Let f:[0,1] rightarrow [0,1] be a continuous | vedclass

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(A) Given that $f:[0,1] \rightarrow [0,1]$ is a continuous function satisfying $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$.This implies $(f(x))^2 \leq

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$\frac{\pi}{12}$

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(A) Given that $f:[0,1] \rightarrow [0,1]$ is a continuous function satisfying $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$.This implies $(f(x))^2 \leq 1-x^2$,so $f(x) \leq \sqrt{1-x^2}$ since $f(x) \geq 0$.Integrating both sides from $0$ to $1$,we get $\int_0^1 f(x) dx \leq \int_0^1 \sqrt{1-x^2} dx$.The integral $\int_0^1 \sqrt{1-x^2} dx$ represents the area of a quarter circle of radius $1$,which is $\frac{1}{4} \pi (1)^2 = \frac{\pi}{4}$.Since it is given that $\int_0^1 f(x) dx = \frac{\pi}{4}$,the equality must hold,which implies $f(x) = \sqrt{1-x^2}$ for all $x \in [0,1]$.Now,we evaluate the integral $\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx = \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2}}{1-x^2} dx = \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1-x^2}} dx$.This is equal to $[\sin^{-1} x]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} = \sin^{-1}(\frac{1}{\sqrt{2}}) - \sin^{-1}(\frac{1}{2}) = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$.

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$ and $\int_0^1 f(x) dx = \frac{\pi}{4}$. Then,$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx$ equals

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