JEE Advanced 2021 Paper 2 · Q16 · Definite Integrals
Which of the following statements is TRUE?
Reveal answer + step-by-step solution
Correct answer:D
Solution
Using the series $e^{-t} = 1 - t + \dfrac{t^2}{2} - \dfrac{t^3}{6} + \cdots$, we get $\sqrt{t}\,e^{-t} \le t^{1/2} - t^{3/2} + \dfrac{t^{5/2}}{2}$ for small $t$ (alternating series upper bound). Actually a cleaner upper bound: $\sqrt{t}\,e^{-t} \le \sqrt{t}(1 - t + \tfrac{t^2}{2})$. Integrating on $[0, x^2]$: $g(x) = \int_0^{x^2}\sqrt{t}\,e^{-t}\,dt \le \dfrac{2}{3}x^3 - \dfrac{2}{5}x^5 + \dfrac{1}{7}x^7$ for $x\in(0,1/2)$. Option (D).
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