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

**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.

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