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 $| x | = \begin{cases} -x & \quad \text{if } x < 0 \\ x & \quad \text{if } x \ge 0 \end{cases}$
Also, $\left|x^{2} - 3x+2\right| = \left|\left(x-1\right)\left(x-2\right)\right| = \begin{cases} (1-x) (2-x ) & \quad \text{if } x < 1 \\ (x - 1)(2-x) & \quad \text{if } 1 \le x < 2 \\ (x-1)(x-2) & \quad \text{if } x \ge 2 \end{cases} $
As $\cos (- \theta ) = \cos \theta \Rightarrow \cos |x| = \cos x$
$\therefore $ Given function can be written as $ f(x) = \begin{cases} (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{cases} $
This function is differentiable at all points except possibly at $x = 1$ and $x = 2$.
$\text{Lf}'\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}'\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 $
$\because \text{Lf}'\left(1\right) = \text{Rf}' \left(1\right)$
$ \therefore $ f is differentiable at $x = 1$
$\text{Lf}'\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 - \sin2$
$ \text{Rf}'\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 -\sin2$
$ \because \text{Lf}'\left(2\right) = \text{Rf}' \left(2\right)$
$ \therefore $ f is not differentiable at $x = 2$.