Question

If the range of $f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta}, \, \theta \in \mathbb{R}$ is $[\alpha, \beta]$, then the sum of the infinite G.P., whose first term is $64$ and the common ratio is $\frac{\alpha}{\beta}$, is equal to ____.

Answer 1

Correct Answer: 96

The range of \( f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta} \) is determined by rewriting it as \(f(\theta) = \frac{x^2 + 3(1-x)}{x^2 + 1-x}\) where \( x = \sin^2\theta \), with \(0 \leq x \leq 1\).

This simplifies to \(f(x) = \frac{x^2 - 3x + 3}{x^2 - x + 1}\).

The range of \( y = \frac{x^2 - 3x + 3}{x^2 - x + 1} \) for \(0 \leq x \leq 1\) is found to be \([1,3]\), so \(\alpha = 1\) and \(\beta = 3\).

The sum of an infinite geometric progression with first term \( a = 64 \) and common ratio \( r = \frac{\alpha}{\beta} = \frac{1}{3} \) is calculated using the formula \( S = \frac{a}{1 - r} \).

Substituting the values, \( S = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \times \frac{3}{2} = 96 \).

The sum of the G.P. is 96.