Let $\psi_{1}:[0, \infty) \rightarrow \mathbb{R}, \psi_{2}:[0, \infty) \rightarrow \mathbb{R}, f:[0, \infty) \rightarrow \mathbb{R}$ and $\mathrm{g}:[0, \infty) \rightarrow \mathbb{R}$ be functions such that $$ \begin{aligned} & f(0)=g(0)=0 \\ & \psi_{1}(x)=e^{-x}+x, \quad x \geq 0 \\ & \psi_{2}(x)=x^{2}-2 x-2 e^{-x}+2, x \geq 0 \\ & f(x)=\int_{-x}^{x}\left(|t|-t^{2}\right) e^{-t^{2}} d t, x>0 \end{aligned} $$ and $$ g(x)=\int_{0}^{x^{2}} \sqrt{t} e^{-t} d t, x>0 $$ Which of the following statements is TRUE ?

(C) For every $\mathrm{x}>0$, there exists a $\beta \in(0, \mathrm{x})$ such that $\Psi_{2}(\mathrm{x})=2 \mathrm{x}\left(\psi_{1}(\beta)-1\right)$

(A) $f(\sqrt{\ln 3})+g(\sqrt{\ln 3})=\frac{1}{3}$

(B) For every $\mathrm{x}>1$, there exists an $\alpha \in(1, \mathrm{x})$ such that $\psi_{1}(\mathrm{x})=1+\alpha \mathrm{x}$

(D) $f$ is an increasing function on the interval $\left[0, \frac{3}{2}\right]$

Application of Derivatives — JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 Question with Solution

Let

ψ_{1} : [0, \infty) \to R, ψ_{2} : [0, \infty) \to R, f : [0, \infty) \to R

and

g : [0, \infty) \to R

be functions such that

f (0) = g (0) = 0 ψ_{1} (x) = e^{- x} + x, x \geq 0 ψ_{2} (x) = x^{2} - 2 x - 2 e^{- x} + 2, x \geq 0 f (x) = \int_{- x}^{x} (∣ t ∣ - t^{2}) e^{- t^{2}} d t, x > 0

and

g (x) = \int_{0}^{x^{2}} t e^{- t} d t, x > 0

Which of the following statements is TRUE ?

(A) $f (ln 3) + g (ln 3) = \frac{1}{3}$
(B)For every $x > 1$ , there exists an $α \in (1, x)$ such that $ψ_{1} (x) = 1 + α x$
(C)For every $x > 0$ , there exists a $β \in (0, x)$ such that $Ψ_{2} (x) = 2 x (ψ_{1} (β) - 1)$ ✓ Correct
(D) $f$ is an increasing function on the interval $[0, \frac{3}{2}]$

Answer: (C)

First, note that

f^{'} (x) = 2 (x - x^{2}) e^{- x^{2}}

for

x \geq 0

. This is positive for

0 < x < 1

and negative for

x > 1

, so (D) is false.

For (A):

f^{'} (x) + g^{'} (x) = 2 x e^{- x^{2}}

, so

f (x) + g (x) = 1 - e^{- x^{2}}

. At

x = ln 3

f (ln 3) + g (ln 3) = 1 - e^{- l n 3} = 1 - \frac{1}{3} = \frac{2}{3}

. So (A) is false.

For (B): Let

H (x) = ψ_{1} (x) - 1 - α x = e^{- x} + x - 1 - α x

for

x \geq 1

and

α \in (1, x)

. We have

H (1) = e^{- 1} - α < 0

and

H^{'} (x) = - e^{- x} + 1 - α

. For

x > 1

e^{- x} < e^{- 1} < 1

, so if

α > 1 - e^{- 1}

, then

H^{'} (x) < 0

and H is decreasing. This doesn't guarantee a root. (B) is false.

For (C): By the Mean Value Theorem applied to

ψ_{2}

[0, x]

ψ_{2}^{'} (β) = \frac{ψ _{2} ( x ) - ψ _{2} ( 0 )}{x}

2 β - 2 + 2 e^{- β} = \frac{ψ _{2} ( x )}{x}

2 (ψ_{1} (β) - 1) = \frac{ψ _{2} ( x )}{x}

ψ_{2} (x) = 2 x (ψ_{1} (β) - 1)

This is exactly the MVT statement, which is true. (C) is correct.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Single correct option