Question

A function $f:R-\left\{\right(2k-1\left)\pi \right\}\to R$ is defined as, $f\left(x\right)=\frac{1}{\sqrt{m^2-n^2}}\ln \left(\frac{\sqrt{m+n}+\sqrt{m-n}\tan \left(x/2\right)}{\sqrt{m+n}-\sqrt{m-n}\tan \left(x/2\right)}\right)$ where $k,m,n\in Z^{+}$ and $n<m$
If $f^{\prime }\left(\frac{\pi }{3}\right)=\frac{1}{9}$, then which of the following ordered pairs of $(m,n)$ is/are correct?

• A. $(4,6)$
• B. $(6,6)$
• C. $(8,5)$
• D. $(10,4)$

Answer 1

Answer: (B), (C), (D)

$(10,4)$

$f\left(x\right)=\frac{1}{\sqrt{m^2-n^2}}\left[\ln \right(\sqrt{m+n}+\sqrt{m-n}\tan \left(x/2\right))-\ln (\sqrt{m+n}-\sqrt{m-n}\tan \left(x/2\right)\left)\right]$
$f^{\prime }\left(x\right)=\frac{1}{\sqrt{m^2-n^2}}[\frac{\frac{1}{2}\sqrt{m-n}{\sec }^2\frac{x}{2}}{\sqrt{m+n}+\sqrt{m-n}\tan \frac{x}{2}}-\frac{-\frac{1}{2}\sqrt{m-n}{\sec }^2\frac{x}{2}}{\sqrt{m+n}-\sqrt{m-n}\tan \frac{x}{2}}]$

$=\frac{\frac{1}{2}\sqrt{m-n}{\sec }^2\frac{x}{2}}{\sqrt{m^2-n^2}}(\frac{1}{\sqrt{m+n}+\sqrt{m-n}\tan \frac{x}{2}}+\frac{1}{\sqrt{m+n}-\sqrt{m-n}\tan \frac{x}{2}})$

$=\frac{{\sec }^2\frac{x}{2}}{2\sqrt{m+n}}\left(\frac{2\sqrt{m+n}}{(m+n)-(m-n){\tan }^2\frac{x}{2}}\right)$

$f^{\prime }\left(\frac{\pi }{3}\right)=\frac{{\sec }^2\left(\frac{\pi }{6}\right)}{(m+n)-(m-n){\tan }^2\left(\frac{\pi }{6}\right)}$

$\Rightarrow \frac{\frac{4}{3}}{(m+n)-\frac{m-n}{3}}=\frac{1}{9}\Rightarrow 36=2m+4n$
$\Rightarrow m+2n=18$

Only $(m,n)\equiv (6,6),(8,5),(10,4)$ satisfied the above equation.