Question

The function $f\left(x\right)=(x^2-1)|x^2-3x+2|+\cos |x|$ is non-differentiable at

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

We know that function $|x|$ is not differentiable at $x=0$
Therefore, $|x^2-3x+2|=|(x-1)(x-2)|$
Hence, it is not differentiable at $x=1$ and $2$
Now, $f\left(x\right)=(x^2-1)|x^2-3x+2|+\cos |x|$ is not differentiable at $x=2$.
For $1<x<2,f\left(x\right)=-(x^2-1)(x^2-3x+2)+\cos x$
For $2<x<3,f\left(x\right)=(x^2-1)(x^2-3x+2)+\cos x$
$L{f}^{\prime }\left(x\right)=-(x^2-1)(2x-3)-2x(x^2-3x+2)-\sin x$
$L{f}^{\prime }\left(2\right)=-3\sin 2$
$R{f}^{\prime }\left(x\right)=(x^2-1)(2x-2)+2x(x^2-3x+2)-\sin x$
$R{f}^{\prime }\left(2\right)=(4-1)(4-3)+0-\sin 2=3-\sin 2$
Hence, $L{f}^{\prime }\left(2\right)\ne R{f}^{\prime }\left(2\right)$.
So, $f\left(x\right)$ is not differentiable at $x=2$.