UPSC NDA II 2015 Mathematics Paper

Concept:A function is differentiable at $x = a$ if the left-hand derivative and right-hand derivative at that point are equal and finite.Explanation:For function 1: $f(x) = \frac{1}{x}$ for $x \neq 0$, $f(0) = 0$.Compute the derivative limit: $\lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{1/h - 0}{h} = \lim_{h \to 0} \frac{1}{h^2}$.This limit is infinite, hence not finite. So the derivative does not exist at $x = 0$.Thus function 1 is not differentiable at $x = 0$.For function 2: $f(x) = 2x+5$ for $x > 0$, $f(x) = x^2 + 2x + 5$ for $x \le 0$, so $f(0) = 5$.Left-hand derivative at $0$: $\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{(x^2+2x+5) - 5}{x} = \lim_{x \to 0^-} (x+2) = 2$.Right-hand derivative at $0$: $\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{(2x+5) - 5}{x} = \lim_{x \to 0^+} 2 = 2$.Both LHD and RHD are equal to $2$, a finite number. Hence function 2 is differentiable at $x = 0$.Answer:Only function 2 is derivable at $x = 0$. So the correct option is B (2 only).