Continuity and Differentiability
Fundamental Theorem of Calculus
JEE Main 2023
Grade 12
Question:
Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 0$, $f(\pi/2) = 3$ and $f'(0) = 1$. If $g(x) = \int_x^{\pi/2} [f'(t) \csc t - \cot t \csc t \cdot f(t)] dt$ for $x \in (0, \pi/2]$, then $\lim_{x \to 0^+} g(x) =$
(1) $0$
(2) $2$
(3) $3$
(4) $-3$
Step-by-Step Solution
Key Concept: Notice $f'(t) \csc t - f(t) \csc t \cot t = \frac{d}{dt} \left[\frac{f(t)}{\sin t}\right] \cdot \sin t \cdot \frac{1}{\sin t}\dots$ actually it equals $\frac{d}{dt} \left[\frac{f(t)}{\sin t}\right]$. So $g(x) = \left[\frac{f(t)}{\sin t}\right]_x^{\pi/2} = 3 - \frac{f(x)}{\sin x} \to 3 - f'(0)/1 = 3 - 1 = 2$.
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Correct Answer: (2)