અહી f:[0,1] rightarrow[0,1] એ સતત વિધેય છે કે | vedclass

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Question

(a)

For a continuous function $f:[0,1] \longrightarrow[0,1]$, it is given that

$x^2+(f(x))^2 \leq 1$

$\Rightarrow(f(x))^2 \leq 1-x^2$

$\le

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(a)

For a continuous function $f:[0,1] \longrightarrow[0,1]$, it is given that

$x^2+(f(x))^2 \leq 1$

$\Rightarrow(f(x))^2 \leq 1-x^2$

$\left.\Rightarrow \quad f(x) \leq \sqrt{1-x^2} \quad \quad \quad \because f(x) \in[0,1]ight\}$

$	herefore \int_0^1 f(x) d x \leq \int_0^1 \sqrt{1-x^2} d x$

$\Rightarrow \int_0^1 f(x) d x \leq\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} xight]_0^1$

$\Rightarrow \int_0^1 f(x) d x \leq \frac{\pi}{4}$

So, $f(x)=\sqrt{1-x^2}$, because it is given that

$lint_0^1 f(x)=\frac{\pi}{4}$

Now,

$\int_{1 / 2}^{1 / \sqrt{2}} \frac{f(x)}{1-x^2} d x=\int_{1 / 2}^{1 / \sqrt{2}} \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} xight]_{1 / 2}^{1 / \sqrt{2}}$

$=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$

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