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JEE Advanced 2021 · Paper 2, Mathematics question 16

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Let ψ1:[0,∞)→R\psi_1: [0, \infty) \to \mathbb R, ψ2:[0,∞)→R\psi_2: [0, \infty) \to \mathbb R, f:[0,∞)→Rf: [0, \infty) \to \mathbb R and g:[0,∞)→Rg: [0, \infty) \to \mathbb R be functions such that f(0)=g(0)=0f(0) = g(0) = 0,
ψ1(x)=e−x+x,x≥0,ψ2(x)=x2−2x−2e−x+2,x≥0,f(x)=∫−xx(∣t∣−t2)e−t2 dt,x>0\begin{aligned}\psi_1(x) &= e^{-x} + x,\quad x \ge 0,\\ \psi_2(x) &= x^2 - 2x - 2e^{-x} + 2,\quad x \ge 0,\\ f(x) &= \int_{-x}^{x}(|t| - t^2)e^{-t^2}\,dt,\quad x > 0\end{aligned}
and
g(x)=∫0x2t e−t dt,x>0.g(x) = \int_0^{x^2}\sqrt t\,e^{-t}\,dt,\quad x > 0.

Which of the following statements is TRUE?

Choose one option
Answer and worked solution

Official answer: D

Check each option.

  • (A) ψ1′=1−e−x>0\psi_1'=1-e^{-x}>0 and ψ1(0)=1\psi_1(0)=1, so ψ1>1\psi_1>1 for x>0x>0. False.
  • (B) ψ2′=2(ψ1−1)>0\psi_2'=2(\psi_1-1)>0 and ψ2(0)=0\psi_2(0)=0, so ψ2>0\psi_2>0. False.
  • (C) f=1−e−x2−2∫0xt2e−t2dtf=1-e^{-x^2}-2\int_0^xt^2e^{-t^2}dt. Since e−t2≥1−t2e^{-t^2}\ge1-t^2, the integral term is at least 23x3−25x5\frac23x^3-\frac25x^5. That gives f≤1−e−x2−23x3+25x5f\le1-e^{-x^2}-\frac23x^3+\frac25x^5: the reverse inequality. False.
  • (D) e−u2≤1−u2+u42e^{-u^2}\le1-u^2+\frac{u^4}2, so
    g=2∫0xu2e−u2du≤2(x33−x55+x714)=23x3−25x5+17x7.g=2\int_0^xu^2e^{-u^2}du\le2\left(\frac{x^3}3-\frac{x^5}5+\frac{x^7}{14}\right)=\frac23x^3-\frac25x^5+\frac17x^7.
    True.

(D)

Topics:MonotonicityProperties of definite integrals
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