JEE Advanced 2021 — Paper 2 — Question 50

Expand $e^{-t} = 1 - t + \frac{t^2}{2!} - \frac{t^3}{3!} + \cdots$ (alternating series for $t>0$). For $0 < t < 1$, the series is alternating with decreasing terms, so $e^{-t} < 1 - t + \frac{t^2}{2}$. Multiply by $\sqrt{t}$: $\sqrt{t}\,e^{-t} < \sqrt{t} - t^{3/2} + \frac{1}{2}t^{5/2}$. Integrate from $0$ to $x^2$ (with $0 < x < \frac{1}{2}$): $g(x) = \int_0^{x^2} \sqrt{t}\,e^{-t}\,dt < \int_0^{x^2} \left(\sqrt{t} - t^{3/2} + \frac{1}{2}t^{5/2}\right)dt$. Compute integrals: $\int_0^{x^2} \sqrt{t}\,dt = \frac{2}{3}x^3$, $\int_0^{x^2} t^{3/2}\,dt = \frac{2}{5}x^5$, $\int_0^{x^2} \frac{1}{2}t^{5/2}\,dt = \frac{1}{7}x^7$. Thus $g(x) < \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7$, which implies $g(x) \le \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7$ for all $x \in (0,\frac{1}{2})$. Hence option D is true.