NIMCET 2020 — Mathematics PYQ

Continuity of a Function:
- A function f(x) is said to be continuous at a point x = a in its domain, if lim f(x) exists or if its graph is a single unbroken curve at that point.
- f(x) is continuous at x = a ⇔ lim f(x) = lim f(x) = lim f(x) = f(a).

Differentiability of a Function:
- A function f(x) is differentiable at a point x = a in its domain if its derivative is continuous at a.
- This means that f'(a) must exist, or equivalently: lim f'(x) = lim f'(x) = lim f'(x) = f'(a).

Maxima/Minima:
- If f(x) has a local maximum or a local minimum at a point x = a, then it must be either a critical point [f'(a) = 0] or a point of non-differentiability.

Calculation
Let us check for the continuity and differentiability (maxima/minima) of the function at x = 0.
Continuity:
$f(x)=\{\begin{array}{llc}{x}^2;& amp;x\le 0& \\ 2\sin x;& amp;x& gt;0\end{array}$
${\lim }_{x\to 0}f(x)={\lim }_{x\to 0}{x}^2=0^2=0.$
${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{+}}2\sin x=2\sin 0=0.$
$f(0)=0^2=0.$

$\because {\lim }_{x\to 0}f(x)={\lim }_{x\to 0^{+}}f(x)=f(0)$, the function $f(x)$ is continuous at $x=0$.

Differentiability:

$f^{\prime }(x)=\{\begin{array}{llc}2x,& amp;x\le 0& \\ 2\cos x;& amp;x& gt;0\end{array}$

${\lim }_{x\to 0^{-}}{f}^{\prime }(x)={\lim }_{x\to 0}2x=2\times 0=0$.

${\lim }_{x\to 0^{+}}{f}^{\prime }(x)={\lim }_{x\to 0^{+}}2\cos x=2\cos 0=2$.

$\because {\lim }_{x\to 0^{-}}{f}^{\prime }(x)\ne {\lim }_{x\to 0^{+}}{f}^{\prime }(x)$, the function is not differentiable at $x=0$.

Since, the function is not differentiable at $x=0$, let us examine the possibility of maximum/minimum at the point.

The function $f(x)=x^2$ is strictly decreasing in $(-\infty ,0]$ and its minimum is $0^2=0$ at $x=0$.

The function $f(x)=2\sin x$ is strictly increasing in $\left(0,\frac{\pi }{2}\right]$ and its minimum is $2\sin 0=0$ at $x=0$.

$\therefore$ The function $f(x)$ has a local minimum at $x=0$.