Question

The function
$f\left(x\right)=(x^2-1)|x^2-3x+2|+\cos \left(|x|\right)$ is not diffentiable at

• A. $x=-1$
• B. $x=0$
• C. $x=1$
• D. $x=2$

Answer 1

Answer: (D)

$x=2$

$f\left(x\right)=(x^2-1)|x^2-3x+2|+\cos \left(|x|\right)$

We have $|x|=\{\begin{array}{ccc}-x& if& x<0\\ x& if& x>0\end{array}$

Also $|x^2-3x+2|=|(x-1)(x-2)|$

$\Rightarrow \{\begin{array}{ccc}(1-x)(2-x)& if& x<1\\ (x-1)(2-x)& if& 1\le x\le 2\\ (1-x)(x-2)& if& x\ge 2\end{array}$

As $\cos (-\theta )=\cos \theta \Rightarrow \cos |x|=\cos x$

Given function can be written as

$f\left(x\right)=\{\begin{array}{ccc}(x^2-1)(x-1)(x-2)+\cos x& if& x\le 1\\ -(x^2-1)(x-1)(x-2)+\cos x& if& 1\le x\le 2\\ (x^2-1)(x-1)(x-2)+\cos x& if& x\ge 2\end{array}$

This function is differentiable at all points except possibly at $x=1$ & $x=2$

$L{f}^{\prime }\left(1\right)=\{\frac{d}{dx}{\left[\right(x^2-1\left)\right(x-1\left)\right(x-2)+\cos x]}_{x=1}$

$=-\sin 1$

f is differentiable at $x=1$

$L{f}^{\prime }\left(1\right)=R{f}^{\prime }\left(1\right)$

$L{f}^{\prime }\left(2\right)=\{\frac{d}{dx}{[-(x^2-1\left)\right(x-1\left)\right(x-2)+\cos x]}_{x=2}$

$=-3-\sin 2$

$R{f}^{\prime }\left(2\right)=\{\frac{d}{dx}{\left[\right(x^2-1\left)\right(x-1\left)\right(x-2)+\cos x]}_{x=2}$

$=3-\sin 2$

$L{f}^{\prime }\left(2\right)\ne R{f}^{\prime }\left(2\right)$

$\therefore$ f is not differentiable at $x=2$.

Option D is correct answer.