JEE Advanced 2014 · Paper 2 · Mathematics Q44

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MathematicsSingle correct+3 / -1Question 44 of 60

Let $f:[0,2]\to\mathbb R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$ . Let
$F(x)=\int_0^{x^2}f(\sqrt t)\,dt$
for $x\in[0,2]$ . If $F'(x)=f'(x)$ for all $x\in(0,2)$ , then $F(2)$ equals

e^2-1

e^4-1

e-1

e^4

Answer and worked solution

Official answer: B

Differentiate $F$ . $F'(x)=f(\sqrt{x^2})\cdot2x=2xf(x)$ for $x\ge0$ .

The condition $F'=f'$ gives $f'(x)=2xf(x)$ , so $f(x)=Ce^{x^2}$ , and $f(0)=1$ gives $f(x)=e^{x^2}$ .

Evaluate.
$F(2)=\int_0^4f(\sqrt t)\,dt=\int_0^4e^t\,dt=e^4-1.$

(B)

Topics:Leibniz rule Variable separable

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JEE Advanced 2014 · Paper 2, Mathematics question 44