Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $f(x) = \int_0^x [f(t)]^{-1} dt$ and $\int_0^1 [f(x)]^{-1} dx = \sqrt{2}$, then:
f(2) = 2
f'(2) = 1/2
f^{-1}(2) = 2
\int_0^1 f(x)dx = \sqrt{2}
Step-by-Step Solution
Key Concept: Use functional equation substitution at shifted arguments to create a system of equations for $f(x)$ and $f(x+\pi)$.
Given $f(x) + \sin x \cdot f(x+\pi) = \sin^2 x$, substitute $x \to x+\pi$ to get $f(x+\pi) - \sin x \cdot f(x) = \sin^2 x$. Solving these two equations simultaneously yields $f(x) = \frac{\sin^2 x(1-\sin x)}{1+\sin^2 x}$. The integral is then evaluated by splitting into partial fractions and standard forms, yielding $x - \frac{1}{2}\sqrt{2}\tan^{-1}(\sqrt{2}\tan x) - \frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{2}+\cos x}{\sqrt{2}-\cos x}\right) + \cos x + C$.
Correct Answer: 2,3