Question

Let $f:[0,1]\to [0,1]$ be a continuous function such that $x^2+\left(f\right(x){)}^2\le 1$ for all $x\in [0,1]$ and $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{\pi }{4}$, then $\underset{\frac{1}{2}}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{f\left(x\right)}{1-x^2}dx$ equals

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{15}$
• C. $\frac{\sqrt{2}-1}{2}\pi$
• D. $\frac{\pi }{10}$

Answer 1

The correct option is A $\frac{\pi }{12}$

Given that $x^2+\left(f\right(x){)}^2\le 1$
$\Rightarrow \left(f\right(x){)}^2\le 1-x^2\Rightarrow -\sqrt{1-x^2}\le f(x)\le \sqrt{1-x^2}\Rightarrow -\underset{0}{\overset{1}{\int }}\sqrt{1-x^2}dx\le \underset{0}{\overset{1}{\int }}f(x)dx\le \underset{0}{\overset{1}{\int }}\sqrt{1-x^2}dx\Rightarrow -\frac{\pi }{4}\le \underset{0}{\overset{1}{\int }}f(x)dx\le \frac{\pi }{4}\Rightarrow f(x)=\sqrt{1-x^2}$

$\underset{\frac{1}{2}}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{f\left(x\right)}{1-x^2}dx=\underset{\frac{1}{2}}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{\sqrt{1-x^2}}{1-x^2}dx=\underset{\frac{1}{2}}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{1}{\sqrt{1-x^2}}dx=[{\sin }^{-1}x{]}_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}=\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}$