JEE Advanced 2018 · Paper 1 · Mathematics Q6

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MathematicsMultiple correct+4 / -2Question 42 of 54

Let $f: [0, \infty) \to \mathbb{R}$ be a continuous function such that

$f(x) = 1 - 2x + \int_0^x e^{x-t} f(t)\, dt$

for all $x \in [0, \infty)$ . Then, which of the following statement(s) is (are) TRUE?

The curve

y = f(x)

passes through the point

(1, 2)

The curve

y = f(x)

passes through the point

(2, -1)

The area of the region

\{(x, y) \in [0, 1] \times \mathbb{R} : f(x) \le y \le \sqrt{1 - x^2}\}

\dfrac{\pi - 2}{4}

The area of the region

\{(x, y) \in [0, 1] \times \mathbb{R} : f(x) \le y \le \sqrt{1 - x^2}\}

\dfrac{\pi - 1}{4}

Answer and worked solution

Official answer: B, C

Differentiate. Write $f(x)=1-2x+e^x\displaystyle\int_0^xe^{-t}f(t)\,dt$ . Then
$f'(x)=-2+e^x\int_0^xe^{-t}f\,dt+f(x)=-2+\big(f(x)-1+2x\big)+f(x).$
So
$f'-2f=2x-3,\qquad f(0)=1.$

Solve. A particular solution is $f=1-x$ (check: $-1-2(1-x)=2x-3$ ). The general solution is $1-x+Ce^{2x}$ , and $f(0)=1$ gives $C=0$ . So $f(x)=1-x$ .

(A) $f(1)=0\ne2$ . False.
(B) $f(2)=-1$ . True.
(C)/(D) On $[0,1]$ , $1-x\le\sqrt{1-x^2}$ . The area is the quarter disc minus the triangle:
$\frac\pi4-\frac12=\frac{\pi-2}4.$
(C) is true and (D) is false.

(B), (C)

Topics:Linear first-order equations Fundamental theorem of calculus Area between curves

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JEE Advanced 2018 · Paper 1, Mathematics question 6