JEE Advanced 2015 · Paper 1 · Mathematics Q48

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MathematicsInteger+4Question 48 of 60

Let $F(x)=\displaystyle\int_x^{x^2+\frac\pi6}2\cos^2t\,dt$ for all $x\in\mathbb R$ and $f:\left[0,\frac12\right]\to[0,\infty)$ be a continuous function. For $a\in\left[0,\frac12\right]$ , if $F'(a)+2$ is the area of the region bounded by $x=0$ , $y=0$ , $y=f(x)$ and $x=a$ , then $f(0)$ is

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Answer and worked solution

Official answer: 3

Set up. Let $A(a)=\int_0^af(x)\,dx=F'(a)+2$ . Differentiating gives $f(a)=F''(a)$ , so $f(0)=F''(0)$ .

Differentiate F. By the Leibniz rule,
$F'(x)=4x\cos^2\left(x^2+\frac\pi6\right)-2\cos^2x.$

Second derivative at 0. Every term except the first vanishes at $x=0$ :
$F''(0)=4\cos^2\frac\pi6=3.$
(Check: $A(0)=F'(0)+2=-2+2=0$ .)

Answer: 3

Topics:Leibniz rule Fundamental theorem of calculus

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JEE Advanced 2015 · Paper 1, Mathematics question 48