Question

Function $f\left(x\right)=\left|x-\left(\frac{1}{2}\right)\right|+|x-1|+\tan x$ how many points are not differentiable in $(0,2)$?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C

$3$

Explanation for the correct option:

Find the number of points:

$f\left(x\right)=\left|x-\left(\frac{1}{2}\right)\right|+|x-1|+\tan x$

Let $\left|x-\left(\frac{1}{2}\right)\right|=g\left(x\right)$

$|x-1|=h\left(x\right)$

and $\tan x=k\left(x\right)$

Now, At $x=\frac{1}{2}$, $g\left(x\right)=0$

$\Rightarrow$$g\left(x\right)$ is not differentiable at $x=\frac{1}{2}$

At $x=1$, $h\left(x\right)=0$

$\Rightarrow$$h\left(x\right)$ is not differentiable at $x=1$

At $x=\frac{\mathrm{\pi }}{2}$, $k\left(x\right)=0$

$\Rightarrow$$k\left(x\right)$ is not differentiable at $x=\frac{\mathrm{\pi }}{2}$

$\therefore$The number of points at which the given function is not differentiable is $\mathbf{3}$

Hence, Option ‘C’ is Correct.