JEE Advanced 2017 Paper 1Question 1
Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $f(0)=0$, $f\left(\frac{\pi}{2}\right)=3$ and $f'(0)=1$. If $$g(x)=\int_x^{\pi/2}\left[f'(t)\csc t-\cot t\csc t\,f(t)\right]dt$$ for $x\in\left(0,\frac{\pi}{2}\right]$, then $\lim_{x\to0}g(x)=$
Definite Integration
Answer$2$
Solution
Concept. The integrand is $\frac{d}{dt}\left(\frac{f(t)}{\sin t}\right)$. Formulas used. (1) $\dfrac{d}{dt}\left(f(t)\csc t\right)=f'(t)\csc t-f(t)\cot t\csc t$(2) $\lim_{x\to0}\dfrac{f(x)}{\sin x}=\lim_{x\to0}\dfrac{f(x)-f(0)}{x}\cdot\dfrac{x}{\sin x}=f'(0)$Step 1 - Integrate. By (1), $$g(x)=\left[\frac{f(t)}{\sin t}\right]_x^{\pi/2}=3-\frac{f(x)}{\sin x}.$$ Step 2 - Limit. By (2), $\dfrac{f(x)}{\sin x}\to1$, so $\lim_{x\to0}g(x)=3-1=2$. Answer: 2