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Functions Solutions

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0% found this document useful (0 votes)
43 views3 pages

Functions Solutions

Uploaded by

smpopade
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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ii 133. 6) =Hos (x +571) ana £9) = log (-r+ vei) fx) + x) =log (x+V? 41) + toe(-x 2 £@) pan odd function. 1M 83) = sin(loe(x-+ V7) fa 2) = sin[ fog (x ex) ] = sng (Sie 4) “*) (vies? +3) 1 Vitex +x = sin log(x+ i? y] = sia{ og(x+ V+) | 7 ~sin[fog(x+ Vir ) =-f(x) (2) is an odd function. = sing 35, f-2)= see|log(—x + i+=F)| se[loe( -=+Vie)] see{fog(vi+ xx | )] 5] mle ger [-! log( Vie +)| = se[log( VF] f(x) is an even function. = sd e{ = sec} 136. Consider option (C), (a* -1) fa)= 4 -x(a*=1 w(i-a')_ x02"=l) ey +a" ated (2) is an even function. 137. Since, f(x) is even f(x) = fle) 138. As f{-x)=f%) VxeR ‘. y remains unchanged when x is changed to ~~. Hence, the graph of y = f(x) is symmetric about Y-axis. 139. f&) = f- x) > (0 + x) = {0 - x is symmetrical about x= 0. . £2 + x) = {(2 —x) is symmetrical about x= 140. Value of the function f(x) does not exist. = 1or—1ie,, not unique) x 143. 3S Gvenesreion= $24] 22a e724 -2(2.4), 8 faa tea] a [33] -on 8 [2 ; 5G] = 66 each term in the summation is one but less than 2 when i = 33, 34, 35 mes 145. (46. 1 =1+ 3 [oseaes on (4 >) ~eoe{ 21+) -e(2}] + arco 20+ 28) -eos{2x4) 3 +) cos 2-2 sin 2008 sin{ = 2 2 6 +1 cos ax-asin Ese) 4 2 2 )2 +} (cos2x—00s2x) 0 3 a 3 a 3 ry 3 4 +2=2 =P=2-2 y= log: (2 - 2°) is defined, if2- 2"> 0 = <2 a2t 0, logs x > 0 and x>0 = logsx > 3°= 1,x>4°= 1 andx>0 =x>4'jx>landx>0 =x>4 domain of fis (4, «). . f(x) is defined, if 1 = logy (2? — 5x + 16) > O and x* = Sx+16>0 => logio@? — 5x +16) <1 and 2 (5 +32 50 for all real x 2) 4 => 5x+16<10'=10 =P -5x+6<0 = (r= 3)(&-2)<0 32 0,14 p> Oand.x£0 z 1 = lon (It fe eh + > 0 andx#0 ¥ z a 1 > (ta}> (4) sche >and x #0 = toot, Ay Ht and 40 wy 2 0-I andx #0 domain of f(x) =(0, 1). Since, e* is defined for all real x. domain is (— 2, 0) |. Since, domain of Va? —x* is [-a, a] domain of V4—x" is [-2, 2]. Since,domain of Vx’ —a” is (=, ~a] U [a, =) domain of Vx? -16 is (~~,-4] U [4, 0) 1 Since, domain of is (-a, a). 1 domain of is (-3, 3). 1 Since, domain of ————$———= is. Vee) (-, a) U (b, «), where a 0 =>xt>(0.5)! Sat21 xl=1=x=0orl Domain of the function jx+2] = isR- f(x) 2 ® {2} |x+¢] isR- «| xt 157. f(x) = log (Ve=4 4 o=x)is defined, when x-4>0and6-x>0 =>x2>4andx<6 = [4, 0) A(-~, 6] domain of f(x) = [4, 6] — = 158. Dr = fF ER: 1-x>0}nfreR ~&ERIX< A Ger: ieys~lorxet = Ee, IV (00, Uli eo oe bse Comune =(,-1] “ett 159. Dr=Dp Dy . 1 where g(x) = Toga(iza othe) = Vex Now, Dp= { € R: 1—x>0, logo (1 —2)0} ={xeRix0} = {xe Rix>2} Dr= [(-~, 1) - {0}] 9-2, «) =[-2, 1) - {0} 160. Let g(x) = sin os 32 ana ne) = “ton x) g(x) is defined, if -1 3 <1520>x<4 domain ofy=[1, 5] 0 [-,4) =[1,4) 161. 5 Let 262) =logi-=7 > and h(x) = fe+5 Now, g(x) is defined, if D,=(4,5)U 6,2) oi) h@) Yx+5 is defined for all real x Dy=R From (i) and (ii), we get domain of f(x) =D, 0 Ds =RA (4,5) V6,2)} = (4,5) (6,%) 162. For domain of f(), x + 3 > 0 and P43 220 =px>—3and (r+ 1) (e+2)#0 = x>-Bandx 4-1, -2 Hence, domain = (-3, ©) ~ {-1,-2} He hive 0 1 3 3 2 D: +) 3| x43 164. = |_—, ®” Va-aG-5 f(x) is defined, if +3) Q-a) (e520 and (2—x) (x5) #0 = (v3) @-2)(e-5) <0 andx #2, 5 domain of ffx) = (20, -3] U 2, 5) 2H 2Oand2- |x| #0 = Uob)(2-b) (2-h\) = ([s|-1) ([x]-2) > 0 andx 4-2, 2 = || s Lor fx] >2 2Oandx#-2,2 > -1Sx<1 or(e<—2orx>2) domain of fix) = [-1, 1] U (0, -2) u (2, 0)

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