Question

The function ${{\sin }^{-1}}(\cos x)$
(A) has infinite number of discontinuities
(B) has finite number of discontinuities
(C) has no discontinuities
(D) has infinite number of points at which the function is not differentiable

Answer 1

Hint: Use RHL and LHL and also Use RHD and LHD and Use $x\to 0$ you will get the solution.
Complete step-by-step answer:
  Let $f(x)={{\sin }^{-1}}(\cos x)$
So first of all,
A function$f(x)$is said to be continuous at $x=a$ if $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=f(a)$
i.e. L.H.L.$=$ R.H.L$=f(a)$ = value of the function at a i.e.$\underset{x\to a}{\mathop{\lim }}\,f(x)=f(a)$
If$f(x)$ is not continuous at $x=a$ , we say that$f(x)$ is discontinuous $x=a$.

So First we will solve Left hand limit(LHL) and Right hand limit(RHL)
However $f(x)$ does not tend to infinity, because it does not stay larger than the number we have chosen, but instead returns to zero. For a similar reason, $f(x)$ does not tend to minus infinity. So we cannot talk about the limit of this function as x tends to infinity.
So we took $x\to 0$,
So for LHL, Here we have took the limit for $x$ tending to ${{0}^{-1}}$,
$\begin{align}
  & \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{\sin }^{-1}}(\cos x)=\underset{h\to 0}{\mathop{\lim }}\,{{\sin }^{-1}}(\cos (0-h)) \\
 & \underset{h\to 0}{\mathop{\lim }}\,{{\sin }^{-1}}(\cos (-h)) \\
 & \\
\end{align}$……………….$(\cos (-h)=\cosh )$
Applying limit
$\begin{align}
  & {{\sin }^{-1}}(\cos (0)) \\
 & =\dfrac{\pi }{2} \\
\end{align}$……….. (1)
For RHL,
$\begin{align}
  & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{\sin }^{-1}}(\cos x)=\underset{h\to 0}{\mathop{\lim }}\,{{\sin }^{-1}}(\cos (0+h)) \\
 & \underset{h\to 0}{\mathop{\lim }}\,{{\sin }^{-1}}(\cos (h)) \\
 & \\
\end{align}$
Applying limit
$\begin{align}
  & {{\sin }^{-1}}(\cos (0)) \\
 & =\dfrac{\pi }{2} \\
\end{align}$………. (2)
So LHL=RHL=$f(0)=\dfrac{\pi }{2}$…………..(from(1) and (2))
So function is continuous.
Consider a continuous function $y=f(x)$,
$y=f(x)$at $x=a$.
So for a function to be differentiable if these limits exist. Then ${{f}^{'}}(a)$exists if and only if these one-sided derivatives exist and are equal.
So LHD and RHD of a function$f(x)$ at point$x=a$ is,
The right hand derivative of a function$f(x)$ at point$x=a$ is,
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(a+h)-f(a)}{h}={{f}^{'}}(a)$
And, the left hand derivative of the function$f(x)$ at point$x=a$ is,

$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(a)-f(a-h)}{h}={{f}^{'}}(a)$

So now applying RHD and LHD,
function$f(x)$ at point$x=0$ is,
$\begin{align}
  & f(x)={{\sin }^{-1}}(\cos x) \\
 & f(0)={{\sin }^{-1}}(\cos 0)={{\sin }^{-1}}(1)=\dfrac{\pi }{2} \\
\end{align}$
First We will find Right hand derivative(RHD)
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$
So at$x=0,$
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(h)-f(0)}{h}$
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(\cos (h))-\dfrac{\pi }{2}}{h}$
So we can see if we apply limit we get,
$\dfrac{{{\sin }^{-1}}(\cos (0))-\dfrac{\pi }{2}}{0}=\dfrac{\dfrac{\pi }{2}-\dfrac{\pi }{2}}{0}=\dfrac{0}{0}$
So for this, Using L ‘hopital Rule, It is given below
If \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{0}{0}or\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\pm \infty }{\pm \infty }\] then \[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\]
So using it in (1), We get
$\dfrac{-\sinh }{\sqrt{1-{{\cos }^{2}}h}}=\dfrac{-\sinh }{\sinh }=-1$
RHD$=-1$
So for LHD, We get,
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x)-f(x-h)}{h}$
function$f(x)$ at point$x=0$ is,
$\begin{align}
  & f(x)={{\sin }^{-1}}(\cos x) \\
 & f(0)={{\sin }^{-1}}(\cos 0)={{\sin }^{-1}}(1)=\dfrac{\pi }{2} \\
\end{align}$
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(0)-f(0-h)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(0)-f(-h)}{h}$
So applying,
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(0)-f(0-h)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(\cos 0)-{{\sin }^{-1}}(\cos (-h))}{h}$
So here$\cos (-h)=\cos (h)$, So replacing in above we get,
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(0)-f(0-h)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{-1}}(\cos 0)-{{\sin }^{-1}}(\cos (h))}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\pi }{2}-{{\sin }^{-1}}(\cos (h))}{h}$
So applying limit we get,
$\dfrac{\dfrac{\pi }{2}-{{\sin }^{-1}}(\cos (0))}{0}=\dfrac{\dfrac{\pi }{2}-\dfrac{\pi }{2}}{0}=\dfrac{0}{0}$
So Using L ‘hopital Rule we get,
$-\left( \dfrac{-\sinh }{\sqrt{1-{{\cos }^{2}}h}} \right)=\dfrac{\sinh }{\sinh }=1$
LHD$=1$
So LHD$\ne $RHD
Hence the function is not differentiable.
So Option (C) is correct.
Answer is Option(C)

Note: Be careful about the limits and points. We should remember how to start with the problem. Read first the Question and see the options thoroughly. While applying L ‘hopital Rule be careful of differentiating numerator and denominator. Don’t jumble yourself with the limits. Don’t get confused between RHD and RHD and LHL and LHL. We should know that when the LHD$\ne $RHD function is not differentiable and that the LHL$=$RHL function is continuous.