Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

We have $cos\theta =cos-\theta$
Hence, $f\left(x\right)=\{\begin{array}{c}(x^2-1)(x^2+3x+2)+cosx,-\infty <x<1\\ -(x^2-1)(x^2+3x+2)+cosx,1<x<2\\ (x^2-1)(x^2+3x+2)+cosx,2<x<\infty \end{array}$

We need to check at points $x=1$ and $x=2.$
Thus, $f^{\prime }\left(1^{+}\right)=\underset{h\to 0}{lim}(h+2)(h^2+5h+6)=12$
$f^{\prime }\left(1^{-}\right)=\underset{h\to 0}{lim}(-h+2)(h^2-5h+6)=12$
$f^{\prime }\left(2^{+}\right)=\underset{h\to 0}{lim}\frac{(h^2+4h+3)(h^2+7h+12)}{h}$ which $\to \infty$
$f^{\prime }\left(2^{-}\right)=\underset{h\to 0}{lim}\frac{(h^2-4h+3)(h^2-7h+12)}{-h}$ which $\to -\infty$
Hence, $h\left(x\right)$ isn't differentiable at $x=2.$