JEE Advanced 2021 · Paper 2 · Mathematics Q15

← JEE Advanced 2021 · Paper 2

MathematicsSingle correct+3 / -1Question 53 of 57

Passage

Paragraph

Let $\psi_1: [0, \infty) \to \mathbb R$ , $\psi_2: [0, \infty) \to \mathbb R$ , $f: [0, \infty) \to \mathbb R$ and $g: [0, \infty) \to \mathbb R$ be functions such that $f(0) = g(0) = 0$ ,
$\begin{aligned}\psi_1(x) &= e^{-x} + x,\quad x \ge 0,\\ \psi_2(x) &= x^2 - 2x - 2e^{-x} + 2,\quad x \ge 0,\\ f(x) &= \int_{-x}^{x}(|t| - t^2)e^{-t^2}\,dt,\quad x > 0\end{aligned}$
and
$g(x) = \int_0^{x^2}\sqrt t\,e^{-t}\,dt,\quad x > 0.$

Which of the following statements is TRUE?

f(\sqrt{\ln 3}) + g(\sqrt{\ln 3}) = \frac13

For every

x > 1

, there exists an

\alpha \in (1, x)

such that

\psi_1(x) = 1 + \alpha x

For every

x > 0

, there exists a

\beta \in (0, x)

such that

\psi_2(x) = 2x(\psi_1(\beta) - 1)

f

is an increasing function on the interval

\left[0, \frac32\right]

Answer and worked solution

Official answer: C

Simplify f and g.

The integrand of $f$ is even, so $f(x)=2\int_0^x(t-t^2)e^{-t^2}dt$ .
Substituting $t=u^2$ in $g$ gives $g(x)=2\int_0^xu^2e^{-u^2}du$ .
So $f+g=2\int_0^xte^{-t^2}dt=1-e^{-x^2}$ .

Check each option.

(A) $f+g$ at $\sqrt{\ln3}$ is $1-\frac13=\frac23$ , not $\frac13$ . False.
(B) $\alpha=\dfrac{\psi_1(x)-1}x=\dfrac{x+e^{-x}-1}x<1$ , so $\alpha\notin(1,x)$ . False.
(C) $\psi_2(0)=0$ and $\psi_2'(x)=2x-2+2e^{-x}=2(\psi_1(x)-1)$ . The mean value theorem gives $\psi_2(x)=x\psi_2'(\beta)=2x(\psi_1(\beta)-1)$ for some $\beta\in(0,x)$ . True.
(D) $f'(x)=2(x-x^2)e^{-x^2}<0$ for $x>1$ , so $f$ is not increasing on $\left[0,\frac32\right]$ . False.

(C)

Topics:Rolle's and mean value theorems Properties of definite integrals Monotonicity

← Previous question Next question →

JEE Advanced 2021 · Paper 2, Mathematics question 15