Mathematics - Continuity - Differentiability Question with Solution | TestHub

Here $\cos (|x|)=\cos (\pm x)=\cos x$.

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

$=\left(x^2-1\right)\left(x^2-3x+2\right)+\cos x,x\le 1orx\ge 2$

Clearly $f(1)=\cos 1,\underset{x\to 1}{\mathrm{Lt}}f(x)=\cos 1$ and $$f(2)=\cos 2,{\mathrm{Lt}}_{x\to 2}f(x)=\cos 2$$.

Hence $f(x)$ is continuous at $x=1,2$.

Now $f^{\prime }(x)=-2x\left(x^2-3x+2\right)-\left(x^2-1\right)(2x-3)-\sin x,1\le x<2$

$$=2x\left(x^2-3x+2\right)+\left(x^2-1\right)(2x-3)-\sin x,x<1\text{ or }x>2$$

$f^{\prime }(1-0)=-\sin 1,f^{\prime }(1+0)=-\sin 1$, $f^{\prime }(2-0)=-3-\sin 2$, $f^{\prime }(2+0)+3-\sin 2$.

Hence $f(x)$ is not differentiable at $x=2$.