UPSC NDA I 2016 Mathematics Solved Paper

Concept:A function is differentiable at a point if the left-hand derivative (LHD) equals the right-hand derivative (RHD).If LHD ≠ RHD, the function is not differentiable at that point.Explanation:The given function is:$f(x) = \begin{cases} x+\pi & \text{for } x \in [-\pi, 0) \\ \pi \cos x & \text{for } x \in [0, \frac{\pi}{2}] \\ (x-\frac{\pi}{2})^2 & \text{for } x \in (\frac{\pi}{2}, \pi] \end{cases}$Check at $x=0$:LHD: $\displaystyle \lim_{h\to 0^-} \frac{f(0-h)-f(0)}{-h} = \lim_{h\to 0^-} \frac{(0-h+\pi) - \pi}{-h} = \lim_{h\to 0^-} \frac{-h}{-h} = 1$RHD: $\displaystyle \lim_{h\to 0^+} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0^+} \frac{\pi \cos(0+h) - \pi}{h} = \pi \lim_{h\to 0} \frac{\cos h - 1}{h} = \pi \cdot 0 = 0$Since LHD $= 1$ and RHD $= 0$, they are not equal.Thus $f(x)$ is not differentiable at $x = 0$.Check at $x = \frac{\pi}{2}$:RHD: $\displaystyle \lim_{h\to 0^+} \frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} = \lim_{h\to 0^+} \frac{(\frac{\pi}{2}+h-\frac{\pi}{2})^2 - 0}{h} = \lim_{h\to 0^+} \frac{h^2}{h} = 0$LHD: $\displaystyle \lim_{h\to 0^+} \frac{f(\frac{\pi}{2}-h)-f(\frac{\pi}{2})}{-h} = \lim_{h\to 0^+} \frac{\pi \cos(\frac{\pi}{2}-h) - 0}{-h} = \lim_{h\to 0^+} \frac{\pi \sin h}{-h} = -\pi \lim_{h\to 0} \frac{\sin h}{h} = -\pi$Since LHD $= -\pi$ and RHD $= 0$, they are not equal.Thus $f(x)$ is not differentiable at $x = \frac{\pi}{2}$.Therefore, both statement (1) and statement (2) are incorrect.