Question

Let $a,b\in R$ and $f:R\to R$ be defined by $f\left(x\right)=a\cos (|x^3-x|)+b|x|\sin (|x^3+x|).$ 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)

The correct options are
A differentiable at $x=0$ if $a=0$ and $b=1$
B differentiable at $x=1$ if $a=1$ and $b=0$

$f\left(x\right)=a\cos (|x^3-x|)+b|x|\sin (|x^3+x|)$

As we know that $\cos \theta =\cos (-\theta )$,
$f\left(x\right)=\{\begin{array}{cc}a\cos (x^3-x)-bx\sin (-x^3-x)& x<0\\ a\cos (x^3-x)+bx\sin (x^3+x)& x\ge 0\end{array}$

$\therefore f\left(x\right)=a\cos (x^3-x)+bx\sin (x^3+x)\forall x\in R$
Hence, $f\left(x\right)$ is differentiable at all $x\in R$ for any $a$ and $b$.