80200 If f(x) and g(x) are two functions with g(x)=x−1x and fog(x)=x3−1x3, then f′(x)=
(A) : Given, g(x)=x−1xAnd, f∘g(x)=x3−1x3Using (a−b)3=a3−b3−3ab(a−b)Then,(x−1x)3=x3−1x3−3x⋅1x(x−1x)x3−1x3=(x−1x)3+3(x−1x)We have, fog(x)=x3−1x3fog(x)=(x−1x)3+3(x−1x)As g(x)=x−1x, this yieldsf(x−1x)=(x−1x)3+3(x−1x)Putting, x−1x=tf(t)=t3+3tThus, f(x)=x3+3xf′(x)=3x2+3
80201 If f(x)={e3x−14x for x≠0k+x4 for x=0 is continuous at x=0,
(B)(x)={e3x−14x, for x≠0k+x4, for x=0}f(x) is continuous at x=0∴limx→0f(x)=limx→0f(x)=f(0)limx→0e3x−14x=k4limh→03e3 h4=k4( Using L− Hospital rule in 00 form )So, k=3
80202 If f(x)={[x]+[−x],x≠2 K,x=2 then f(x) is continuous at x=2, provided K is equal to
(C) : Given, f(x) is continuous at x=2limx→2+f(x)=[2+]+[−2+]=2−3=−1limx→2−f(x)=[2−]+[−2−]=1−2=−1For f(x) to be continuouslimx→2+f(x)=limx→2−f(x)=f(x)=k∴k=−1
80203 If f(x)={xsin1/x,x≠0k,x=0 is continuous at x=0 , then the value of k will be
(C) : Since f(x) is continuous at x=0, sof(0)=limx→0f(x)k=limx→0(xsin1/x)=0×(a finite quantity )=0