Let f: [0, ∞) → R be a continuous function such that f(x) = 1 - 2x + ∫ _ 0 ^ x e ^ x - t f(t)dt for all x ∈ [0, ∞) . Then, which of the following statement(s) is (are) TRUE?

Correct answer: B. The curve y = f(x) passes through the point (2, - 1) f(x) = 1 - 2x + ∫ _ 0 ^ x e ^ x - t f(t)dt ⇒ e ^ - x f(x) = e ^ - x (1 - 2x) + ∫ _ 0 ^ x e ^ - t f(t)dt Differentiate w.r.t. x e ^ - x f(x) + e ^ + x f ^ ′ (x) = - e ^ - x (1 - 2x) + e ^ - x (- 2) + e ^ - x f(x) ⇒ - f(x) + f ^ ′ (x) = - (1 - 2x) - 2 + f(x) ⇒ f ^ ′ (x) - 2f(x) = 2x - 3 Integrating factor = e ^ - 2x f(x) . e ^ - 2x = ∫ e ^ - 2x (2x - 3)dx = (2x - 3) ∫ e ^ - 2x dx - ∫ ((2) ∫ e ^ - 2x dx)dx = (2x - 3) e ^ - 2x - 2 - e ^ - 2x 2 + c f(x) = (2x - 3)/(- 2) - (1)/(2) + c e ^ 2x f(0) = (3)/(2) - (1)/(2) + c = 1 ⇒ c = 0 ∴ f(x) = 1 - x Area = (π)/(4) - (1)/(2) = (π - 2)/(4)

JEE Advanced 2018Paper 1

Exam: JEE Advanced
Year: 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Differential Equations

JEE Advanced 2018 · Paper 1 · Mathematics · Differential Equations

Let f: [0, ∞) → R be a continuous function such that f(x) = 1 - 2x...

Let

f: \left[0, \infty\right) \to R

be a continuous function such that

f\left(x\right) = 1 - 2x + {∫}_{0}^{x}{e}^{x - t}f\left(t\right)dt

for all

x \in \left[0, \infty\right) .

Then, which of the following statement(s) is (are) TRUE? Let f: [0, ∞) → R be a continuous function such that f(x) = 1 - 2x + ∫ _ 0 ^ x e ^ x - t f(t)dt for all x ∈ [0, ∞) .

Options

AThe curve $y = f\left(x\right)$ pass through the point \left(1, 2\right)
BThe curve $y = f\left(x\right)$ passes through the point \left(2, - 1\right)
CThe area of the region $\left\{\left(x, y\right) \in \left[0,1\right] \times R: f\left(x\right) \le y \le \sqrt{1 - {x}^{2}}\right\}is\dfrac{\pi - 2}{4}$
DThe area of the region $\left\{\left(x, y\right) \in \left[0,1\right] \times R: f\left(x\right) \le y \le \sqrt{1 - {x}^{2}}\right\}is\dfrac{\pi - 1}{4}$

Correct answer

B. The curve $y = f\left(x\right)$ passes through the point \left(2, - 1\right)

Step-by-step solution

f\left(x\right) = 1 - 2x + {∫}_{0}^{x}{e}^{x - t}f\left(t\right)dt

⇒ {e}^{- x}f\left(x\right) = {e}^{- x}\left(1 - 2x\right) + {∫}_{0}^{x}{e}^{- t}f\left(t\right)dt

Differentiate w.r.t.

x

{e}^{- x}f\left(x\right) + {e}^{+ x}{f}^{′}\left(x\right) = {- e}^{- x}\left(1 - 2x\right) + {e}^{- x}\left(- 2\right) + {e}^{- x}f\left(x\right)

⇒ - f\left(x\right) + {f}^{′}\left(x\right) = - \left(1 - 2x\right) - 2 + f\left(x\right)

⇒ {f}^{′}\left(x\right) - 2f\left(x\right) = 2x - 3

Integrating factor

= {e}^{- 2x}

f\left(x\right) . {e}^{- 2x} = ∫ {e}^{- 2x}\left(2x - 3\right)dx

= \left(2x - 3\right) ∫ {e}^{- 2x}dx - ∫ \left(\left(2\right) ∫ {e}^{- 2x}dx\right)dx

= \dfrac{\left(2x - 3\right){e}^{- 2x}}{- 2} - \dfrac{{e}^{- 2x}}{2} + c

f\left(x\right) = \dfrac{2x - 3}{- 2} - \dfrac{1}{2} + c{e}^{2x}

f\left(0\right) = \dfrac{3}{2} - \dfrac{1}{2} + c = 1 ⇒ c = 0

∴ f\left(x\right) = 1 - x

Area

= \dfrac{\pi}{4} - \dfrac{1}{2} = \dfrac{\pi - 2}{4}

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