Question

Let $a, b \in R$ and $f: R \rightarrow R$ be defined by $f(x)=a \cos \left(\mid x^{3}-x\right]+b|x| \sin \left(\left|x^{3}+x\right|\right)$. Then $f$ is

• A. differentiable at $x=0$ if $a =0$ and $b =1$
• B. differentiable at $x=1$ if $a =1$ and $b =0$
• C. NOT differentiable at $x =0$ if $a =1$ and $b =0$
• D. NOT differentiable at $x =1$ if $a =1$ and $b =1$

Answer 1

Answer: (A), (B)

$ f : R \rightarrow R $
$ f ( x )= a \cos \left( x ^{3}- x \right)+ b | x | \sin \left(\mid x ^{3}+ x \right)$
(A) at $a=0 \& b=1$
$f(x) =0+|x| \sin \left(\left|x^{3}+x\right|\right)$
$= \begin{cases}x \sin \left(x^{3}+x\right) & x<0 \\ x \sin \left(x^{3}+x\right) & x \geq 0\end{cases}$
So $f(x)$ differentiable at $x=0$
(B) at $a =1 \& b =0$
$f ( x )=\cos \left( x ^{3}- x \right)$ is differentiable at $x =1$
(C) at $a =1 \& b =0$
$f(x)=\cos \left(x^{3}-x\right)$ is differentiable at $x=0$
(D) at $a=1 \& b=1$
$f ( x )=\cos \left( x ^{3}- x \right)+ x \sin \left( x ^{3}+ x \right)$ is always differentiable