Integral Calculus
Integral inequality; monotonicity argument
Grade Class 12
Question:
Let $f$ be continuous and differentiable in $(x_1, x_2)$. If $f(x)f'(x) \geq x\sqrt{1-[f(x)]^4}$ and $\lim_{x\to x_1}(f(x))^2=1$, $\lim_{x\to x_2}(f(x))^2=\frac{1}{2}$. Then minimum value of $\left[x_1^2 - x_2^2\right]$ is ........... (where $[\cdot]$ denotes GIF)
Step-by-Step Solution
Key Concept: Rewrite the inequality as $\frac{2f(x)f'(x)}{\sqrt{1-(f(x))^4}} \geq 2x$, which is $\frac{d}{dx}\left[\sin^{-1}(f(x))^2\right] \geq \frac{d}{dx}[x^2]$. So $g(x)=\sin^{-1}(f(x))^2 - x^2$ is non-decreasing.
$g(x)=\sin^{-1}((f(x))^2)-x^2$ is non-decreasing. $g(x_1)=\pi/2-x_1^2$, $g(x_2)=\pi/6-x_2^2$. So $x_1^2-x_2^2 \geq \pi/3$, minimum $[\pi/3]=1$.
Correct Answer: 1