Question

A function f is defined as follows: $f(x) =$

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

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 =

\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.