Question

The number of points at which the function $f\left(x\right)=|x-0.5|+|x-1|+\tan x$ does not have a derivative in the interval $(0,2)$ is/are?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

The given function can be defined as $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{3}{2}-2x+\tan x;& 0\le x\le \frac{1}{2}\end{array}\\ \begin{array}{cc}\frac{1}{2}+\tan x;& \frac{1}{2}\le x\le 1\end{array}\\ \begin{array}{cc}2x-\frac{3}{2}+\tan x& 1\le x<\frac{\pi }{2}\end{array}\\ \begin{array}{cc}2x-\frac{3}{2}+\tan x& \frac{\pi }{2}<x<2\end{array}\end{array}$

$L.H.D.\Rightarrow f^{\prime }\left(\frac{1}{2}\right)=-2+{\sec }^2\left(\frac{1}{2}\right)$

$R.H.D.\Rightarrow f^{\prime }\left(\frac{1}{2}\right)={\sec }^2\left(\frac{1}{2}\right)\ne L{f}^{\prime }\left(\frac{1}{2}\right)$

$\therefore f\left(x\right)$ is not differendiable at $x=\frac{1}{2}$

and $L{f}^{\prime }\left(1\right)={\sec }^{2}1$

$R{f}^{\prime }\left(1\right)=2+{\sec }^{2}1\ne L{f}^{\prime }\left(1\right)$

$\therefore f\left(x\right)$ is also non differentiable at $x=1$

Finally as $f\left(x\right)$ is not defined at $x=\frac{\pi }{2},$

So $f\left(x\right)$ is discontinovs at $x=\frac{\pi }{2}$ and so non-differentiable at $x=\frac{\pi }{2}$.

Here there are $3$ points at which $f\left(x\right)$ does not have derivative in $(0,2)$.

which given option (C)