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
Verified664.5k+ views
Hint: We will split the function f(x) in four integrals as, $f\left( x \right)=\left\{ \begin{align}
& -x+\dfrac{1}{2}-x+1+\tan x;0\le x\le \dfrac{1}{2} \\
& x-\dfrac{1}{2}-x+1+\tan x;\dfrac{1}{2}\le x\le 1 \\
& x-\dfrac{1}{2}+x-1+\tan x;1\le x\le \dfrac{\pi }{2} \\
& x-\dfrac{1}{2}+x-1+\tan x;\dfrac{\pi }{2}\le x\le 2 \\
\end{align} \right.$
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,
$f\left( x \right)=\left\{ \begin{align}
& -x+\dfrac{1}{2}-x+1+\tan x;0\le x\le \dfrac{1}{2} \\
& x-\dfrac{1}{2}-x+1+\tan x;\dfrac{1}{2}\le x\le 1 \\
& x-\dfrac{1}{2}+x-1+\tan x;1\le x\le \dfrac{\pi }{2} \\
& x-\dfrac{1}{2}+x-1+\tan x;\dfrac{\pi }{2}\le x\le 2 \\
\end{align} \right.$
On further solving f(x) in different intervals, we get,
$f\left( x \right)=\left\{ \begin{align}
& \dfrac{3}{2}-2x+\tan x;0\le x\le \dfrac{1}{2} \\
& \dfrac{1}{2}+\tan x;\dfrac{1}{2}\le x\le 1 \\
& 2x-\dfrac{3}{2}+\tan x;1\le x\le \dfrac{\pi }{2} \\
& 2x-\dfrac{3}{2}+\tan x;\dfrac{\pi }{2}\le x\le 2 \\
\end{align} \right.$
Now, we will check the differentiability of f(x) for different values of x.
We have,
$\begin{align}
& LHD=\underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=\dfrac{3}{2}-2x+\tan x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=-2+{{\sec }^{2}}x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=-2+{{\sec }^{2}}\left( \dfrac{1}{2} \right) \\
\end{align}$
And we have,
$\begin{align}
& RHD=\underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=\dfrac{1}{2}+\tan x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=0+{{\sec }^{2}}x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,={{\sec }^{2}}\left( \dfrac{1}{2} \right) \\
\end{align}$
Thus, we get,
$LHD\ne RHD$
Thus, the function f(x) is not differentiable at $x=\dfrac{1}{2}$.
Now, we will consider,
\[\begin{align}
& LHD=f\left( {{1}^{-}} \right)=\dfrac{1}{2}+\tan x \\
& f’\left( {{1}^{-}} \right)={{\sec }^{2}}x \\
& f’\left( {{1}^{-}} \right)={{\sec }^{2}}\left( 1 \right) \\
& RHD=f\left( {{1}^{+}} \right)=2x-\dfrac{3}{2}+\tan x \\
& f'\left( {{1}^{+}} \right)=2+{{\sec }^{2}}x \\
& f'\left( {{1}^{+}} \right)=2+{{\sec }^{2}}\left( 1 \right) \\
& So,LHD\ne RHD \\
\end{align}\]
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.
& -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{align} \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,
$f\left( x \right)=\left\{ \begin{align}
& -x+\dfrac{1}{2}-x+1+\tan x;0\le x\le \dfrac{1}{2} \\
& x-\dfrac{1}{2}-x+1+\tan x;\dfrac{1}{2}\le x\le 1 \\
& x-\dfrac{1}{2}+x-1+\tan x;1\le x\le \dfrac{\pi }{2} \\
& x-\dfrac{1}{2}+x-1+\tan x;\dfrac{\pi }{2}\le x\le 2 \\
\end{align} \right.$
On further solving f(x) in different intervals, we get,
$f\left( x \right)=\left\{ \begin{align}
& \dfrac{3}{2}-2x+\tan x;0\le x\le \dfrac{1}{2} \\
& \dfrac{1}{2}+\tan x;\dfrac{1}{2}\le x\le 1 \\
& 2x-\dfrac{3}{2}+\tan x;1\le x\le \dfrac{\pi }{2} \\
& 2x-\dfrac{3}{2}+\tan x;\dfrac{\pi }{2}\le x\le 2 \\
\end{align} \right.$
Now, we will check the differentiability of f(x) for different values of x.
We have,
$\begin{align}
& LHD=\underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=\dfrac{3}{2}-2x+\tan x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=-2+{{\sec }^{2}}x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=-2+{{\sec }^{2}}\left( \dfrac{1}{2} \right) \\
\end{align}$
And we have,
$\begin{align}
& RHD=\underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=\dfrac{1}{2}+\tan x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,=0+{{\sec }^{2}}x \\
& \underset{x\to \dfrac{1}{2}}{\mathop{f\left( x \right)}}\,={{\sec }^{2}}\left( \dfrac{1}{2} \right) \\
\end{align}$
Thus, we get,
$LHD\ne RHD$
Thus, the function f(x) is not differentiable at $x=\dfrac{1}{2}$.
Now, we will consider,
\[\begin{align}
& LHD=f\left( {{1}^{-}} \right)=\dfrac{1}{2}+\tan x \\
& f’\left( {{1}^{-}} \right)={{\sec }^{2}}x \\
& f’\left( {{1}^{-}} \right)={{\sec }^{2}}\left( 1 \right) \\
& RHD=f\left( {{1}^{+}} \right)=2x-\dfrac{3}{2}+\tan x \\
& f'\left( {{1}^{+}} \right)=2+{{\sec }^{2}}x \\
& f'\left( {{1}^{+}} \right)=2+{{\sec }^{2}}\left( 1 \right) \\
& So,LHD\ne RHD \\
\end{align}\]
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.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE
Draw a labelled diagram of the human male reproductive class 12 biology CBSE
Describe the experiment to compare the emf of two cells class 12 physics CBSE
What is standard hydrogen electrode
What is conventional current and electric current class 12 physics CBSE
2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE
Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE
Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?