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 $\min_{0<\theta<\frac{\pi}{2}} f(\theta) = \max_{0<\theta<\pi} g(\theta)$. 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 ________.

Hint:

Apply the AM-GM inequality to find the minimum of f(theta) and analyze the range of sin^2(theta) to find the maximum of g(theta). Then use the GP sum formula.

Answer 1

Correct Answer: 1279

The problem involves finding the minimum of a trigonometric function $f(\theta)$ and the maximum of another function $g(\theta)$, then relating them to a Geometric Progression (G.P.).

Step 1: Find the minimum value of $f(\theta) = \alpha \tan^{2}\theta + \beta \cot^{2}\theta$ for $0<\theta<\frac{\pi}{2}$.
Since $\alpha, \beta, \tan^{2}\theta$, and $\cot^{2}\theta$ are all positive in the given interval, we can use the AM-GM (Arithmetic Mean - Geometric Mean) inequality:
$$\frac{\alpha \tan^{2}\theta + \beta \cot^{2}\theta}{2} \geq \sqrt{(\alpha \tan^{2}\theta) (\beta \cot^{2}\theta)}$$
$$\frac{f(\theta)}{2} \geq \sqrt{\alpha \beta (\tan \theta \cot \theta)^2}$$
Since $\tan \theta \cot \theta = 1$, we have $f(\theta) \geq 2\sqrt{\alpha \beta}$.
Thus, $\min f(\theta) = 2\sqrt{\alpha \beta}$.

Step 2: Find the maximum value of $g(\theta) = \alpha \sin^{2}\theta + \beta \cos^{2}\theta$ for $0<\theta<\pi$.
We can rewrite $g(\theta)$ as:
$g(\theta) = \alpha \sin^{2}\theta + \beta (1 - \sin^{2}\theta) = (\alpha - \beta) \sin^{2}\theta + \beta$.
Since $\alpha>\beta$, the term $(\alpha - \beta)$ is positive. The maximum value of $\sin^{2}\theta$ in the interval $(0, \pi)$ is 1 (which occurs at $\theta = \frac{\pi}{2}$).
Thus, $\max g(\theta) = (\alpha - \beta)(1) + \beta = \alpha$.

Step 3: Equate the two values as given in the question:
$\min f(\theta) = \max g(\theta) \implies 2\sqrt{\alpha \beta} = \alpha$.
Squaring both sides gives $4\alpha \beta = \alpha^{2}$.
Since $\alpha>0$, we divide by $\alpha$ to get $\alpha = 4\beta$.
This implies $\frac{\alpha}{\beta} = 4$.

Step 4: Determine the properties of the G.P.
First term $a = \frac{\alpha}{2\beta} = \frac{1}{2} \left( \frac{\alpha}{\beta} \right) = \frac{4}{2} = 2$.
Common ratio $r = \frac{2\beta}{\alpha} = 2 \left( \frac{\beta}{\alpha} \right) = 2 \left( \frac{1}{4} \right) = \frac{1}{2}$.

Step 5: Calculate the sum of the first 10 terms ($S_{10}$).
The sum formula for a G.P. is $S_n = a \frac{1 - r^n}{1 - r}$.
$$S_{10} = 2 \frac{1 - (1/2)^{10}}{1 - 1/2} = 2 \frac{1 - 1/1024}{1/2} = 4 \left( \frac{1023}{1024} \right) = \frac{1023}{256}$$
Given $S_{10} = \frac{m}{n}$ and $\gcd(m, n) = 1$.
$m = 1023$, $n = 256$.
Since $1023 = 3 \times 11 \times 31$ and $256 = 2^8$, their greatest common divisor is indeed 1.

Step 6: Calculate $m + n$.
$m + n = 1023 + 256 = 1279$.