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

(C) The area of the region x,y∈0,1×R:fx≤y≤1-x2 isπ-24

(A) The curve y=fx pass through the point 1,2

(B) The curve y=fx passes through the point 2,-1

(C) The area of the region x,y∈0,1×R:fx≤y≤1-x2 isπ-24

(D) The area of the region x,y∈0,1×R:fx≤y≤1-x2 isπ-14

JEE Advanced · Mathematics · 29. Differential Eqns

Let $f : [0, \infty) \to R$ be a continuous function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty) .$ Then, which of the following statement(s) is (are) TRUE?

A The curve $y = f (x)$ pass through the point $(1, 2)$
B The curve $y = f (x)$ passes through the point $(2, - 1)$
C The area of the region $\{(x, y) \in [0,1] \times R : f (x) \leq y \leq \sqrt{1 - x^{2}}\} i s \frac{π - 2}{4}$
D The area of the region $\{(x, y) \in [0,1] \times R : f (x) \leq y \leq \sqrt{1 - x^{2}}\} i s \frac{π - 1}{4}$

Verified Solution

Answer & Solution

Correct Answer

(C) The area of the region $\{(x, y) \in [0,1] \times R : f (x) \leq y \leq \sqrt{1 - x^{2}}\} i s \frac{π - 2}{4}$

Step-by-step Solution

Detailed explanation

f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t

\Rightarrow e^{- x} f (x) = e^{- x} (1 - 2 x) + \int_{0}^{x} e^{- t} f (t) d t

Differentiate w.r.t.

x

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

\Rightarrow - f (x) + f^{'} (x) = - (1 - 2 x) - 2 + f (x)

\Rightarrow f^{'} (x) - 2 f (x) = 2 x - 3

Integrating factor

= e^{- 2 x}

f (x) . e^{- 2 x} = \int e^{- 2 x} (2 x - 3) d x

= (2 x - 3) \int e^{- 2 x} d x - \int ((2) \int e^{- 2 x} d x) d x

= \frac{(2 x - 3) e^{- 2 x}}{- 2} - \frac{e^{- 2 x}}{2} + c

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

f (0) = \frac{3}{2} - \frac{1}{2} + c = 1 \Rightarrow c = 0

∴ f (x) = 1 - x

Area

= \frac{π}{4} - \frac{1}{2} = \frac{π - 2}{4}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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