Question

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

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

Answer 1

Answer: (C), (D)

The correct options are
C $f\left(\alpha \right)f^{\prime }\left(\alpha \right)=\frac{1}{\pi }$ for at least one $\alpha \in (0,\frac{\pi }{2})$
D $f^{\prime }\left(\alpha \right)=\frac{8\alpha }{{\pi }^2}$ for at least one $\alpha \in (0,\frac{\pi }{2})$

$\to$
Let $g\left(x\right)={\sin }^{-1}\left(f\right(x\left)\right)-x$
$g\left(0\right)=g(\frac{\pi }{2})=0$
$g^{\prime }\left(\alpha \right)=\frac{f^{\prime }\left(\alpha \right)}{\sqrt{1-f^2\left(\alpha \right)}}-1=0$
So $f^{\prime }\left(\alpha \right)=\sqrt{1-f^2\left(\alpha \right)}$ for at least one value of $\alpha$ but may not be for all $\alpha \in (0,\frac{\pi }{2})$

$\to$
Let $g\left(x\right)=f\left(x\right)-\frac{2x}{\pi }$
$g\left(0\right)=g(\frac{\pi }{2})=0$
$g^{\prime }\left(\alpha \right)=f^{\prime }\left(\alpha \right)-\frac{2}{\pi }=0$
$\Rightarrow f^{\prime }\left(\alpha \right)=\frac{2}{\pi }$ for at least one value of $\alpha$ but may not be for all $\alpha \in (0,\frac{\pi }{2})$

$\to$
Let $g\left(x\right)=f^2\left(x\right)-\frac{2x}{\pi }$
$g\left(0\right)=g(\frac{\pi }{2})=0$
$g^{\prime }\left(\alpha \right)=2f\left(\alpha \right)f^{\prime }\left(\alpha \right)-\frac{2}{\pi }=0$
$\Rightarrow f\left(\alpha \right)f^{\prime }\left(\alpha \right)=\frac{1}{\pi }$ for at least one $\alpha \in (0,\frac{\pi }{2})$


$\to$

Let $g\left(x\right)=f\left(x\right)-\frac{4x^2}{{\pi }^2}$
$g\left(0\right)=g(\frac{\pi }{2})=0$
$g^{\prime }\left(\alpha \right)=f^{\prime }\left(\alpha \right)-\frac{8\alpha }{{\pi }^2}=0$
$\Rightarrow f^{\prime }\left(\alpha \right)=\frac{8\alpha }{{\pi }^2}$ for at least one $\alpha \in (0,\frac{\pi }{2})$