Let $F(x)=\int_{x}^{x^{2}+\frac{\pi}{6}} 2 \cos ^{2} t d t$ for all $x \in \mathbb{R}$ and $f:\left[0, \frac{1}{2}\right] \rightarrow[0, \infty)$ be a continuous function. For $a \in\left[0, \frac{1}{2}\right]$, if $F^{\prime}(a)+2$ is the area of the region bounded by $x=0, y=0, y=f(x)$ and $x=a$, then $f(0)$ is

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Definite Integrals

Let

F (x) = \int_{x}^{x^{2} + \frac{π}{6}} 2 cos^{2} t d t

for all

x \in R

and

f : [0, \frac{1}{2}] \to [0, \infty)

be a continuous function. For

a \in [0, \frac{1}{2}]

, 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)

Answer: 3

By Leibniz rule:

F^{'} (x) = 2 cos^{2} (x^{2} + \frac{π}{6}) \cdot 2 x - 2 cos^{2} (x)

Given that

F^{'} (a) + 2 = \int_{0}^{a} f (x) d x

, differentiating both sides with respect to

a

\frac{d}{d a} [F^{'} (a) + 2] = f (a)

F^{''} (a) = f (a)

Compute

F^{''} (x)

F^{''} (x) = 4 cos^{2} (x^{2} + \frac{π}{6}) + 4 x \cdot 2 cos (x^{2} + \frac{π}{6}) \cdot (- sin (x^{2} + \frac{π}{6})) \cdot 2 x + 4 cos x sin x

x = 0

f (0) = F^{''} (0) = 4 cos^{2} (\frac{π}{6}) + 0 + 0 = 4 (\frac{3}{2})^{2} = 4 \cdot \frac{3}{4} = 3

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer