Let $a, b \in \mathbb{R}$ and $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=a \cos \left(\left|x^{3}-x\right|\right)+b|x| \sin \left(\left|x^{3}+x\right|\right)$. 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 $b=1$

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Continuity and Differentiability

Let

a, b \in R

and

f : R \to R

be defined by

f (x) = a cos (x^{3} - x) + b ∣ x ∣ sin (x^{3} + x)

. Then

f

Answer: (A), (B)

For

x > 0

f (x) = a cos (x^{3} - x) + b x sin (x^{3} + x)

, which is differentiable everywhere. The function simplifies to

f (x) = a cos (x^{3} - x) + b x sin (x (x^{2} + 1))

, which is a composition of differentiable functions for all

x \in R

. At

x = 0

f (0) = a

. For differentiability at

x = 0

, we need

lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{h}

to exist. When

a = 0, b = 1

, this limit exists. At

x = 1

, since

x^{3} - x = 0

and the function is smooth, it is differentiable when

a = 1, b = 0

. Options (C) and (D) are incorrect because the function is actually differentiable in those cases.

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · One or more correct options