JEE Challenger
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MathHardAdv 2021 • Paper 2

Evaluation of Integrals and Mean Value Theorem on Functions

Comprehension Passage

Consider real-valued functions ψ1:[0,∞)→R\psi_1 : [0,\infty) \rightarrow \mathbb{R}, ψ2:[0,∞)→R\psi_2 : [0,\infty) \rightarrow \mathbb{R}, f:[0,∞)→Rf : [0,\infty) \rightarrow \mathbb{R} and g:[0,∞)→Rg : [0,\infty) \rightarrow \mathbb{R} satisfying f(0)=g(0)=0f(0) = g(0) = 0, defined as: ψ1(x)=e−x+x,x≥0,\psi_1(x) = e^{-x} + x, \quad x \ge 0, ψ2(x)=x2−2x−2e−x+2,x≥0,\psi_2(x) = x^2 - 2x - 2e^{-x} + 2, \quad x \ge 0, f(x)=∫−xx(∣t∣−t2)e−t2dt,x>0f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} dt, \quad x > 0 and g(x)=∫0x2te−tdt,x>0.g(x) = \int_0^{x^2} \sqrt{t} e^{-t} dt, \quad x > 0.

Which among the given statements is TRUE?

Options

A

f(ln⁡3)+g(ln⁡3)=13f(\sqrt{\ln 3}) + g(\sqrt{\ln 3}) = \frac{1}{3}

B

For every x>1x > 1, there exists some α∈(1,x)\alpha \in (1, x) satisfying ψ1(x)=1+αx\psi_1(x) = 1 + \alpha x

C

For every x>0x > 0, there exists β∈(0,x)\beta \in (0, x) such that ψ2(x)=2x(ψ1(β)−1)\psi_2(x) = 2x(\psi_1(\beta) - 1)

Correct
D

ff is an increasing function over the interval [0,32]\left[0, \frac{3}{2}\right]

Topics & Concepts

Application of DerivativesMonotonicityDefinite IntegrationProperties of Definite Integrals

Step-by-Step Solution

To determine which of the given statements is TRUE, let us analyze each option individually.

1. Analysis of Option (A):

The given functions are: f(x)=∫−xx(∣t∣−t2)e−t2dtfor x>0f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} dt \quad \text{for } x > 0 g(x)=∫0x2te−tdtfor x>0g(x) = \int_0^{x^2} \sqrt{t} e^{-t} dt \quad \text{for } x > 0

Since the integrand of f(x)f(x), namely (∣t∣−t2)e−t2(|t| - t^2) e^{-t^2}, is an even function of tt, we can rewrite f(x)f(x) as: f(x)=2∫0x(t−t2)e−t2dt=2∫0xte−t2dt−2∫0xt2e−t2dtf(x) = 2 \int_0^x (t - t^2) e^{-t^2} dt = 2 \int_0^x t e^{-t^2} dt - 2 \int_0^x t^2 e^{-t^2} dt

For g(x)g(x), substitute t=u2  ⟹  dt=2u dut = u^2 \implies dt = 2u\,du. As tt ranges from 00 to x2x^2, uu ranges from 00 to xx: g(x)=∫0xu2e−u2(2u du)=2∫0xu2e−u2du=2∫0xt2e−t2dtg(x) = \int_0^x \sqrt{u^2} e^{-u^2} (2u\,du) = 2 \int_0^x u^2 e^{-u^2} du = 2 \int_0^x t^2 e^{-t^2} dt

Adding f(x)f(x) and g(x)g(x): f(x)+g(x)=2∫0xte−t2dtf(x) + g(x) = 2 \int_0^x t e^{-t^2} dt

Evaluating this integral: f(x)+g(x)=[−e−t2]0x=1−e−x2f(x) + g(x) = \left[ -e^{-t^2} \right]_0^x = 1 - e^{-x^2}

For x=ln⁡3x = \sqrt{\ln 3}, we have x2=ln⁡3x^2 = \ln 3: f(ln⁡3)+g(ln⁡3)=1−e−ln⁡3=1−13=23f(\sqrt{\ln 3}) + g(\sqrt{\ln 3}) = 1 - e^{-\ln 3} = 1 - \frac{1}{3} = \frac{2}{3}

Since 23≠13\frac{2}{3} \neq \frac{1}{3}, Option (A) is FALSE.

2. Analysis of Option (B):

We are given ψ1(x)=e−x+x\psi_1(x) = e^{-x} + x. The equation ψ1(x)=1+αx\psi_1(x) = 1 + \alpha x gives: e−x+x=1+αx  ⟹  αx=e−x+x−1  ⟹  α=1+e−x−1xe^{-x} + x = 1 + \alpha x \implies \alpha x = e^{-x} + x - 1 \implies \alpha = 1 + \frac{e^{-x} - 1}{x}

For any x>1x > 1, we have e−x<1e^{-x} < 1, which implies e−x−1<0e^{-x} - 1 < 0. Thus, e−x−1x<0\frac{e^{-x} - 1}{x} < 0, leading to: α<1\alpha < 1

Therefore, there exists no α∈(1,x)\alpha \in (1, x) satisfying the equation. Hence, Option (B) is FALSE.

3. Analysis of Option (C):

Consider the function ψ2(x)=x2−2x−2e−x+2\psi_2(x) = x^2 - 2x - 2e^{-x} + 2 defined on the interval [0,x][0, x] for x>0x > 0.

  • ψ2(0)=02−2(0)−2e0+2=0\psi_2(0) = 0^2 - 2(0) - 2e^0 + 2 = 0.
  • ψ2(t)\psi_2(t) is continuous on [0,x][0, x] and differentiable on (0,x)(0, x).

By Lagrange's Mean Value Theorem (LMVT) applied to ψ2(t)\psi_2(t) on [0,x][0, x], there exists at least one β∈(0,x)\beta \in (0, x) such that: ψ2′(β)=ψ2(x)−ψ2(0)x−0=ψ2(x)x\psi_2'(\beta) = \frac{\psi_2(x) - \psi_2(0)}{x - 0} = \frac{\psi_2(x)}{x}

Differentiating ψ2(t)\psi_2(t) with respect to tt: ψ2′(t)=ddt(t2−2t−2e−t+2)=2t−2+2e−t=2(e−t+t−1)\psi_2'(t) = \frac{d}{dt}\left(t^2 - 2t - 2e^{-t} + 2\right) = 2t - 2 + 2e^{-t} = 2\left(e^{-t} + t - 1\right)

Since ψ1(t)=e−t+t\psi_1(t) = e^{-t} + t, we can express ψ2′(t)\psi_2'(t) as: ψ2′(t)=2(ψ1(t)−1)\psi_2'(t) = 2(\psi_1(t) - 1)

Evaluating at t=βt = \beta: ψ2′(β)=2(ψ1(β)−1)\psi_2'(\beta) = 2(\psi_1(\beta) - 1)

Equating the two expressions for ψ2′(β)\psi_2'(\beta): ψ2(x)x=2(ψ1(β)−1)  ⟹  ψ2(x)=2x(ψ1(β)−1)\frac{\psi_2(x)}{x} = 2(\psi_1(\beta) - 1) \implies \psi_2(x) = 2x(\psi_1(\beta) - 1)

Thus, for every x>0x > 0, there exists β∈(0,x)\beta \in (0, x) such that ψ2(x)=2x(ψ1(β)−1)\psi_2(x) = 2x(\psi_1(\beta) - 1). Hence, Option (C) is TRUE.

4. Analysis of Option (D):

Differentiating f(x)=2∫0x(t−t2)e−t2dtf(x) = 2 \int_0^x (t - t^2) e^{-t^2} dt using the Leibniz Rule: f′(x)=2(x−x2)e−x2=2x(1−x)e−x2f'(x) = 2(x - x^2) e^{-x^2} = 2x(1 - x) e^{-x^2}

For x∈(1,32]x \in \left(1, \frac{3}{2}\right], we have (1−x)<0(1 - x) < 0, which implies f′(x)<0f'(x) < 0. Since f′(x)<0f'(x) < 0 on (1,32]\left(1, \frac{3}{2}\right], f(x)f(x) is strictly decreasing on this sub-interval, and therefore ff is not an increasing function on [0,32]\left[0, \frac{3}{2}\right]. Hence, Option (D) is FALSE.

Conclusion:

The correct and TRUE statement is (C).

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