• # question_answer Let$(0,\,1)\to \,(0,1)$be a differential function such that $f'(x)\ne 0$ for all $x\,\in (0,\,\,1)$and $f\,\left( \frac{1}{2} \right)=\frac{\sqrt{3}}{2}$. Suppose for all $x$, $\underset{t\to x}{\mathop{\lim }}\,\,\frac{\int\limits_{0}^{t}{\sqrt{1-{{\{f(S)\}}^{2}}}\,dS-\int\limits_{0}^{x}{\sqrt{1-{{\{f(S)\}}^{2}}}dx}}}{f(t)-f(x)}$ Then, the value of $f\left( \frac{1}{4} \right)$ belongs to A)  $\left\{ \sqrt{7},\,\,\sqrt{15} \right\}$             B)  $\left\{ \frac{\sqrt{7}}{2},\,\,\frac{\sqrt{15}}{2} \right\}$ C)  $\left\{ \frac{\sqrt{7}}{3},\,\,\frac{\sqrt{15}}{3} \right\}$               D)  $\left\{ \frac{\sqrt{7}}{4},\,\,\frac{\sqrt{15}}{4} \right\}$

Given $\Rightarrow$ $v'\frac{v}{\sqrt{2}}$ for all $E=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}(2m){{v}^{2}}$ $=mv{{'}^{2}}+m{{v}^{2}}$ $\Rightarrow$ By L ?Hospital rule, $m{{\left( \frac{v}{\sqrt{2}} \right)}^{2}}+m{{v}^{2}}=\frac{3}{2}m{{v}^{2}}$ ${{T}_{1}}-{{T}_{2}}=12a$         ${{T}_{2}}=3a$ ${{T}_{2}}$ $Eq$               ${{T}_{1}}=15a$ $\therefore$    $\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{15a}{3a}=\frac{5}{1}$ $\Rightarrow$            ${{T}_{1}}:{{T}_{2}}=5:1$ $T=M\,\left( g-\frac{g}{4} \right)=\frac{3Mg}{4}$  $W=\mathbf{T}\cdot \mathbf{d}\Rightarrow \,W=Td$ $\Rightarrow$            $W=-Td=-\frac{3Mgd}{4}$ $\Sigma mvr=\,({{l}_{system}})\omega$    $\Rightarrow$ $mv\frac{l}{2}=\frac{(2m)\,{{(2l)}^{2}}}{12}\omega =\frac{2m(4{{l}^{2}})}{12}\omega$            $\Rightarrow$