Question

The function $f(x)=(x^2-1)|x^2-3x+2|+\cos (|x|)$ is NOT differentiable at

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (D)

We have $|x|=\{\begin{array}{ll}-x& \text{if }x<0\\ x& \text{if }x\ge 0\end{array}$
Also, $\left|x^2-3x+2\right|=\left|\left(x-1\right)\left(x-2\right)\right|=\{\begin{array}{ll}(1-x)(2-x)& \text{if }x<1\\ (x-1)(2-x)& \text{if }1\le x<2\\ (x-1)(x-2)& \text{if }x\ge 2\end{array}$
As $\cos (-\theta )=\cos \theta \Rightarrow \cos |x|=\cos x$
$\therefore$ Given function can be written as $f(x)=\{\begin{array}{l}(x^2-1)(x-1)(x-2)+\cos x\text{if}x\le 1\\ -(x^2-1)(x-1)(x-2)+\cos x\text{if }1\le x<2\\ (x^2-1)(x-1)(x-2)+\cos x\text{if }x\ge 2\end{array}$
This function is differentiable at all points except possibly at $x=1$ and $x=2$.
${\text{Lf}}^{\prime }\left(1\right)={\left\{\frac{d}{dx}\left[\left(x^2-1\right)\left(x-1\right)\left(x-2\right)+\cos x\right]\right\}}_{x=1}$
$=-\sin 1$
${\text{Rf}}^{\prime }\left(1\right)={\left\{\frac{d}{dx}\left[-\left(x^2-1\right)\left(x-1\right)\left(x-2\right)+\cos x\right]\right\}}_{x=1}$
$=-\sin 1$
$\because {\text{Lf}}^{\prime }\left(1\right)={\text{Rf}}^{\prime }\left(1\right)$
$\therefore$ f is differentiable at $x=1$
${\text{Lf}}^{\prime }\left(2\right)={\left\{\frac{d}{dx}\left[-\left(x^2-1\right)\left(x-1\right)\left(x-2\right)+\cos x\right]\right\}}_{x=2}$
$=-3-\sin 2$
${\text{Rf}}^{\prime }\left(2\right)={\left\{\frac{d}{dx}\left[\left(x^2-1\right)\left(x-1\right)\left(x-2\right)+\cos x\right]\right\}}_{x=2}$
$=3-\sin 2$
$\because {\text{Lf}}^{\prime }\left(2\right)={\text{Rf}}^{\prime }\left(2\right)$
$\therefore$ f is not differentiable at $x=2$.