Question

The number of points at which the function $f\left( x \right)=\left| x-0.5 \right|+\left| x-1 \right|+\tan x$ does not have a derivative in the interval (0,2) is/are?
a. 1
b. 2
c. 3
d. 4

Answer 1

We will split the function f(x) in four integrals as, f\left( x \right)=\left\\{ \begin{aligned} & -x+\dfrac{1}{2}-x+1+\tan x;0\le x\le \dfrac{1}{2} \\\ & x-\dfrac{1}{2}-x+1+\tan x;\dfrac{1}{2}\le x\le 1 \\\ & x-\dfrac{1}{2}+x-1+\tan x;1\le x\le \dfrac{\pi }{2} \\\ & x-\dfrac{1}{2}+x-1+\tan x;\dfrac{\pi }{2}\le x\le 2 \\\ \end{aligned} \right.
We will then check the value of f(x) in different intervals by differentiating f(x) with respect to x and then finally compare both the sides. If both the sides are equal, then the function is differentiable in (0,2) at that point and if not then the function is not differentiable at that point.

Complete step-by-step answer :
It is given in the question that we have to find the number of points at which the function $f\left( x \right)=\left| x-0.5 \right|+\left| x-1 \right|+\tan x$ does not have a derivative in the interval (0,2).
We have $f\left( x \right)=\left| x-0.5 \right|+\left| x-1 \right|+\tan x$. We can break the interval of (0,2) into four parts as, $0\le x\le \dfrac{1}{2},\dfrac{1}{2}\le x\le 1,1\le x\le \dfrac{\pi }{2},\dfrac{\pi }{2}\le x\le 2$. This is because, we have $f\left( x \right)=\left| x-0.5 \right|+\left| x-1 \right|+\tan x$, where, in $\left| x-0.5 \right|$, x cannot be $\dfrac{1}{2}$, then in $\left| x-1 \right|$, x cannot be 1 and in tan x, x cannot be $\dfrac{\pi }{2}$.
Now, if we put any value of x from $\left( 0to\dfrac{1}{2} \right)$ in function, we get,
f\left( x \right)=\left\\{ -x+\dfrac{1}{2}-x+1+\tan x \right\\}
Now, if we put any value of x from $\left( \dfrac{1}{2}to1 \right)$, we get,
f\left( x \right)=\left\\{ x-\dfrac{1}{2}-x-1+\tan x \right\\}
Now, if we put any value of x from $\left( 1to\dfrac{\pi }{2} \right)$, we get,
f\left( x \right)=\left\\{ x-\dfrac{1}{2}+x-1+\tan x \right\\}
Similarly if we put any value of x from $\left( \dfrac{\pi }{2}to2 \right)$, we get,
f\left( x \right)=\left\\{ x-\dfrac{1}{2}+x-1+\tan x \right\\}
So, we can from the above observations, we can write,
f\left( x \right)=\left\\{ \begin{aligned} & -x+\dfrac{1}{2}-x+1+\tan x;0\le x\le \dfrac{1}{2} \\\ & x-\dfrac{1}{2}-x+1+\tan x;\dfrac{1}{2}\le x\le 1 \\\ & x-\dfrac{1}{2}+x-1+\tan x;1\le x\le \dfrac{\pi }{2} \\\ & x-\dfrac{1}{2}+x-1+\tan x;\dfrac{\pi }{2}\le x\le 2 \\\ \end{aligned} \right.
On further solving f(x) in different intervals, we get,
f\left( x \right)=\left\\{ \begin{aligned} & \dfrac{3}{2}-2x+\tan x;0\le x\le \dfrac{1}{2} \\\ & \dfrac{1}{2}+\tan x;\dfrac{1}{2}\le x\le 1 \\\ & 2x-\dfrac{3}{2}+\tan x;1\le x\le \dfrac{\pi }{2} \\\ & 2x-\dfrac{3}{2}+\tan x;\dfrac{\pi }{2}\le x\le 2 \\\ \end{aligned} \right.
Now, we will check the differentiability of f(x) for different values of x.
We have,
$\begin{aligned} & LHD=\underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=\dfrac{3}{2}-2x+\tan x \\\ & \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=-2+{{\sec }^{2}}x \\\ & \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=-2+{{\sec }^{2}}\left( \dfrac{1}{2} \right) \\\ \end{aligned}$
And we have,
$\begin{aligned} & RHD=\underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=\dfrac{1}{2}+\tan x \\\ & \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=0+{{\sec }^{2}}x \\\ & \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,={{\sec }^{2}}\left( \dfrac{1}{2} \right) \\\ \end{aligned}$
Thus, we get,
$LHD\ne RHD$
Thus, the function f(x) is not differentiable at $x=\dfrac{1}{2}$.
Now, we will consider,

& LHD=f\left( {{1}^{-}} \right)=\dfrac{1}{2}+\tan x \\\ & f’\left( {{1}^{-}} \right)={{\sec }^{2}}x \\\ & f’\left( {{1}^{-}} \right)={{\sec }^{2}}\left( 1 \right) \\\ & RHD=f\left( {{1}^{+}} \right)=2x-\dfrac{3}{2}+\tan x \\\ & f'\left( {{1}^{+}} \right)=2+{{\sec }^{2}}x \\\ & f'\left( {{1}^{+}} \right)=2+{{\sec }^{2}}\left( 1 \right) \\\ & So,LHD\ne RHD \\\ \end{aligned}$$ Thus, f(x) is not differentiable at x = 1. As $\tan \dfrac{\pi }{2}$ is not defined so, we will not check the function differentiability at $\dfrac{\pi }{2}$ because the function is not defined at $x=\dfrac{\pi }{2}$. Thus, f(x) is discontinuous at $\dfrac{\pi }{2}$. So, f(x) is not differentiable at $x=\dfrac{\pi }{2}$. Therefore, there are 3 points at which f(x) do not have derivative in interval (0,2). So, option (c) is the correct answer. **Note** :Most of the students make mistakes in opening the modulus. They may directly open the modulus assuming that all values will be positive and exactly same, as a result they may get $f\left( x \right)=\left\\{ x+\dfrac{1}{2}+x+1+\tan x \right\\}$ in all the four cases, which is not correct at all. We have to take care of signs while opening the modulus.