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Functions (4 12)

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71 views37 pages

Functions (4 12)

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Pradip Pal
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Session 4 Piecewise Functions Piecewise Functions (i) Absolute Value Function (Or Modulus Function) ¢ modulus function is defined as viste| reo. Hx, x<0 It is the numerical value of x. ” Figure 9.21 Geometrical Interpretation of Modulus of a Function Geometrically, | x | represents the distance of the point P(x,0) or Q(x, 0) from origin. i: ea FF a cD 2ex0) Oo 2) Figure 3,22 ie. (0, P) = y(x -0)? +(0-0)? = Properties of Modulus a (xP =x? (i Vx? =|.x| ii) |] =|] (iv) | x |= max {=x, x} (¥) 1x |= min {=x, x} (vi) max (a6) = 22% ~x =] x] (vii) min (a, 6) = 2+ _ (viii)| x +y|<| x] +] y| (ix) |x +y|=|x]+]y | iffxy 20 (9) lx—y]=|x]+]y| iff xy OOO SOO 2 x€[-2n,-n]U[-1, solve |_X aa Sal, Let f(x) = and g(x) = x Then, f(x) + g(x) = We know,| f(x)! +] g(2)|=[ f(x) + a(x], iff F(x): a(2) 20 Sor205 x-1 a eae = XE MUU) [Example 27 Find domain for y rs Sal, is defined, i(|x|~ x) >0 =3]x|> x, which is true for negative x only. Hence, domain € (— ©, 0). J Example 28 Find domain for Jeter a1 tk, 3 > -3s1-2\x|s3 3 -4s-2/x/s2 > 2B|xlz-1 = -2S x82 ‘Also, log, > exist, if x0, [x=1]>0and|x—1]#1 x50, xe R-{i}andx #02 Hence, from Eqs. i) and (i), we get €(0,1) U(,2) Chap 03 Functions 111 T Example 29 Domain of the function fxs) _ ig [4x |x? -10x+9) ©) 7 ~Vi0,7 + 140) 0) 0.7 + Lad) © 0-V8,=) (@ None ofthese Sol. Here, f(x) = 1 would exists, if ex -|x* = 10x +9) 4x=[x7-10e+9|>0 ie [x10 49 /<4x, where : Pauet9, xStorxz |x? -10x-49]=4 5 : (a? ~105+9), 10 = (x~3) >0, whichis always true except at x =3 * x (1,9)—(3} sii) From Eqs. (i) and (i), domain of f(x)is (= V0.7 a8) ~ (3) Hence, (isthe correct answer. 1 Example 30 The domain of the function F(x) = yIsin“ (sin x)| = cos (cos x) in [0,272] is Cy) [s$]-[=*] (©) [nan] @ wan-{ a ©, n-{3} a Sol, As,|sin“* (sin x)| could be sketched, as “ 112. Textbook of Differential Calculus and cos" (cos x) could be sketched as Wt sco xt +1=0 txt +1>0 ¥ soe ; ree | sin“ (sin x)] > cos" (cos x) is not possible. ‘Only equality holds as SlaynnxerR ii) f(x) = sgn (loge x) 1, loge x0 fs Pa ale, dope x<0 2. flt)=]- 0eze; a lg x=0 ee) Graph for, f(x) = sgn (log, x) is “x Y i eee ‘Thus, | sin“ (sin x)| = cos (cos x) Ot When xefo$]u[ an] 47 , a)*Le I So, domain for renis[az] Uf, 22 Ree elt eed 2)"L2 1 sinz>0 2mm < x <(2n+1)n Hence, (a) is the correct answer. flat sinxe (n+ 1)R 0 SSO 4 if x <0 ae 0 if x=0 cos x >0 cosx <0 cos x=0 Onn nD < x < 2mm + 2 Qn + HID < x < On + 3H x=(2n+1)n2 Figure 3.23, [Example 31. sketch the graph of (0) flo =sgn (x? +1), (i) flx)=sgn (loge x). (il) fe) =sgn (sin x). (iv) flx)=sgn (cos x). Chap 03 Functions 113 (ix) [x]> neo x 2n+ hme Integer greatest Integer Function ‘il + ) or Step Function (0) [x]sneo x [075]+...+ [099] + [10] +..+ [1.74] 75 Olx]sx<[x]+1 = ‘TSterms are each ‘equal to zero ‘equal to 1 (x-t<[x] nyne Integer Hence, [x +y]=[4+ f +1l)=[15+ f]=15 114 Textbook of Differential Calculus J Example 35 Find domain for ” flo=sin.x}e05| (e-1] except when [x-1]=0=90 x-1<1 = Isx<2 Hence, domain of f(x)is R— [1,2 Sa ams aye dein T Example 36 The domain of the function _0g4(5—(x-11-Ex7?) - xP+x-2 C (where [x] denotes greatest integer function) ‘Sol, For domain of f(x), S-[x-1]-[xP >0 and xitx-240 = (x42)(x-1) 20 = xeL-2 Now, S-[x]+1-[x} >0 = [xP +[x]-6<0 - (Cx) +3((x]-2) <0 = ~3<[x]<2 = -2Sx<2 xeL-2 ++ Domaine (~2, 1) (1,2) Example 37 Let (yn? +1]=C/n? +1, where 1, XEN, Show that A can have 2n different values. ‘Sol. We have, n? + 1=(n +1)? -2n<(n+1);nEN ie. yrPticns1 Wit ey=n Wh +a)=n 2 neqnthcnti or n(n? +A)< (n+) = Ox domain f(x)=6 H Example 39 if domain for y = f(x) is [- 3,21 find the Hence, exists, if [xh x>0 [x]>x . (as x= x}s45 tay domain of g (x)= f {|Cx]]}. ‘Sol, Here, f(x) is defined in [-3.2} = re(-32]. (Le. we can substitute only those values of x, which lie between [-3.2]) For g(x) = f{[x])} to be defined, we must have -3s|[x]s2 = 0s/ [x52 {s| x| 20forall x] = -2s[x]s2 [asx] Sa-asxsg 3 -28x<3 [by definition of greatest integral function] Hence, domain of g(x) is {-23[or[-23). TExample 40 Find the domain of function fid= ———E Tecapeeroarce Me C+ denotes te greatest integer function, Sol. f (x) is defined when = > Ox-1)+[|7-x[]-6#0 fa x]+(7-x] #6, when x1 ) [x-1]+[7~x] #6, when 1sx<7 i) [x-t]+[x-7] #6, when x27 ) (i x]}+[7- x] #6 1+[-x]+7+[-x]#6 = 2[-x]Je-2 [-x]e-1 x€(0,1] 4) From Eq. (ii), we have = = (x-1]+07-x] 46 (e]-147+ [x] #6 []+L-x] + 0 xe Integer x6 (1,2,3,4,5,6,7) ) From Eq, (i), we have = = (x-1]+[x-7] 46 [x]-1+[x]-746 2[x]eu4 = [x]e7 xe (7,8) ( Hence froms (a), (0) and (9), we get Domain f (2) is R-K0.1)(1.2.3,4,5,67}0(7.8). example 41 ifthe function f(x) =[3.54bsinx] (where (-] denotes the greatest (feger function) isan even function, the complete set of values of Bis (a) 05,05) ) &05,051 @on @ tn Sol We have f(x) =[35+bsin x] For f(x) to be an even funtion 3<85+bsin v4, YaeR 4 -050, te (ees 205x275 Tous (2? + 10x +25)>0 = veo tas It1et4 So,1=3-5 and f=1 aa 1 f=0,N=4 So.T=+4 and f=0 ‘Thus, number of solutions are 1-4 solutions. ( Least Integer Function y=(x) =[x}[x]or (x) indicates the integral part of ‘which is the nearest and greater to x. It is known as ceiling of x. Figure 327 (0.23) =1 (- 8.0725) =- 8(-28) «: ‘Thus, [2.3203)=3, Ingeneral, n n-ne! (il) @)>nexonnel (ix) (&) Sneoxsn nel 0) (x) 25, then x belongs to Exercise for Session 4 —— a = Sol. Let x=1+ f, where Ie integer, f € fractional part such that os f <1. * Lx} + (98 > 25, = + sP e+ fas = P+ Usa >2s = Par 42r41>25 = ar +21-24>0 = P+1-12>0 > (r+ 4yr~3)>0 I<-4 or I>3 Here, xelt+f So, x<-4tf or x>atf ( Now, let x = I, then x? +x? > 25 = x? > 125 = xS-4or x24 Form Eqs. (i) and (ii), we get x€ (~ =, ~ 4] U [4,) = Directions (Q. Nos. 1 to 18) Find the domain of the following. 1. (x)= yx =| 1-2 100) = 2-181 + TEDL F(x) =10ge [ogo x | 22 ABN 1 (O" eF=0d-6 N which [-] denotes the greatest integer function. f(x) =log (x ~ [x]), where [-] denotes the greatest integer function. , where [-] denotes the greatest integer function. f(x) =cosec™' [1+ sin? x], where [-] denotes the greatest integer function. 8. (x)= cos jlogix) [21 where [-] denotes the greatest integer function. cu Textbook of Differential Calculus 9. 10. 17. 12. 13. 14. Kx 15. 16. 17. 18. F(x)= where {-} denotes the fractional part function. x -~2{x}" (x)= sin-t( L 2). where [-] and {-} denotes the greatest integer and fractional part function. f(x) = sin” [2x? — 3), where [-] denotes the greatest integer function. 2 Nx)= onlin, (2 ) Jer donctes the grote integer function. x} +4, x €L-1, 1], where {-} denotes the fractional part function. Hx) = 2 (x)? 1 “ITx- ~21]+ [1x ~10)}~a Where [-] denotes the greatest integer function. =Isin x |+sin x, 9(x)=sin x +08 x,0 0 ie, (-4)'- 43\5-e7)20 => 12k B44 soe'2 ay» hy) soos} ‘liter Since, log san increasing function and3 Axis sminimumat x Fie loge? — 4x +5)isminimomat 2 oy +2 andminimam ve ofy «loge (loge! sy (x= 4) a(x? -4x45)20 Jogs (logy (x? + 4x + 4) exists, if = ax? 440 Tog. .2(x? + 4x +.4)>0 > (-A6r-2)50 = Hrarsae(}) a 2 + xe [2/32] (orig bogs > bose # 0< <1] Hence, (isthe correct answer. = x'+4xe¢4e1 . aes V Example 57 the range of the function = (etNK49<0 3 tered fb jeosa® (3) (22,9) (b) (2, 2V5)(€) (0, 202) ¢) (2V, 4) 1 and x? $4x44>0 (x42) >0, Sel F(x) cos x] which is always true except for x =~2. lil) Using AM> GM, we ‘From Eqs. (i) and (ii), we have coe te 3-2) U-2-1) #1 Teor] fe Thus, domain is (-3,~2) U(-2.-1) eee (aia ‘Now, we find out the range. 1 1 Since, 0< logya(x? + 4x4 4) ete [Example 55 Range ofthe function ee fl=tos"t-e jis ~ Range of f(x) is 202, wfe " ' [ ‘ 0 ae Hence, a) isthe conrect answer. [24] [0] oom a(S.) 2 oe 1 Example 58 if 2 = x-+iy and x? + y2=16, then the Sol. Here, (x)= cos""|1~ x*| would exists, only when range of ||x|—Ly I]is [a-x?]s1 (a) (0,4) (b) (0,2) = -1Si-x"S1 = 22x*20 O24 (@) None ofthese gol et x = 4.605 0,9 =4 sin, then | 4|608 0 | 4|sin 8 ||= 4 cos 8|— I = 4 YI=2] cos 0] sin] = 4Yi=[sin 28] +, Range is (0, 4} Hence, ( is the correct answer Example 59 The range of fod= = sin x tant x)+ aah att is Xo 42x45 3 53 o[ts] obey 35 3 o[-34] @[-3] 4 gin ot sal Here, x)= 2 sin“! x + tan 1 : = (x) +), ey where domain of g(x) is [1,1] and minimum value of g(x) = g(~1) Also, maximum value of h(x) occurs, when (x + 1) +—4 is minimum at x = 1. Gey = Range o 0) 6[- 3.1] Hence, (d) is the correct answer. 1 Example 60 The range of the function flx)=sin? x - 5sin x ~6 is, a) [10,0] (b) 1,11 (0,21 (a) Sol, Here, f(x) =sin* x ~5sin x-6 [int xsi 2) 4 where 2 ¢(inx-3) <% 4) 2) <4 From Eqs. (i) and (ii), we get ~10 < f(x) $0 = Range of f(*) i [-10,0} Hence, (a) is the correct answer. {Example 61 if f(x) =(x21—[x1?, where E] denotes the greatest integer function and x€[0, n], n€ N, then the number of elements in the range of f(x) is (a) (2n+1) (b) 4n—3 (c) 32-3 (d) 2-1 1 Chap 03. Functions Sol. When x When nu syetn no-lexen, yf -(n- 1 =0 [Jena (n= Sp] sn? 1 OS[x*]- [xf Sn? =1- (n=) => 0S f(x) S2n—2 but f(x) has to be an inleger. The set of values of f(x) 18 {0, 1,2, 20 = 2) Hence, (is the correct answer. and 1 Example 62 Range of the function Fx) = ffsin™ [sin x|] cos" Joos x, is @o} fo & | (0 (0,4) (@) None ofthese ‘Sol. We know that, | sin“! sin x || = cos" | cos x |, x € domain 2. f(x) = yjsin™ | sin x || - cos | cos x | =0, Vx domain Range of f() is (0) Hence, (a) is the corsect answer. LExample 63 The number of values of y in (~27, 2n] satisfying the equation | sin 2x|+| cos 2x| =|sin y| is fa)3 (b) 4 (5 (d) 6 Sol, Here, 1 sin 2x |+| cos 2x |< ¥/2 and |sin y|< So, solution is possible only when| sin y| =. am = siny=t1 = y= 2.82 1 Number of values of yis 4 Hence, (b) isthe correct answer. 1 Example 64 Let f(x)=cot™ (x? -4x+ 5), then range of f(x) is equal to x wo, | 7 (043) (d) None of these ele Sol. Here, x ~ 4x +5=9(x-2)' +121 Isx?-4xt5<0 = O cot" x3, since cot” x is decreasing) Range of f(x)is (4 Hence, (b) is the correct answer. 122 Textbook of Differential Calculus Case 00, then range of f(x) is R-('1.92)- x? + 14x49 . So sin? x+sin x -1 x? + 2K+3 Sol, Let y 1 Example 65 Find the range of f(x) where xER sin? x—sin x +2 : =12=-8, B=12+36-56=—-8 C=160 Let tesinx = -1stStand y=Sttt Pata Sol. Here, A= Now -B+ yB’ -4AC > (y-t -(y +1) t y+ = Ai) ae here A<0 sincetisres, 2-211 g, <34aVHL 7 Range is [- 5.4] CaseI If both roots of Eq, (i) are greater than 1, ie.t)>1 1 Example 66. For what real values of a does the ee! met = hetp>2 ; atte range of f(x) =~" contains the interval [0,1 = eae x+1 = wtts2 ae a ay+i_yti = pojext and -2**+1>0 ylatxtyaxtt y= = yx? — x + (ay ~ 1) = Ohas real roots for every y€ [0,1] = itory< = — day? + 4y 20 holds for00 = Yee co and VEAP a1>0 yor 7 F- Chap 03 Functions 123 1 p<% <1 and y> tory. <1 m BHU YL cy ggg DEL tt . eh : ee Case Wh <~Vandts >—1,t, 19, (x)= 25.420 20. f(x) ={}sinx| + |cos x |]. where [J denotes the greatest integer function. xa 3 3 bin (an 0-9) 22. Find the image ofthe following sets under the mapping f(x) = x* ~ |, where [] denotes the greatest integer function 21. tx)= Bx? 4 22x? —24x +10 (i) (-=1) (iif, 2 1 23. Find the domain and range of f(x) oa s08 [ele i 24. Find the domain and range of (x)= sin“ (lg [x}) + log (sin [x]), where f] denotes the greatest integer function 25, Find the domain and range of (x) =[log (sin yx? + Sx +2)} where] denotes the greatest integer function Session6 Odd and Even Functions Odd and Even Functions Odd Functions ¥s std to be an odd function. if for ax Graph of an odd function is Symmetrical m opposite quadrants. ic. the curve in first guadbent s adenocal to the curve im the third quadrant and 18 identical to the curve in gzaphs which are symmetrical in spposte quadrants (or shout origin) are 4 Samcton fix Finer ice Figure 3.30 Figure 331 Even Functions 4 funcoon fix) 1s said to be an even, if f(-x) = f(x) for all saph is always symmetrical about Y-axis, ic. the raph on left hand side of Y-axis is the mirror image of the eve on its nght hand side Some graphs which are symmetrical about Y-axis are Figure 3.53 Properties of Odd and Even Functions () Product of two odd functions or two even functions an yn function (i) Product of odd and even function isan odd function, i) very Function y = f(x) can be expressed asthe sum ofan even and odd function. (i) The derivative ofan odd function isan even function ang Cervative ofan even function isan odd function (W) Afunction which is even or odd, when squared becomes an even function, (4) The only function which s both even and odd is f(x) =a i 2ero function. L Example 68 if fis an even function, then find the real values of x satisfying the equation (x+1 for 07 3 199600) Sol. Since, f(x) is even, so f(-x) = f(x) {Example 69 Find out whether the given function is even, odd or neither even nor odd; (ll el where fla) = lene xd+(1=x) » clex then f(x) is periodic. if flx+T)= f(z). sainterval o = wee [ Example 73 Le: f(x) be periodic and k be a posse + k)+ fle) =0 for all But period of f(x) smallest postive real number. real number such that flx=k)~ flx Thus, periad of f(2)is2 Prove that flx) isa periodic with period 2k Aliter f(x) = sin x could be expressed graphically as ‘Sol. We have. shown in figure. flx-k)+fiz)20¥ xeR > (fix+k)=-f(x),¥ xe R, puts = +k) =-flx+k.¥ xR [as f(x+k)=- f(x) = flx+2)= fix,¥ eR ‘which clearly shows that f(x) periodic with period 2 Some Standard Results on Periodic Functions Here, graph zepeats at an interval of 2m Fanesons ‘Thus, f(x)is periodic with period 2x. @ atx costs % fniseven ieee 2m, ifmis odd or tation 1 Example 72 Prove that f(x) = x- [x1 is periodic function. Also, find its period. (@) tan'x, cot" mis even or odd. Sol Let T >a Then, fle4T)=flW ER a = (xt T)-[x+T]=x-[xh¥ eR &) 1 = [x+T]-[]=T.¥xeR (0) Algebraic functions — = TH1234, eg VER +5. ete (sine subwaction of two integers] 7) onsant Periodic with no fandane™™ ‘The smallest value of T satisfying f(x +T) = f(x)is 1 period. ‘Thus, it s periodic with period 1 j Example 74 Find periods for (cost. (isin? x. (i) cosv. (w) JOS. ol ()co8*x has a period 7 as nis even, (ji) sin® x has a period 2x as mis odd (ii) cos Vx is not periodic, as for no value of T, Sle+T)= f(x) = 008 xFT = cos (Vz) Thus, there exists no value of T for which f(x +T)= f(x). Hence, cos vx is not periodic. (iv) f(x) = [008 « has the period 2n as a is in fraction, Aliter f(x-+T) = f(x)= feos (+ T) = feos (=) = T=2n, 4x, But Tis the least positive value, hence f(x) is periodic ae Flx)is period Properties of Periodic Functions (i IFfla)is periodic with period 7, then (a) c-flx)is periodic with period T (b) f(x + c) is periodic with period T (0) f(x) 4 cis periodic with period T, where cis any constant. We know, sn has period 2. Then, f(x) =5(sin x) + 4 is also periodic with period 2. ie. “If constant is added, subtracted, multiplied or divided in 2 pend function, ts perind remains he ame" (i) #f(xis periodic with period, then hf (cx + d) has period i.e. period is only affected by coefficient of x, where, k, c,d € constant. Weknow. fix) -f sin(ac 2} ~ has the prod = as sinss poi with petod 2 i) Ife), fb) are periodic functions with period TT respectively, then f(x) = fx) + fle) has period LM of {7 Tif hes not an even function or 7°) FLOM (1 Fife) and fats ate complementary pair-wise comparable functions. While taking LCM we should always remember (ahem of{ 2, £, £) = UemoF (ace) ' a5 Sy HeFof Od) LOW of (2n, m,n) _ 2m HCF of (3,612) 3 voor, 2,2) <7 36m) 3 (0) LOM of rational with rational is possible. LH of irrational wth rationals possible. But LOM of rational and rational isnot possible 8 LCM of (2x, U6) isnot possible, as 2, 6m irrational anid 1e rational eg LMof=: Chap 03 Functions 127 Periodicity of Constant Function ‘The LCM rule is not applicable, if function reduces to constant, eg. fx) =sin? x +08? x. Since, period of sin’ x and cos? x are r. e.Period of f(x) ; LCM {n,n} = 5. which is not correct. is a constant function and Whereas, sin’ x +cos* period is undetermined. L Example 75 Find the period, if f(x) =sin x-+{x}, where {x} is fractional part of x Sol, Here, sin x is periodic with period 2x and {1x} is periodic with period 1. Thus, LCM of 2m and 1 =» Does not exist. Hence, f(x) is not periodic. LExample 76 Find period of f(x)= tan 3x+sin (3) st ma reas u evo forsin sa ‘Thus, LCM of = * 3 a aH 14 Hence, f(x) is periodic with period 6x. [Example 77 Find the period of Bis said to be many-one function, if ‘© or more elements of set A have the same image in. B. ‘nother words; f:A— Bis a many-one function, ifit is "ol one-one function. Chap 03 Functions 131 Verbal Description Let f:A—> Band g:X > be two. functions represented by tia a Figure 338 Figure 3.39 Clearly, f and g both are many-one as there are two elements x5, x, which correspond to the same image ie. flas)=fle4)=ys. Thus, many-one. Method to Check Many-one ‘They are same as for one-one because, if mapping is not ‘one-one itis many-one. L Example 91 show f: RR defined by fix)=x? +x for all xeR is many-one Sol. By graph flx)= x7 + x can be represented graphically, figure, ‘where the straight line parallel to X-axis meets the curve at two points (i, more than one point). shown in, ‘Thus, itis not one-one itis many-one. Alter By calculus, fayev tx > fi(a)= 2041 where "(x)>0 if xt 2 and F'(x)<0 if x<-} which shows f(x)is neither increasing nor decreasing i.e. not monotonic, Hence, many-one. 132 Textbook of Differential Calculus 3. Onto Mapping or Surjective A function f:A — Bis an onto function, if such that each element of B is the f image of atleast one element in A It is expressed as f:A >. Here, range of f = codomain. ie, f(A)=B Method to show onto or surjective Find the range of y= f(x) and show range of f(x) = codomain of f(x). Remark \trange = codomain, then fx) onto, Any poly agree nas range al eal numbers angi oro or 1 Example 92 show f:2 > defined by fox) = (x-1)(x-2)(x—3) is surjective but not injective ‘Sol. We have, al ofode lx) = (x-x-20-3) fl= fQ)= f3)=0 Hence, itis not injective itis many-one function. Now, f(x)is a polynomial of degree 3, ie. odd. Hence, f(x)is surjective. Number of Onto Functions If Aand B are two sets having m and n elements respectively, such that 1< n Sm then number of onto functions from A to Bis Coefficient of x™ in m!(e* - 1)" = Coefficient of x” in Ose "Cet mif"Coe™ ~ "Ce tte" Cy} =yovrrc,r® Or Consider the set of all possible functions from A to B, ie (f:AB}. ‘Now, let us define a subset Q such that A/= (fe Q|ie Range of f},ien UA f is injective} To find number of onto functions = Total number of functions ~ Number of injective functions where O A\= 314 [= 3 AVA) + ene(n— 1" = "Cy(n—2)" eM Cyl Number of onto functions ant (a(n 1)" = "Cal —2)" AC, “Sars en 1 Example 93 1f f aE 7 is an onto function, the set of values of is w[-4-1 aed (d) None of these se @ ¢ Ors) Sol. Here, f(x) is onto. Hence, (c) is the correct answer, 4, Into Mapping ‘A function f:A— Bis an into function, if there exists an element in B having no pre-image in A. In other words, f : A > B is imto function, if it is not onto function (mapping). T Example 94 show f:R +2 defined by fl)= x? +4x+5 is into. Sol, We have, f(x)= x7 +4x+5 S(x)= (x42) 41 Since, the codomain of f is R but the range of f is[}.=} Hence, f is into. One-one Onto Mapping or Bijective ‘A function is one-one onto or bijective, if itis ‘one-one and onto. A function is bijective if and 00 every possible image is mapped to by exactly ® argument. ‘The function f:A— Bis bijective iff forall y€ 8, the" a unique x € A such that f(x) =y. example 95 Let A={x:-1 defined as Flo) =loge (oer x4 VWF 41x) is (a) one-one and onto both {(b) one-one but not onto (©) onto but not one-one (d) Neither one-one nor onto Sol. (2) Here, f(x) Bog Ware Wr 1x) oh Were Wms) shad ET aie Hod fer oie] = Floeteyx? +1 +2) Now, f(-x)= f(x) 2. f(x)is even function. Hence, neither one-one nor onto. 134 Textbook of Differential Calculus Quick Review [ie.nelements] lie. elements} Figure 3.40 (a) Total numberof functions = 1 umber of elements in codomain)"™™" sens # oma (b) Total number of one to one functions = more (Total number of many-one functions =|" oe ren (6) Total umber of constant functions (e) Total number of ono functions PFC (r= If + Clr 2P G03) +... Fem - ten q pa (9 Total number of ino functions Se 1s J Example 97 1f X={1,2,3,4,5}and Y and f.:X— ¥, find the total number of {i functions (i) one to one functions {i many-one functions (jv) constant functions (\)onto functions (wv) into functions Sol, (i) Total number of functions =6° = 7776 (i) Total numberof one to one functions = "C55! = 61= 720 (ii) Total number of many-one functions = (iv) Total numberof constant functions = 6 (@) Total numberof onto functions = 0(8sr > n) (i) Total number of into functions = 6° = 7776 Abd p = 105 [Example 98 Find the number of surjections from « to.B, where A={1,2,3,4} B=(0D}., MT JEE24y Sol. Number of surjctions from A to B= Y(-1)°°" 70," (1)? Cay +1? #202)! = -2 + 16-14 ‘Therefore, number of onto mappings from A to B= i [liter Total number of mapping from A to Bis2of whi two functions f(x) =afor all xe A.and g(x} = bforall x4 ae not surjective. Thus, total numberof surjections from ‘AtoB =2'—2=14 Exercise for Session 8 7. There are exactly two distinct linear functions, which map —11] onto [0,3}. Find the point of intersection of he two functions, 2. Letbe one-one function with domain (x, y.2) and ange (1,2, 3). Its given that exactly one of the folowing statement is true and remaining two are false, 10x) = My) #412) #2. Determine 1), 3. LetA=R-@).B=R {I} andt A+B defined by (x) ="=2, is ¥" bijective? Give reasons. 4. Letf:R +R defined byfx)= Prove that fis neither injective nor surjective. x 5. the function :R +A, given by x)= ¥ Tar e 6. Ifthe function f: > A, given by (x). e “7/8 surection, find A ix is surjecton, find A. 7. Letf(x)=ax? + bx? + 0x +d sin x. Find the condition that f(x) fs always one-one function. 8. Lett XY bo a function defined by) =a sn(x+ 2] bcos x +0 If fis both one-one and ont, find sets X and Y. 4 Session 9 — Identical (or Equal) Functions |dentical (or Equal) Functions wo functions f and g are said to be identical (or equal) functions, if (i) the domain of f =the domain of g, (i) the range of f =the range of g, and to fe) =a2hV x6 domain Examples of Equal or Identical Functions (i) flx)=Inx?, g(x) =2In x (wp fx) =cosee x, g(x) = © sinx (i) fl) =cot (cot™ x), g(x) =x 0 — tan Glabrae CS) @x-4x")g(x)=3sin'x (ND) n(x? +1), g(x)=sin’ x-teos? x) * x sin” x, g(x) =tan? x-sin? x — () (ci) fle) =see? x — tan? x, g(x) =1 wy (a fle) lots 80)= o (viii) f(x) = tan (cot x), g(x) =cot (tan™ x) o (60) flx) =)? 1, g(x) = fe-1- fe wy ) flx)=tan x-cot x, g(x) =sin x -cosee x (NI) (ai) fix) =e", g(x) =e* 0 (xi) fx) = poets, ax) sin x (cit) f(a) =x", g(x) =(Vx)? (xiv) f(x) = log (x +2) + log (x -3), (x) = log (x? - x 6) (NI) (ev) ffx) =x |x) g(x) =x sgn x o (ev) f(a) = tim 2°=" gx) =sgnqxi-2) noe a (evil) F(x) =sin sin x), g(x) = cos (cos™! x) o oe (xviii) fone OO arr (NI) x (xix) fx) =e", g(x) =sec™ x. wp Identical, if x € (-%,— 1] [1=) (ex) Fl) =(fog)(x), G(x) = (gof (x), where f(x) =e*, ax) =Inx wy T Example 99 if f(x) =log 2 25 and g(x) =log, 5, then Fx) = g(x) holds, now find the interval for x. Sol. Domain of fe R-{#1,0} Domain of g(0,<=)~(1} For ie, flx)= g(x), if x2(0.00)- (1) 1 Example 100 Let A={1,2}, 8={3,6} and f:A +8 given by f(x)=x? +2 and g:A—»B given by g(x)=3x. Find whether they equal or not. to FA)=3, f2)=6 #(1)=3,£)=6 which shows f(x) and g(+)have same domains and range, & f= T Example 101 Which pair of functions is identical? (a) sin (sin x) and sin (sin~ x) (b) loge e%, elt * (0) log, x?, 2log, x (d) None of the above Sol. Here, (a) sin“ (sin x) is defined for xe 4; While sin (sin“ x) is defined only for x € (1, 1} (©) log. e* is defined for all x, while * is defined for x >a. (6) log, x* is defined forall x € R~ {0}, while 2 log. xis defined for x >a, s+ None is identical Hence, (A) isthe correct answer. Exercise for Session 9 identical (or equal) functions? = Directions (Q. Nos. 1 to 10) Which 1 2. Pan w L. f(x) =cot® x cos? x, g(x) =00t® |. f(x) = san (cot x), g(x) = san (x? ~ of the following are" f(x) =Ine*’,g(x)=0"" fod= =- (x) =sec x, g(x) : x F(x) =sec* x + cosec'x, 9(X)= > x cos? x 4x +5) (x) logs x, a0) = seg fx) = 17g) = VTE TX (x)= gfx) =v Ix!’ f(x) = [09], g(x) = {Lc} [Note that f(x) and g(x) are constant functions] A(x) =0*"* g(x) =cot" x Session10 Composite Functions Composite Functions et us consider two functions, f:X >, and g:¥, Y. We define function h: X — Y, such that, ft 9 560 To obtain h(x), we first take f-image of an element x € X so that f(x) Y;, which is the domain of g(x).Then, we take g-image of f(x), Le. g(f(x)) which would be an clement of Y. ‘The diagram below shows the steps to be taken. yo =and g:B > be two bijections. Then, gf exists such that gof : AC We have to prove that gof is one-one and onto ‘One-One Leta; a; € A such that (Gof Ka) = (gof Xa), then (gof) (a3) = alflay)} = aff(or)] (c;) fr gis one-one} 2 [ef is one-one} ». gof is also one-one function. Onto Let c EC, then c eC = 3b eBsuch that lb) =c legis onto) and b eB = 3a €Asuch that f(a) =b ke fis onto] ‘Therefore, we see that eC = 3a eAsuch that (gof a) = df(a)] = og) =¢ ie. Every element of Cis the gof image of some element ofA. AS ‘such gof isan onto function. Hence, gof being one-one and onto is a bjection, (W) fis even, gis even =sfogis even function. (vi) fis odd, gis 046 = fogis odd function. (vi) is even, gis odd = fogis even function. (vill) f is odd, gis even => fogis even function. Example 102 Let f:A—8B and g:B->C be functions and gof : A—> C. Which of the following, statements is true? (a) IF gofis one-one, then f and g both are one-one (b) IF gofis one-one, then fis one-one (6) If gofisa bijection, then fis one-one and gis onto (@) IF f and gare both one-one, then gof is one-one Sol. (a) As shown gof is one-one, but g is many-one. = (a) is not correct. 138 Textbook of Differential Calculus OIF gof tone one, then fs also one-one, Bex, Osxst Aff is many-one, then gof cannot be one-one. 2-2, lexea Moe deren Alter x) ean be expressed grape 8 shown ni, below: 7 =3 (¢) and (¢) are obviously true Hence, (b),(c) and (d) are correct answers. J Example 103 if f-R8, f(x)= sand g:R9R; g(x) =2x-+ 1 Find fog and gof, also show fog + gof. Sol (sof Mx) = gtf(x)} = gtx} wn (gof Xx) =2x? +1 and OS f(xy 0 (opiay=l24x, osxst Thus, for all xeR. 2-x, 1 sin (H{A(R (x). )) = sin" <2 Domain of sin” (H(H(A(R 2) ))) 8-161) Hence, (a) isthe correct answer LExample 108 A function f: R92 satisfies sin xcos y (f(2x-+ 2y)~ fl2x -2y)) = cos xsin y (fl2x+ 2y)+ flex -2y)) id 1< g(x) $2; V x: . oe ena Ff (0)=4, then acen-| th -isect 2 ax?41, 1< x <2 (@) fx) = fx) =0 (b) 4f"(x) + fly =0 : © F%a+ FQ)=9—@ AF")= fO=0 example 107 IF f(x)=2x-+| x], g(x)= = (2x—-|x1) and sot. We have flex +2y) _ sin(x+y) A(x) = (g(x), domain of sin“ (h(h(h(h.....h(x)...)))) is F(@x—2y) sin (xy) SNe) sm C40 (b) [+-3] fz ‘| ce a = flay= Kin ® © [>-4] @ [24] . in 2 2x +x, 220. x20 Sol. Since, soo={ ax-x, x<0 |x, x<0 and £"x) ipx-x, x20_[¥, x20 =. 4p") flxy=o ae =|} ~~ Hee x<0 { a Hence, (bis the correct answer. Exercise for Session 10 1. Consider the real-valued function satisfying 2f(sin x) + f(cos x) = x. Find the domain and range of f(x). 2. Iff(x)is defined in [- 3, 2], find the domain of definition of (Lx [})andf({2x + SD) 3. f(x) is 1, aa and g(x) =sin x. Find (x) =f(19001) + (900) 41, -2sx<0 oa (x) =F(lx 1) + Lf) find 9 (2) 4. Let (x) be defined on [-2,2] and is given by (x) = {i 18% <2 ang (x)= A such that. e ats linear in x Wao reece whi i cet beste itis Then, g is said to be inverse off. Malta a Thus, g= f':B— A=((f(x),x)|(x, fla) € f) 7 Pesta Let us consider a one-one function with domain A and 5 range B. ‘i = POE teseryorryy 4} and B=(2,4,6,8} and f:A—>B is as then write fand f~! asa set of a eo ae 110 If f:[1,2e)— [2,9) is given by A 8 a Foo=) xed, find f(x), (assume bijection). 1 2 ‘IT SEE 200,20 2 ‘ Sol. Let y= f(x) a 6 = x? -xyt1=0 4 8 Figure 343 2 Beat aA = fo)= las f(x)= ys x= fy) 2 1 aie = eal : = et i‘ : feeeeee as [e) therefore : i neglecting the negative sign, we have F)= Figure 3.44 a Here, member y¢ B arises from one and only one member Example 111 Let f(x) = x3 +3 be bijective, then find xeA its inverse So, F={(L2)(2, 4)(3,6)(4,8)} Sol. Let yers3 fie y= flall and 4(2,1)(4,2)(6,3)(8,4)} > ayes In above function, = y - 39" Domain of f ={1,2,3,4}= Range of f~! = 9-3)" y= flx) = f1y)=4) Range of f ={2, 4,6,8}= Domain of f~' 7 (x -3)" which represents for a function to have its inverse, it must Thus, F1G)= (x3)? be one-one onto or bijective. when f(x) = x? + 3is bijective

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