Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0, f\left(\frac{\pi}{2}\right)=3$ and $f^{\prime}(0)=1$. If \[ g(x)=\int_{x}^{\frac{\pi}{2}}\left[f^{\prime}(t) \operatorname{cosec} t-\cot t \operatorname{cosec} t f(t)\right] d t \] for $x \in\left(0, \frac{\pi}{2}\right]$, then $\lim _{x \rightarrow 0} g(x)=$

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Definite Integrals

Let

f : R \to R

be a differentiable function such that

f (0) = 0, f (\frac{π}{2}) = 3

and

f^{'} (0) = 1

. If

g (x) = \int_{x}^{\frac{π}{2}} [f^{'} (t) cosec t - cot t cosec t f (t)] d t

for

x \in (0, \frac{π}{2}]

, then

lim_{x \to 0} g (x) =

Answer: 2

The integrand can be written as

\frac{d}{d t} [f (t) cosec t]

. Thus

g (x) = [f (t) cosec t]_{x}^{π /2} = f (π /2) cosec (π /2) - f (x) cosec x = 3 - \frac{f ( x )}{s i n x}

. Therefore

lim_{x \to 0} g (x) = 3 - lim_{x \to 0} \frac{f ( x )}{s i n x} = 3 - lim_{x \to 0} \frac{f ( x ) - f ( 0 )}{x} \cdot \frac{x}{s i n x} = 3 - f^{'} (0) \cdot 1 = 3 - 1 = 2

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer