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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(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

Similar Questions

The value of $\sum\limits_{r = 2}^{16} {\int\limits_r^{r + 1} {\frac{{dx}}{{\left( {2r - x} \right)\left( {2r + 2 - x} \right)}}} }$ is equal to

The value of $\int_3^5 {\frac{{{x^2}}}{{{x^2} - 4}}\,dx} $ is

$\int_{\pi / 4}^{\pi / 3} \frac{\cos x-\sin x}{\sin 2 x} d x=$

$\int_0^1 (1+x) \log (1+x) \, dx =$

If $[.]$ represents the greatest integer function,then the value of $\int_{0}^{\sqrt{\pi / 2}}\left(\left[ x ^{2}\right]+[-\cos x ]\right) d x$ is.............

Vedclass Test Series

Exam Paper Generator

Online Exam Module