UPSC NDA 2 2016 Mathematics Solved Paper

Concept:A function $f(x)$ is strictly increasing in a domain if its derivative $f'(x) > 0$ for all points in that domain.Explanation:We have $f(x) = \frac{e^x - 1}{x}$ for $x > 0$ and $f(0) = 0$. Only the part $x > 0$ matters for monotonicity.Differentiate using quotient rule:$f'(x) = \frac{x \cdot e^x - (e^x - 1) \cdot 1}{x^2} = \frac{x e^x - e^x + 1}{x^2} = \frac{(x-1)e^x + 1}{x^2}$.For $x > 0$, denominator $x^2 > 0$. So sign of $f'(x)$ depends on numerator $g(x) = (x-1)e^x + 1$.Differentiate $g(x)$: $g'(x) = (x-1)e^x + e^x = x e^x$. For $x > 0$, $g'(x) > 0$, so $g(x)$ is strictly increasing for $x > 0$.At $x = 0$, $g(0) = (0-1)(1) + 1 = 0$. Since $g(x)$ is increasing, for any $x > 0$, $g(x) > g(0) = 0$.Thus numerator $g(x) > 0$ for all $x > 0$, making $f'(x) > 0$ for $x > 0$. Hence $f(x)$ is strictly increasing in $(0, \infty)$.Answer:$f(x)$ is a strictly increasing function in $(0, x)$ (option B).