Question

Let $f:[0,\frac{\pi }{2}]\to [0,1]$ be a differentiable function such that $f\left(0\right)=0,f\left(\frac{\pi }{2}\right)=1$, then

• A. $f^{\prime }\left(\alpha \right)=\sqrt{1-\left(f\right(\alpha ){)}^2}$ for all $\alpha ϵ(0,\frac{\pi }{2})$
• B. $f^{\prime }\left(\alpha \right)=\frac{2}{\pi }$ for all $\alpha ϵ(0,\frac{\pi }{2})$
• C. $f\left(\alpha \right)f^{\prime }\left(\alpha \right)=\frac{1}{\pi }$ for at least one $\alpha ϵ(0,\frac{\pi }{2})$
• D. $f^{\prime }\left(\alpha \right)=\frac{8\alpha }{{\pi }^2}$ for at least one $\alpha ϵ(0,\frac{\pi }{2})$

Answer 1

The correct option is A $f^{\prime }\left(\alpha \right)=\sqrt{1-\left(f\right(\alpha ){)}^2}$ for all $\alpha ϵ(0,\frac{\pi }{2})$

$f:[0,\frac{\pi }{2}]\to [0,1]$be a.....
(A) Consider $g\left(x\right)=si{n}^{-1}f\left(x\right)-x$
since $g\left(0\right)=0,g\left(\frac{\pi }{2}\right)=0$
$\therefore$There is at least one value of $\alpha ϵ(0,\frac{\pi }{2})$such that
$g^{\prime }\left(\alpha \right)=\frac{f^{\prime }\left(\alpha \right)}{\sqrt{1-\left(f\right(\alpha ){)}^2}}-1=0$
$i.e.f^{\prime }\left(\alpha \right)=\sqrt{1-\left(f\right(\alpha ){)}^2}$ for atleast one value of $\alpha$ but may not be for all $\alpha ϵ(0,\frac{\pi }{2})$
​$\therefore$ false
(B) Consider $g\left(x\right)=f\left(x\right)-\frac{2x}{\pi }$
Since $g\left(x\right)=f\left(x\right)-\frac{2x}{\pi }$
$\therefore$ there is at least one value of $\alpha ϵ(0,\frac{\pi }{2})$ such that
$g^{\prime }\left(\alpha \right)=f^{\prime }\left(\alpha \right)-\frac{\pi }{2}=0$
i.e$f^{\prime }\left(\alpha \right)=\frac{2}{\pi }$ for atleast one value of $f^{\prime }\left(\alpha \right)=\frac{2}{\pi }$ for all $\alpha ϵ(0,\frac{\pi }{2})$
$\therefore$ false
(C) Consider $g\left(x\right)=\left(f\right(x){)}^2-\frac{2x}{\pi }$
Since $g\left(0\right)=0,g\left(\frac{\pi }{2}\right)=0$
$\therefore$There is at least one value of $\alpha ϵ(0,\frac{\pi }{2})$such that
$g^{\prime }\left(\alpha \right)=2f\left(\alpha \right)f^{\prime }\left(\alpha \right)-\frac{\pi }{2}=0\therefore f\left(\alpha \right)f^{\prime }\left(\alpha \right)=\frac{1}{\pi }$
$\therefore$ True
(D) Consider $g\left(x\right)=f\left(x\right)-\frac{4x^2}{{\pi }^2}$
Since $g\left(0\right)=0,g\left(\frac{\pi }{2}\right)=0$
$\therefore$ there is at least one value of $\alpha ϵ(0,\frac{\pi }{2})$ such that
$g^{\prime }\left(\alpha \right)=f^{\prime }\left(\alpha \right)-\frac{8\alpha }{{\pi }^2}=0\therefore f^{\prime }\left(\alpha \right)=\frac{8\alpha }{{\pi }^2}\therefore$ True