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# A function f is defined as follows: $f(x) =$$\left\{ \begin{gathered} 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for - \infty < x < 0 \\ 1 + \sin x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;0 \leqslant x < \dfrac{\pi }{2} \\ 2 + {(x - \dfrac{\pi }{2})^2}\;\;\;\;\;\;\;\;\;\;for\dfrac{\pi }{2} \leqslant x < + \infty \\ \end{gathered} \right.$Discuss the continuity and differentiability at $x = 0$ and $x = \dfrac{\pi }{2}$This question has multiple correct optionsContinuous but not differentiable at $x = 0$Differentiable and continuous at $x = \dfrac{\pi }{2}$Neither continuous nor differentiable at $x = 0$Continuous but not differentiable at $x = \dfrac{\pi }{2}$

Last updated date: 08th Sep 2024
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Hint: A function f(x) is said to be continuous at $x = a$ if $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$
That is,
Left hand limit $= {\text{ }}f\left( a \right){\text{ }} =$ Right hand limit
$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)$.
If $f'({a^ + }) = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(a + h) - f(a)}}{h}$ is the right hand derivative at $x = a$and,
$f'({a^ - }) = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(a - h) - f(a)}}{{ - h}}$ is the left hand derivative at $x = a$ then,
The function f(x) is said to be differentiable at $x = a$ if $f'({a^ + }) = f'({a^ - })$

Let us check continuity at $x = 0$
Now as per question, for x→0
Left hand limit = $\mathop {\lim }\limits_{x \to 0 - h} f(x) = \mathop {\lim }\limits_{x \to 0 - h} 1$
$\Rightarrow$$\mathop {\lim }\limits_{h \to 0} (1) = 1$

Right hand limit =
$\begin{gathered} \mathop {\lim }\limits_{x \to 0 + h} f(x) = \mathop {\lim }\limits_{x \to 0 + h} 1 + \sin (x) \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} 1 + \sin (x) = \mathop {\lim }\limits_{h \to 0} [1 + \sin (0 + h)] \\ \Rightarrow \mathop {\lim }\limits_{h \to 0} [1 + \sin (0 + h)] = \mathop {\lim }\limits_{h \to 0} [1 + \sin 0\cos (h) + \cos 0\sin (h)] \\ \therefore \mathop {\lim }\limits_{h \to 0} [1 + \sin (0 + h)] = 1 + 0 = 1 \\ \end{gathered}$
Also at $x = 0$
f(x) $= 1 + \sin 0 = 1$
Hence clearly we see that left hand limit = right hand limit =f (0)
Therefore, Function is continuous at $x = 0$
Now let us check differentiability at $x = 0$
For $x = {0^ - }$ , $f'(x) = \dfrac{{d(1)}}{{dx}} = 0$
And at $x = {0^ + }$, $f'(x) = \dfrac{{d(1 + \sin x)}}{{dx}} = \cos (x) = \cos 0 = 1$
Clearly we see that $f'({0^ - }) \ne f'({0^ + })$
Hence f(x) is not differentiable at $x = 0$

Now let us check continuity at $x = \dfrac{\pi }{2}$
Now as per question, for $x \to \dfrac{\pi }{2}$
Left hand limit = $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2} - h} f(x) = \mathop {\lim }\limits_{x \to \dfrac{\pi }{2} - h} [1 + \sin (x)]$
$\begin{gathered} \mathop { \Rightarrow \lim }\limits_{x \to \dfrac{\pi }{2} + h} 1 + \sin (x) = \mathop {\lim }\limits_{h \to 0} [1 + \sin (\dfrac{\pi }{2} + h)] \\ \Rightarrow \mathop {\lim }\limits_{h \to 0} [1 + \sin (\dfrac{\pi }{2} + h)] = \mathop {\lim }\limits_{h \to 0} [1 + \sin \dfrac{\pi }{2}\cos (h) + \cos \dfrac{\pi }{2}\sin (h)] \\ \therefore 1 + \sin \dfrac{\pi }{2}\cos (0) + \cos \dfrac{\pi }{2}\sin (0) = 1 + 1 + 0 = 2 \\ \end{gathered}$

Right hand limit =
$\begin{gathered} \mathop {\lim }\limits_{x \to \dfrac{\pi }{2} + h} f(x) = \mathop {\lim }\limits_{x \to \dfrac{\pi }{2} + h} 2 + {(x - \dfrac{\pi }{2})^2} \\ \Rightarrow \mathop {\lim }\limits_{x \to \dfrac{\pi }{2} + h} 2 + {(x - \dfrac{\pi }{2})^2} = \mathop {\lim }\limits_{h \to 0} [2 + {(\dfrac{\pi }{2} + h - \dfrac{\pi }{2})^2}] \\ \therefore \mathop {\lim }\limits_{h \to 0} [2 + {(\dfrac{\pi }{2} + h - \dfrac{\pi }{2})^2}] = 2 \\ \end{gathered}$
Also at $x = \dfrac{\pi }{2}$
f(x) $= 2 + {(x - \dfrac{\pi }{2})^2} = 2 + {(\dfrac{\pi }{2} - \dfrac{\pi }{2})^2} = 2$
Hence clearly we see that left hand limit = right hand limit =f($\dfrac{\pi }{2}$)
∴ Function is continuous at $x = \dfrac{\pi }{2}$
Now let us check differentiability at $x = \dfrac{\pi }{2}$
For $x = {\dfrac{\pi }{2}^ - }$ , $f'(x) = \dfrac{{d(1 + \sin x)}}{{dx}} = \cos x = \cos \dfrac{\pi }{2} = 0$
And at $x = {\dfrac{\pi }{2}^ + }$, $f'(x) = \dfrac{{d[2 + {{(x - \dfrac{\pi }{2})}^2}]}}{{dx}} = 0 + 2(x - \dfrac{\pi }{2}) = 0 + 2(\dfrac{\pi }{2} - \dfrac{\pi }{2}) = 0$
Clearly we see that $f'({\dfrac{\pi }{2}^ - }) = f'({\dfrac{\pi }{2}^ + })$
Hence f(x) is differentiable at $x = 0$

Therefore, f (x) is continuous but not differentiable at $x = 0$

And f(x) is continuous as well as differentiable at $x = \dfrac{\pi }{2}$
Hence option (A) and (B) are the correct options.

Note: If any function f(x) is differentiable at $x = a$ then it must be continuous at $x = a$ but the converse is not true.
A function f(x) is said to be continuous in an open interval (a, b), if f(x) is continuous at every point of the interval.