Question

The function $f\left( x \right) = \left\{ \begin{gathered}
  \dfrac{\pi }{4} + {\tan ^{ - 1}}x,\left| x \right| \leqslant 1 \\
  \dfrac{1}{2}\left( {\left| x \right| - 1} \right),\left| x \right| > 1 \\
\end{gathered} \right.$ is:
1. both continuous and differentiable on $R - \left\{ { - 1} \right\}$
2. continuous on $R - \left\{ { - 1} \right\}$and differentiable on $R - \left\{ { - 1,1} \right\}$
3. continuous on $R - \left\{ 1 \right\}$and differentiable on $R - \left\{ { - 1,1} \right\}$
4. both continuous and differentiable on $R - \left\{ 1 \right\}$

Answer 1

Hint: In this question, we are given a function $f\left( x \right) = \left\{ \begin{gathered}
  \dfrac{\pi }{4} + {\tan ^{ - 1}}x,\left| x \right| \leqslant 1 \\
  \dfrac{1}{2}\left( {\left| x \right| - 1} \right),\left| x \right| > 1 \\
\end{gathered} \right.$ and we have to check on which interval or point the function is continuous and differentiable. First step is to open the interval in the function then check the function is continuous at $x = 1, - 1$ or not by proving $f\left( 1 \right) = f\left( {{1^ + }} \right)$ and $f\left( { - 1} \right) = f\left( { - {1^ - }} \right)$. Similarly for differentiability, first differentiate the given function with respect to $x$ and check the condition same a continuity but this time using the first derivate.

Formula Used:
Differentiation formula –
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step Solution:
Given that,
$f\left( x \right) = \left\{ \begin{gathered}
  \dfrac{\pi }{4} + {\tan ^{ - 1}}x,\left| x \right| \leqslant 1 \\
  \dfrac{1}{2}\left( {\left| x \right| - 1} \right),\left| x \right| > 1 \\
\end{gathered} \right.$
It can also be written as,
$f\left( x \right) = \left\{ \begin{gathered}
  \dfrac{\pi }{4} + {\tan ^{ - 1}}x,x \in \left[ { - 1,1} \right] \\
  \dfrac{1}{2}\left( {x - 1} \right),x > 1 \\
  \dfrac{1}{2}\left( { - x - 1} \right),x < - 1 \\
\end{gathered} \right.$
For continuity,
At $x = 1$,
$f\left( 1 \right) = \dfrac{\pi }{4} + {\tan ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{4} + {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) = \dfrac{\pi }{4} + \dfrac{\pi }{4}$
$ \Rightarrow f\left( 1 \right) = \dfrac{\pi }{2}$
And $f\left( {{1^ + }} \right) = \dfrac{1}{2}\left( {1 - 1} \right) = 0$
$f\left( 1 \right) \ne f\left( {{1^ + }} \right)$
$\therefore f\left( x \right)$ is not continuous at $x = 1$
$ \Rightarrow $$f\left( x \right)$ is not differentiable at $x = 1$
Now at $x = - 1$,
$f\left( { - 1} \right) = \dfrac{\pi }{4} + {\tan ^{ - 1}}\left( { - 1} \right) = \dfrac{\pi }{4} + {\tan ^{ - 1}}\left( {\tan \left( { - \dfrac{\pi }{4}} \right)} \right) = \dfrac{\pi }{4} - \dfrac{\pi }{4}$
$ \Rightarrow f\left( { - 1} \right) = 0$
And $f\left( { - {1^ - }} \right) = \dfrac{1}{2}\left( { - \left( { - 1} \right) - 1} \right) = 0$
$f\left( { - 1} \right) = f\left( { - {1^ - }} \right) = 0$
$\therefore f\left( x \right)$ is continuous at $x = - 1$
For differentiability,
Differentiate the given function with respect to $x$,
$f'\left( x \right) = \left\{ \begin{gathered}
  \dfrac{1}{{1 + {x^2}}},x \in \left[ { - 1,1} \right] \\
  \dfrac{1}{2},x > 1 \\
  \dfrac{{ - 1}}{2},x < - 1 \\
\end{gathered} \right.$
At $x = - 1$.
$f'\left( { - 1} \right) = \dfrac{1}{{1 + {{\left( { - 1} \right)}^2}}} = \dfrac{1}{2}$
And $f'\left( { - {1^ - }} \right) = - \dfrac{1}{2}$
$f'\left( { - 1} \right) \ne f'\left( { - {1^ - }} \right)$
$\therefore f\left( x \right)$ is not differentiable at $x = - 1$

Hence, the correct option is 3.

Note: The branch points of the function are the points at which the value of the function changes. A function's differentiability and continuity in an interval are only checked at its branch points. Students should remember that the value of $\left| x \right|$ is determined by the value of $x$ contained within the modulus; if $x > 0$, it is positive; if $x < 0$, it is negative; therefore, don't assume its value is always positive. Also, when defining the function of $x$ for various intervals of $x$; most mistakes are made here. Here, in the option read carefully it is written that continuous at$R - \left\{ 1 \right\}$ it means function is continuous except at $x = 1$. So, don’t get confused about this.