Question

Let $f:(0,1)\to (0,1)$ be a differential function such that $f^{\prime }\left(x\right)\ne 0$ for all $xϵ(0,1)$ and $f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}.$ Suppose for all $x,\underset{t\to x}{lim}\left(\frac{{\int }_0^1\sqrt{1-\left(f\right(s){)}^2}ds-{\int }_0^x\sqrt{1-\left(f\right(s){)}^2}ds}{f\left(t\right)-f\left(x\right)}\right)=f\left(x\right).$ Then the value of $f\left(\frac{1}{4}\right)$ belongs to :

• A. $\{\frac{\sqrt{7}}{4},\frac{\sqrt{15}}{4}\}$
• B. $\{\frac{\sqrt{7}}{3},\frac{\sqrt{15}}{3}\}$
• C. $\{\frac{\sqrt{7}}{2},\frac{\sqrt{15}}{2}\}$
• D. $\{\sqrt{7},\sqrt{15}\}$

Answer 1

The correct option is A $\{\frac{\sqrt{7}}{4},\frac{\sqrt{15}}{4}\}$

$\underset{t\to x}{lim}\frac{\left({\int }_0^t\sqrt{1-{\left(f\left(s\right)\right)}^2}ds-{\int }_0^x\sqrt{1-{\left(f\left(s\right)\right)}^2}ds\right)}{f\left(t\right)-f\left(x\right)}=\frac{0}{0}$ form

L-Hospita's Rule

$\underset{t\to x}{lim}\frac{\left(\sqrt{1-{\left(f\left(t\right)\right)}^2}\right)\cdot 1}{f^{\prime }\left(t\right)}=\frac{\sqrt{1-{\left(f\left(x\right)\right)}^2}}{f^{\prime }\left(x\right)}=f\left(x\right)$

$\Rightarrow \sqrt{\frac{1-{\left(f\left(x\right)\right)}^2}{{\left(f\left(x\right)\right)}^2}}=f^{\prime }\left(x\right)$

$\sqrt{\frac{1-y^2}{y^2}}=y^{\prime }=\frac{dy}{dx}$

$\int dx=\int \frac{y}{\sqrt{1-y^2}}dy1-y^2=t;2ydy=dt$

$x=-\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$x=-\sqrt{t}+c$

$x=-\sqrt{1-y^2}+c$

$1-y^2={(c-x)}^2$

$\sqrt{1-{(c-x)}^2}=y=f\left(x\right)$

Given $f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$

$\Rightarrow \frac{\sqrt{3}}{2}=\sqrt{1-{(c-1/2)}^2}$

$\Rightarrow \frac{1}{4}={(c-1/2)}^2\Rightarrow c=1,0$

$f\left(x\right)=\sqrt{1-{(1-x)}^2},\sqrt{1-x^2}$

$f\left(\frac{1}{4}\right)=\sqrt{\frac{7}{16}},\frac{\sqrt{15}}{4}$