Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade None
Question:
If $f(x) = \left(\sin^{-1}x\right)^3 + \left(\cos^{-1}x\right)^3$, then :
Minimum value of $f(x) = -\frac{\pi^3}{8}$
Minimum value of $f(x) = \frac{\pi^3}{32}$
Maximum value of $f(x) = -\frac{\pi^3}{8}$
Maximum value of $f(x) = \frac{7\pi^3}{8}$
Step-by-Step Solution
Key Concept: Expand using the sum and difference of cubes, then express the result as a sum of constant and quadratic terms in $\sin^{-1}x$.
Let $u = \sin^{-1}x - \pi/4$. Then $(\sin^{-1}x + \cos^{-1}x)^3 + (\sin^{-1}x - \cos^{-1}x)^3 = (\pi/2)^3 - 3\sin^{-1}x\cos^{-1}x(\sin^{-1}x - \sin^{-1}x) = \pi^3/8 + \frac{3\pi}{2}(\sin^{-1}x - \pi/4)^2 - \frac{3\pi^3}{32}$. The expression simplifies to $\frac{\pi^3}{8} + \frac{3\pi}{2}(\sin^{-1}x - \pi/4)^2 - \frac{3\pi^3}{32}$. The minimum value occurs when $\sin^{-1}x = \pi/4$, giving $\frac{\pi^3}{32}$; the maximum is $\frac{\pi^3}{32} + \frac{9\pi^2}{16} - \frac{3\pi}{2} - \frac{7\pi^3}{8}$.
Correct Answer: 2,4