertices (,y ), (,y ) and (,y ) is ½ y y y 6
Here (, y ) = (,0) (,y ) = (6,0) (, y ) = (4,) Area ½ 6 4 0 0 ½ (0 ) 0 (8 0) ½ 8 ½ (5) 5 Show that the points A(a,b+c), B(b, c+a), C(c, a+b) are collinear. Solution Three points (,y ), (,y ) and (,y ) are collinear if y y y = 0 Here (, y ) = (a, b+c) (, y ) = (b, c+a) (, y ) = (c, a+b) y y y = a b c b c c a a b = a b c b c a c a b b c c a a b C C +C = (a+b+c) b c c a a b = (a+b+c)0 (C =C ) Bb-Xp-sIm v = 0 The given points are collnear. 6
Find the equation of the line joining (,) and (,6) using determinants. Solution Let (, y) be any point in the line joining (,) and (,6) then, y 6 = 0 ie, (-6) - y (-) + (6-6) = 0 ie, (-4) -y (-) + 0 = 0 ie, -4 + y = 0 ie, -y = 0 ]cn-io-e\w I Find the determinant of the following matrices.. 4.. 5 4 6 5 4. 4 5 6 7 5. 0 5 II Find the areas of the triangle whose vertices are given as a) (, ), (, 4), (, 6) b) (0, 0), (, ), (, ) c) (-, ), (-, 4), (0, 5) 6
III Find the equation of the line joining (,) and (9,) Eg:- Write the minor and cofactors of the elements of the determinants 0 4 M = A = (-) + M = M = 0 A = (-) + M = 0 M = -4 A = (-) + M = (-) (-4) = 4 M = A = (-) + M = Eg:- Write the minors and cofactors of the elements of the determinant. 0 0 5 4 M = M = M = M = M = M = 5 0 0 0 0 0 0 ( ) A = (-)+ M = 6 0 6 A = (-)+ M = (-)6=-6 5 0 A = (-) + M = 4 0 4 4 A = (-) + M - (-) (-4)=4 4 0 + A = (-) M = ()() = 0 0 A = (-) + M = (-) () = - M = 0 5 4 A = (-) + M = -0 M = 4 A = (-) + M = (-) (-) = 64
Adjoint of a Matri The adjoint of a Square Matri A=(a ij ) nn is defind as the transpose of the matri (Aij) nn. Above Aij is the cofactor of element a ij. Adoint of the matri. It is denoted by adj.a. Let a a a A a a a a a a then adj A = transpose of A A A A A A A A A A A A A adj A A A A A A A Eg: Find adj A for 4 A = (-) + 4=4, A = (-) + =- A = (-) +. = -, A = (-) +. = A Cofactor Matri = A A A 4 adja 4 a a Remark: For a square matri of order, given by A a a The adj A can also be obtained by interchanging a and a and by changing the signs of a and a. a Adj A a a a eg: If A 4 then 4 adj A Theorem If A is any square matri of order n, then A.adj A = adj of A = A I. Above I is the square matri. 65
Singular and Non Singular Matri A square matri A is said to be singular if A 0 eg:- A 4 6 hence 0 A A is non singular. Non Singular A singular matri A is said to be non singular of A 0. Eg:- A 4 then A 4 6 = - 0 Hence A is non Singular. Inverse of a Matri Let A be a square matri of order m. If we can find a square matri B of order m such that AB = BA = I then B is called the inverse of A and it is denoted by A -. Inverse of matri is unique. Necessary and sufficient condition for Inverse A square matri has inverse iff it is non singular ie, A eists A 0 Let A is square matri. adja Then A A Eamples ) Find A -, for Solution adj A = We have = 4 6 = - 4 A. adja A A 4 A 4 66
= 4 - - 4 ) If Solution A 4 4 A = (6-9)- (4-)+(-4) then verify that A.adjA = A. I. Also find A -. 0 Now A =7, A =-, A =-, A = -, A =, A =0, A =-, A =0. A = 7 A 0 0 A.adj A = 7 4 0 4 0 = 7 0 0 7 4 4 0 0 7 4 0 0 4 0 0 0 0 0 0 0 0 = 0 0 0 0 A - = adja A 7 0 0 = 7 0 0 67
Application of Determinants and Matrices Consistent Equation A system of equations is said to be consistent if it has a solution. Inconsistent System A system of equations is said to be inconsistent if its solution does not eist. Solution of system of Linear Equations Consider the system of Equations a b y c z d a b y c z d a b y c z d The system of equations can be represented as A=B where, a b c d A a b c, X y, B d a b c z d If A is no non singular then we have A - eists. AX = B Multiplying on both sides to the left. A - (AX)=A - B (A - A)X = A - B IX = A - B X = A - B By using A - we can solve the system of equations. This method of finding solutions of system of equations is called matri method. Eamples Solve the system of Equations using matri method 5y y 7 5 then A, X, B y 7 Then system can be represented as 68
AX = B Then X= A - B adja A A 5 A 4 5 Adj.A = 5 5 A - = X = A - B 5 = 7 = 5 4 = ie, y, y Eample Solve the system of Equations 0 4 y z 4 6 5 y z 6 9 0 y z The system of equations can be represented as AX=B 69
0 4 A 4 6 5 X B y 6 9 0 z X = A - B A = 75 A = 0 A = 7 A =50 A = -00 A = 0 A =-45 A =50 A =-4 75 50 45 adj. A 0 00 50 7 0 4 A adja A A 0 45 80 0 06 6 =75 - -0+70 = 540 = X = A - B A 75 50 45 0 00 50 540 7 0 4 = = 75 50 45 4 0 00 50 540 7 0 4 5, y, z 5 70
UNIT TEST Time : 40 mts Ma.Marks : 0 7 5 0 ) If the matri 0 4 is not inversible then... ) Le A be a matri with A then A... ) If 5 5 then... 4) Let A be a Non Singular Matri of order. Then adj A is equal to... (a. A b. A c. A d. A ) 5) 7 5 6 5... 5 9 86 (one each) 6) Let A 5 a) Find A b) Find adj A () c) Find A - () d) Using A - solve the system equations () 5y y 7 7) Using properties of determinant () a) Show that a ab ac ba b bc 4a b c ca cb c () b) Using determinant, Find the equation of line joining (,) and (,6) () 8) Let 5 A 4 a) Find A () 7
b) Find adj A. () c) Find A - () d) Solve the system of Equations y 5 y 4z 5 y z () *** 7
- 5. FUNCTIONS LIMITS & CONTINUITY Domain and Range Consider the function y=f(), the set of all possible -values is called the domain of f and set of y-values that results when -values over the domain is called range of f. Eg:- Find the domain of (i) f() =, (ii) f() = ( Ans: i) The function f is a polynomial. Hence the domain is the interval, or R set of reals. ii) The domain = R- {:(-) (-) = 0} = R - {,} Limits : If the values of f() can be made as close to L, by taking the values of '' sufficiently close to 'a' (bu t a) then we write lim it f() a Eg:- i) lim it = =4 (ie, when, 4) ii) lim it ( -+5)= -+5 = 9-6+5 = 8 (ie, when, -+5 8) Right and left hand limits If the values of f() can be made as close to R, by taking values of sufficiently close 'a' (but greater than a), then we write a + lim it f ( ) R called the right hand limit. If the values of f() can be made as close to L by taking values of sufficiently close to a (but less than a) then limit a f() =L called the left hand limit. Eg: f() =, 0 then, when 0+ 7 a n d * Limit eist only when the right hand limit and left hand limit are coincide. ie, limit + a f() = limit a- f() Results limit k k ( where k is constant) a { 0, =0 * limit f() g() limit f() limit g() a a a -, when o-
lim it Sin Cos lim it Sin lim it Cos 0 Eg:- * lim it f() g() lim it f() lim it g() a a a lim it Sin tan lim it Sin lim it tan Eg:- * limit f().g() 4 4 4 limit f().limit g() a a a Eg:- limit ( 5) lim it( )limit 5 ( )( 5) 6.( ) * f() limit f() limit a a g() lim it g() a Eg:- lim it( ) 4 5 5 lim it( 5) 4 5 9 4 4 4 limit limit k. f ( ) k.limit f ( ); k is co ns tan t * a a lim it Sin.lim it Sin.Sin. Eg:- Sin Computing limit for the rational function p() q() a. lim it 0 b. lim it 0 9 0 Hence it is evaluated as follows. lim it (factorising the denominator) 74
c. = lim it 6 4 4 4 4 0 limit 5 6 5 6 4 0 6 0 form. Hence it is evaluated as follows. ( )( ) lim it ( )( ) (factorising both numerator and denomenator) lim it 4 Special Limits a) i) n a lim it a a n na n Eg:- lim it (). ii) ( ) ( ) 5 5 5 4 lim it lim it 5( ) 5( ) 5 - - iii) 7 lim it lim it () 9 7 b) c) e lim it 0 Sin lim it 0 Sina Sina lim it lim it.a Sina = a.lim it a a0 a, 0 0 = a.lim it a0 = a. = a Sina a d) log( ) lim it 0 75
Continuity The continuity of a function y=f() is easily calculated by drawing its graph. If the graph has no break or jump, the function is continuous on the given interval. Eg:- a) f() =, absolute value function, from the graph, is continuous. b) f() = Sin, (0, ) The graph has no jump or break, it is continuous. ( the graph has no jump or break) O Fig () Definition: A function f() is continuous at the point = a if limit f() f(a) or limit a f ( ) limit ( ) ( ) a f f a a Problem, 0 Eamine the continuity of the function f() at =, 4 f( ) lim + it f() lim it 5 Ans: f( ) lim - it f( ) lim it 5 f() 5, Hence f( ) f( ) f() f is continous at = Fig () Problem k, Find the value of k if f() is continuous k, lim it f ( ) lim it ( k ) 4 k + li m i t ( ) l i m i t ( ) 4 - f k k k Since f() is continuous at =, we have limit f ( ) limit f ( ) 4+k=4k 4 k 4 76
Problem Find the value of a and b if f() ; a b; 5 ; 5 is continuous at = and =5. limit f ( ) limit( a b ) a b + lim it f ( ) lim it () = since f() is continuous at =, to get, - f( ) f( ) a b...() lim it f ( ) lim it () =, again + 5 5 limit f ( ) limit( a b ) 5 a b 5-5 Since f is continuous at =5, to get limit f ( ) limit f ( ) 5-5 5a+6=...() Solving () & () a+b= 5a+b= Put a=, in () to get +b= +b= b=- a = & b = - Hence the result. a+b =...() 5a+b =...( ) ()-() a = a= Functions limits and continuity BZy-ambn y=f() F GIZw (^-Mvj³) ]cn-k-wn- p-i. kzoi-cn- p hne-i-fpss KW- ns\ GI-Z- nsâ sumssa³ F v hnfn- p- p. hyxykvx hne-iä kzoi-cn- p-t¼mä y bv v e`n- p hne-i-fpss KWw {]kvxpx GI- Z- nsâ tdbv v F -dn-b-s -Sp- p. Hcp ^Mvjsâ enanäv FIvknÌv sn -W-sa- nâ AXnsâ ssdäv sskuv enanäpw se^väv sskuv enanäpw Xpey-am-bn-cn- -Ww. 77
Transformation of functions ) f() transforms to f()+a If a>0 then f()+a shift the graph of f() 'a' unit upwards. ) f() transforms to f() - a If a>0 then f()-a shift the graph of f() 'a' unit downward. ) f() transforms to f(+a) If a>0 then f(+a) shift the graph 'a' unit towards left. 4) f() transforms to f(-a) If a>0 then f(-a) shift the graph 'a' unit towards right. Illustration Consider the function f() = Then the graph of the function g() = f()+ = +. It is obtained by shifting the graph of f()= unit upward as shown below f()+ f() fig. fig. f() f()- is obtained as shown below f()- - fig. p() = f(+) = (+) is obtained as shown below. - f(+) fig.4 q() = f(-) = (-) is obtained as shown below f(-) 78 fig.5
5) f() transforms to a f(); a> If we multiply f() by a>, the graph of f() is streched 'a' times vertically Illustration Consider f() = Sin fig.6 g() = f() = Sin Struch the graph twice vertically fig.7 6) f() transforms to f(a) If a> f(a) shrink the graph of f() a lines horizontally. Illustration Consider f() - Sin h() = f() - Sin shrink the graph of f() 'a' times unit horizontally. fig.8 79
Alternate method to find left hand and right hand limit limit ( ) limit ( ) 0 a a f f a h where h h0 limit f ( ) limit f ( a h) where h 0 h0 Eg:- Consider if f ( ) if limit f ( ) limit f ( h ) h0 h0 limit h = - 0 = limit f ( ) limit f ( h ) h0 limit h, as h h0 = +0+ = Continuity How to check whether a functions f() is continuous at =a Step - Find limit f ( ) a Step - Find limit f ( ) a Step - Find f ( a) If limit f ( ) limit f ( ) f ( a ) a then the function is continuous at =a a Otherwise f is discontinuous at =a A function fails to continous at =a due to following reasons. ) limit f a ( ) does not eist ) limit a f ( ) does not eist limit f ( ) limit f ( ) ) a a 4) limit a f ( ) f ( a) 80
Score 0 UNIT TEST Time : 40 minutes I. Choose the correct answer from the given alternatives. (Score each). log( ) limit is equal to 0 Sin a) b) c) e d) none of these ) Let a f ( ) a then f() is continuous at = for a) a=0 b) a= c) for all a d) none of these ) If f ( ) then f (0) is a) 0 b) c) - d) none of these 4) If y ( y) then d p q p q dy a) y b) y c) y d) y y 5) If m m d y y a e b e then d a) m y b) -m y c) my d) -my II Answer the following 6) Discuss the continuity of f() at =0 If if 0 f ( ) if 0 7) Differentiate log (Score ) Sin w.r. to (Score ) Sin 8) Differentiate e (tan ) w.r.t (Score ) 9) If a( Sin) Find y a( Cos) dy at d (Score ) 8
0) Verify Languages Mean Value theorem for the function f ( ) 4 in the interval (,4) ) Differentiate the function w.r.to Sin (Score ) Answer and hints. b) log( ) Sin log( ) Sin. c) limit f ( ) a and limit f ( ) a b f () a. d) f ( ) limit f ( ) limit f ( ) f () for any a is not differentiable at 0 4. b) Take log on both sides and differentiate 5. (a) limit f ( ) and limit f ( ) 6. 0 0 f is not differentiable at =0 7. Let y logsin dy ( ) d Sin sin 8. Let y e (tan ) Let u e Sin and (tan ) dy du dv d d d Sin By logarithemic differentiation find du d d 9. a( Cos ) d and dv d 8
dy a( Sin ) d dy dy d a Sin d d d a( Cos) Sin = Cos at Sin dy, d Cos 0 0. Given f ( ) 4 f is continuous on,4 and differentiable on,4 By Lagranges mean value theorem, there eist c (, 4) such that f (4) f () f ( c) 4 f f ( ) 4 ( ) 4 4 f () 44 0 f (4) f () 4 = f ( ) 4 f ( ) 4 4 f ( c) c c 4 f f (4) f () c ( c) c c 4 c ( c 4) 4 c 4 c c 6 c 6 (, 4) Hence Lagrange's mean value theorem is verified. 8
. Let y Sin Put tan Then tan y Sin Sin tan = Sin ( Sin ) = = tan ( ). *** 84
6. DIFFERENTIATION The derivative is a mathematical tool, which is used to study the rate at which the quantities change. - Rate of change of quantities s\ p-dn v ]Tn- phm³ klm-bn- p KWn-X-im-kv{X-kw-ÚbmWv differentiation. Definition ( ) ( ) The derivative of f() at =a is denoted as f '(a) and is defined as f '(a) lim it f f a a a - f() derivative =a F t]mbnânâ f'(a) F v kqnn- n- -s -Sp- p. Xmsg-]-dbpw {]Imcw AXns\ \nàæ-nn- mw. f ( ) f ( a) f '( a) lim it a a Remarks Geometrical meaning of the derivative is slope of tangent. The right hand derivative of f(a) is f(a h ) f(a ) lim it h 0 h Problem The left hand derivative of f(a) is limit h0 f ( a) f ( a h) h Find the right hand and left hand derivative of f() = at = o f(o h) f(o) f(h) f(a) h Right hand derivative at = o is lim it lim it h 0 h h 0 h h f ( o) f ( o h) h Left hand derivative at =0 is lim it lim it h0 h h0 h Note: Every differentiable function is continuous. Some Standard results d n n d * n * d d d d * log d e e d * d d * d d * 85
* d a a log a d d Sin Cos d d Cos Sin d * * d tan Sec d d Cot Co sec d * * * d Sec SecTan d d Co sec Co sec Cot d * d d * Sin d d * Cos d d * tan * Cot * Sec d d d d l d * Co sec d d d * ( log ) d k 0, d d d * k- constant * d f ( ) g( ) d f ( ) d g( ) d d d * Sum rule : d k.f() k. f(), k constant d - Sum rule c v ^Mvj-\p-I-fpsS XpI-bpsS sudn-th-äohv Hcp ^Mvj-sâbpw sudn-th-äo-hnsâbpw XpIbv v Xpey-am-bn-cn- pw. d f ( ) g( ) d f ( ) d g( ) d d d ie, Eg:- Find The Derivative of (Sin+Cos) Solution: d Sin Cos d (Sin) d (Cos) d d d Cos Sin Eg:- Find the derivative of Sin Cos with respect of Solution: d d d Sin Cos Sin Cos d d d.. 86
0 Product Rule d f().g() f(). d g() g() d f() d d d (Derivative product of two functions = I st function derivative of second function + second function derivative of I st function) þ Product rule : c v ^Mvj-\p-I-fpsS KpW-\- nsâ sudn-th-äohv = H mw ^Mvj³ (derivative of c mw ^Mvj³) + (c mw ^Mvj³) (derivative of H mw-^-mvj³) Eg:-(i)Find the derivative of.e with respect of d d d d d d.e ). (e ) e ().e e. e ( ) Eg:(ii) Find the derivative of Sin.Cos with respect of. d d d (Sin.Cos) Sin (Cos) Cos (Sin) d d d Sin. Sin Cos.Sin Sin Cos Cos Sin Eg:- (iii) Find the derivative of e.log with respect of. d d d d d d e.log e (log ) log. (e ) e. log e e ( log ) Rational function A function is of the form p(), q() 0 is known as rational function. g() -þ- tdj-wâ (Hcp ^Mvj³ p() q() F t^manemsw- nâ B ^Mvj-s\ tdj-wâ ^Mvj³ F v hnfn- p- p) q ( ) 0 87