Question

Let f(x) = \left\\{ {x^2}\left| {\cos \dfrac{\pi }{2}} \right|;x \ne 0 \\\ 0;x = 0 \\\ \right. $x \in \mathbb{R}$,

$$then f is$$

A) Differentiable both at $x = 0$ and at $x = 2$.
B) Differentiable at $x = 0$ but not differentiable at $x = 2$.
C) Not differentiable at $x = 0$ but differentiable at $x = 2$.
D) Differentiable neither at $x = 0$ nor at $x = 2$.

Answer 1

Function $f$ is differentiable at any point $x$ if and only if left hand derivative is equal to right hand derivative. First check for $x = 0$ that left hand derivative is equal to right hand derivative or not, and then check for $x = 2$ similarly as $x = 0$. You can also use the graph method for differentiability.

Complete step-by-step answer:
For $x = 0$,$f$ is differentiable if and only if $LHD = RHD$ (left hand derivative = right hand derivative).
$LHD = f'({0^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 - h) - f(0)}}{{ - h}}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( { - h} \right)}^2}\left| {\cos \left( {\dfrac{\pi }{{ - h}}} \right)} \right| - 0}}{{ - h}}$ $= \mathop {\lim }\limits_{h \to {0^{}}} \left( { - h} \right)\left| {\cos \left( {\dfrac{\pi }{{ - h}}} \right)} \right| = \left( { - 0} \right).\left[ { - 1,1} \right] = 0$
$RHD = f'({0^ + }) = \mathop {\lim }\limits_{h \to {0^{}}} \dfrac{{f(0 + h) - f(0)}}{{ - h}}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{h^2}\left| {\cos \dfrac{\pi }{h}} \right| - 0}}{h}$ $= \mathop {\lim }\limits_{h \to 0} h\left| {\cos \dfrac{\pi }{h}} \right| = 0$
Hence $LHD = RHD$ and $f$ is differentiable at $x = 0$.
Similarly, we check for $x = 2$.
$LHD = f'({2^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(2 - h) - f(2)}}{{ - h}}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 - h)}^2}\left| {\cos \left( {\dfrac{\pi }{{2 - h}}} \right)} \right| - {2^2}\cos \dfrac{\pi }{2}}}{{ - h}}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 - h)}^2}\cos \left( {\dfrac{\pi }{{2 - h}}} \right) - 0}}{h}$, convert $\cos$ into $\sin$ form
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 - h)}^2}}}{h}.\sin \left[ {\dfrac{\pi }{2} - \dfrac{\pi }{{2 - h}}} \right]$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 - h)}^2}}}{h}.\sin \left[ {\dfrac{{ - h\pi }}{{2(2 - h)}}} \right]$, multiply and divide by $\dfrac{{ - \pi h}}{{2(2 - h)}}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 - h)}^2}}}{h} \times \dfrac{{ - \pi h}}{{2(2 - h)}}.\left[ {\dfrac{{\sin \left[ {\dfrac{{ - \pi h}}{{2(2 - h)}}} \right]}}{{\dfrac{{ - \pi h}}{{2(2 - h)}}}}} \right]$, as we know $\mathop {\lim }\limits_{a \to 0} \dfrac{{\sin a}}{a} = 1$
$LHD = - \pi$
$RHD = f'({2^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(2 + h) - f(2)}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 + h)}^2}\left| {\cos \left( {\dfrac{\pi }{{2 + h}}} \right)} \right| - {2^2}\cos \dfrac{\pi }{2}}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 + h)}^2}\cos \left( {\dfrac{\pi }{{2 + h}}} \right) - 0}}{h}$, convert $\cos$ into $\sin$ form
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 + h)}^2}}}{h}.\sin \left[ {\dfrac{\pi }{2} - \dfrac{\pi }{{2 + h}}} \right]$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 + h)}^2}}}{h}.\sin \left[ {\dfrac{{h\pi }}{{2(2 + h)}}} \right]$ , multiply and divide by $\dfrac{{\pi h}}{{2(2 + h)}}$
$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(2 + h)}^2}}}{h} \times \dfrac{{\pi h}}{{2(2 + h)}}.\left[ {\dfrac{{\sin \left[ {\dfrac{{\pi h}}{{2(2 + h)}}} \right]}}{{\dfrac{{\pi h}}{{2(2 + h)}}}}} \right]$, as we know $\mathop {\lim }\limits_{a \to 0} \dfrac{{\sin a}}{a} = 1$
$RHD = \pi$
Hence $LHD \ne RHD$, then $f$ is not differentiable at $x = 2$.

Then the correct answer is option B.

Note: Graph method for checking differentiability: A function $f$ is differentiable at $x = a$ whenever $f'(a)$ exists, which means that $f$ has a tangent line at $(a,f(a))$ and thus $f$ is locally linear at the value $x = a$. Informally, this means that the function looks like a line when viewed up close at $(a,f(a))$ and that there is not a corner point or cusp at $(a,f(a))$.To use graph first need to draw accurate graph of given function.