Advanced 2016 | JEE 2016 | Discussion Page

Let $a,b\epsilon R$ and f:R→ R be defined by $f(x)=a\cos(|x^{3}-x|)+b|x|\sin(|x^{3}+x|)$ . 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

Option A,B

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

if $x^{3}-x\geq0$

$\Rightarrow \cos|x^{3}-x|=\cos(x^{3}-x)$

$x^{3}-x\leq0$

$\Rightarrow \cos|x^{3}-x|=\cos(x^{3}-x)$

$\therefore \cos(|x^{3}-x|)=\cos(x^{3}-x), \forall x\epsilon R$ .........(i)

Again , if $x^{3}+x\geq0$

$\Rightarrow |x|\sin(|x^{3}+x|)=x\sin(x^{3}+x)$

$x^{3}+x\leq 0$

$\Rightarrow |x|\sin(|x^{3}+x|)=-x\sin(-(x^{3}+x))$

$\therefore$ $|x|\sin(|x^{3}+x|)=x\sin(x^{3}+x), \forall x\epsilon R$ .......(ii)

$\Rightarrow f(x)=a\cos(|x^{3}-x|)+b|x|\sin(|x^{3}+x|)$

$\therefore$ $f(x)= a\cos (x^{3}-x)+bx \sin (x^{3}+x)$ ........(iii)

which is clearly sum and composition of differential functions.

Hence, f(x) is always continuous and differentiable

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