Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
Let $f: \left[0, \frac{\pi}{2}\right] \to \mathbb{R}$ be such that $f(0) = 3$ and $f'(x) = \frac{1}{1 + \cos x}$. If $a < f\left(\frac{\pi}{2}\right) < b$, then $a$ and $b$ can be
\frac{\pi}{2}, \pi
3, 4
3 + \frac{\pi}{4}, 3 + \frac{\pi}{2}
3 + \frac{3}{2}, 3 + \frac{3\pi}{4}
Step-by-Step Solution
Key Concept: Convert $\frac{1}{1+\cos x}$ to secant squared form and integrate, then apply initial conditions to find the constant.
Given $f'(x) = \frac{1}{1+\cos x} = \frac{1}{2\cos^2(x/2)} = \frac{1}{2}\sec^2(x/2)$, integrating gives $f(x) = \tan(x/2) + c$. Using $f(0) = 0$, we get $c = 3$, so $f(x) = \tan(x/2) + 3$. Then $f(\pi/2) = \tan(\pi/4) + 3 = 4$ and $3 + \pi/4 = 53/14$ and $3 + \pi/4 = 11/7$.
Correct Answer: 3