Question
For the functions $f(\theta) = \alpha\tan^{2}\theta + \beta\cot^{2}\theta$, and
$g(\theta) = \alpha\sin^{2}\theta + \beta\cos^{2}\theta,\alpha > \beta > 0$, let $\min_{0 < \theta < \frac{\pi}{2}}\mspace{2mu} f(\theta) = \max_{0 < \theta < \pi}\mspace{2mu} g(\theta)$. If the first term of a G.P. is $\left( \frac{\alpha}{2\beta} \right)$, its common ratio is $\left( \frac{2\beta}{\alpha} \right)$ and the sum of its first 10 terms is $\frac{m}{n},gcd(m,n) = 1$, then $m + n$ is equal to $\_\_\_\_$ .
Solution
Sol. [Key Concept: By AM-GM $min\ f = 2\sqrt{\alpha\beta}$ and $max\ g = \alpha$; equating gives $\alpha = 4\beta$, so the G.P. has first term $2$ and ratio $\frac{1}{2}$; then $S_{10} = \frac{1023}{256} = \frac{m}{n}$.]
$${\left. \ f(\theta) \right|_{\min} = 2\sqrt{\alpha\beta}(AM \geq GM) }{g(\theta) = \alpha\sin^{2}\theta + \beta\left( 1 - \sin^{2}\theta \right) = (\alpha - \beta)\sin^{2}\theta + \beta }{\left. \ \therefore g(\theta) \right|_{\max} = \alpha }$$Equation $2\sqrt{\alpha\beta} = \alpha$
$$\begin{matrix} & 4\alpha\beta = \alpha^{2} \\ & \alpha = 4\beta(\alpha \neq 0) \end{matrix}$$
Now first term $\left( \frac{\alpha}{2\beta} \right) = 2$ and common ratio
$${\left( \frac{2\beta}{\alpha} \right) = \frac{1}{2} }{S_{10} = \frac{a\left( 1 - r^{10} \right)}{1 - r} = \frac{1023}{256} = \frac{m}{n} }{m + n = 1279 }$$
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