JEE-Advanced 2015 (8/40)

Calculus Level 4

Let $F(x)=\displaystyle \int_{x}^{x^2+\frac{\pi}{6}} 2\cos^2t \, dt$ for all $x\in \mathbb{R}$ and $f:\left[0,\frac{1}{2} \right] \to [0,\infty)$ be a continuous function. For $a \in \left[0,\frac{1}{2} \right]$ , if $F'(a)+2$ is the area of the region bounded by $x=0,y=0 ,y=f(x)$ and $x=a$ , then what is the $f(0)$ ?

The answer is 3.

1 solution

Aditya Agarwal
Aug 31, 2015

$F'(a)+2=\int^{a}_0 f(x)dx$ And, because they both are constants, Differentiating w.r.t $a$ By fundamental theorem of calculus, $F''(a)=f(a)$ $F'(x)=2\cos^2(x^2+\frac{\pi}6).2x-2\cos^2x$ $F''(x)=4\cos^2(x^2+\frac{\pi}6)-16x^2\cos(x^2+\frac{\pi}{6})\sin(x^2+\frac{\pi}6)+4\cos x\sin x$ $F''(0)=f(0)=4\cos^2\frac{\pi}{6}=3$

Same method

Abhinav Shripad - 1 year, 4 months ago

JEE-Advanced 2015 (8/40)

The answer is 3.

1 solution

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