JEE Challenger
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MathMedAdv 2018 • Paper 1

Continuous Function Defined by Integral Equation and Enclosed Area

Let f:[0,∞)→Rf: [0, \infty) \to \mathbb{R} be a continuous function satisfying f(x)=1−2x+∫0xex−tf(t)dtf(x) = 1 - 2x + \int_0^x e^{x-t} f(t) dt for all x∈[0,∞)x \in [0, \infty). Then, which of the following statement(s) is (are) TRUE?

Options

A

The curve y=f(x)y = f(x) passes through the point (1,2)(1, 2)

B

The curve y=f(x)y = f(x) passes through the point (2,−1)(2, -1)

Correct
C

The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{(x, y) \in [0, 1] \times \mathbb{R} : f(x) \leq y \leq \sqrt{1 - x^2}\} is π−24\frac{\pi - 2}{4}

Correct
D

The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{(x, y) \in [0, 1] \times \mathbb{R} : f(x) \leq y \leq \sqrt{1 - x^2}\} is π−14\frac{\pi - 1}{4}

Topics & Concepts

Differential EquationsDifferential EquationsLinear Differential EquationsArea Under CurvesDefinite Integration

Step-by-Step Solution

To determine the correct statements, we first find the explicit form of the function f(x)f(x).

Step 1: Formulating and Solving the Differential Equation

The given integral equation is: f(x)=1−2x+∫0xex−tf(t) dtfor all x∈[0,∞)f(x) = 1 - 2x + \int_0^x e^{x-t} f(t) \, dt \quad \text{for all } x \in [0, \infty)

We can rewrite the integral by factoring out exe^x: f(x)=1−2x+ex∫0xe−tf(t) dtf(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) \, dt

Multiplying both sides by e−xe^{-x}: e−xf(x)=(1−2x)e−x+∫0xe−tf(t) dte^{-x} f(x) = (1 - 2x)e^{-x} + \int_0^x e^{-t} f(t) \, dt

Differentiating both sides with respect to xx using the product rule and Leibniz Rule: e−xf′(x)−e−xf(x)=−2e−x−(1−2x)e−x+e−xf(x)e^{-x} f'(x) - e^{-x} f(x) = -2e^{-x} - (1 - 2x)e^{-x} + e^{-x} f(x)

Since e−x≠0e^{-x} \neq 0 for all real xx, dividing the entire equation by e−xe^{-x} yields: f′(x)−f(x)=−3+2x+f(x)f'(x) - f(x) = -3 + 2x + f(x) f′(x)−2f(x)=2x−3f'(x) - 2f(x) = 2x - 3

This is a first-order linear differential equation of the form dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x) with: Integrating Factor (I.F.)=e∫−2 dx=e−2x\text{Integrating Factor (I.F.)} = e^{\int -2 \, dx} = e^{-2x}

Multiplying both sides by the integrating factor: ddx[f(x)e−2x]=(2x−3)e−2x\frac{d}{dx} \left[ f(x) e^{-2x} \right] = (2x - 3)e^{-2x}

Integrating both sides with respect to xx: f(x)e−2x=∫(2x−3)e−2x dxf(x) e^{-2x} = \int (2x - 3)e^{-2x} \, dx f(x)e−2x=(2x−3)(e−2x−2)−∫2(e−2x−2) dxf(x) e^{-2x} = (2x - 3) \left( \frac{e^{-2x}}{-2} \right) - \int 2 \left( \frac{e^{-2x}}{-2} \right) \, dx f(x)e−2x=−12(2x−3)e−2x−12e−2x+Cf(x) e^{-2x} = -\frac{1}{2}(2x - 3)e^{-2x} - \frac{1}{2}e^{-2x} + C f(x)e−2x=(1−x)e−2x+Cf(x) e^{-2x} = (1 - x)e^{-2x} + C

Multiplying through by e2xe^{2x}: f(x)=1−x+Ce2xf(x) = 1 - x + C e^{2x}

Step 2: Determining the Constant CC

From the original integral equation, substituting x=0x = 0: f(0)=1−2(0)+∫00e0−tf(t) dt=1f(0) = 1 - 2(0) + \int_0^0 e^{0-t} f(t) \, dt = 1

Using our solution for f(x)f(x): f(0)=1−0+Ce0=1+Cf(0) = 1 - 0 + C e^0 = 1 + C

Setting this equal to 11: 1+C=1  ⟹  C=01 + C = 1 \implies C = 0

Thus, the continuous function is: f(x)=1−xf(x) = 1 - x

Step 3: Checking the Options

  1. Option (A): f(1)=1−1=0≠2f(1) = 1 - 1 = 0 \neq 2 Hence, the curve does not pass through (1,2)(1, 2). (Option A is FALSE)

  2. Option (B): f(2)=1−2=−1f(2) = 1 - 2 = -1 Hence, the curve passes through the point (2,−1)(2, -1). (Option B is TRUE)

  3. Options (C) and (D): We find the area of the region bounded by f(x)≤y≤1−x2f(x) \leq y \leq \sqrt{1 - x^2} for x∈[0,1]x \in [0, 1]. Since for all x∈[0,1]x \in [0, 1], 1−x≤1−x21 - x \leq \sqrt{1 - x^2}, the enclosed area AA is given by: A=∫01(1−x2−(1−x))dxA = \int_0^1 \left( \sqrt{1 - x^2} - (1 - x) \right) dx A=∫011−x2 dx−∫01(1−x) dxA = \int_0^1 \sqrt{1 - x^2} \, dx - \int_0^1 (1 - x) \, dx

    • ∫011−x2 dx\int_0^1 \sqrt{1 - x^2} \, dx represents the area of a quarter of a circle of radius 11: ∫011−x2 dx=π(1)24=π4\int_0^1 \sqrt{1 - x^2} \, dx = \frac{\pi (1)^2}{4} = \frac{\pi}{4}
    • ∫01(1−x) dx\int_0^1 (1 - x) \, dx: ∫01(1−x) dx=[x−x22]01=1−12=12\int_0^1 (1 - x) \, dx = \left[ x - \frac{x^2}{2} \right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2}

    Therefore: A=π4−12=π−24A = \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}

    Hence, Option C is TRUE and Option D is FALSE.

Conclusion:

The correct statements are B and C.

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