Question

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 options
Continuous 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 \[x = \dfrac{\pi }{2}\]

Answer 1

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 \[f'({a^ + }) = f'({a^ - })\]

Complete step-by-step answer:
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.