QJEEM24Mad1be: JEE Main 2024 Mathematics - Class 11 JEEM24Mad1be Solution: If the range of is , then the sum of the infinite G.P., whose…

To determine the range of the function $f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}$ , let's start by simplifying the expression. Let $\sin^2 \theta = x$ , so $\cos^2 \theta = 1 - x$ . The function then transforms into:

$f(x) = \frac{x^2 + 3(1-x)}{x^2 + (1-x)}$

Simplify the numerator and denominator separately:

Numerator: $x^2 + 3 - 3x$

Denominator: $x^2 + 1 - x$

Thus, the function becomes:

$f(x) = \frac{x^2 + 3 - 3x}{x^2 + 1 - x} = \frac{x^2 - 3x + 3}{x^2 - x + 1}$

Next, we need to find the range of this function. Let's analyze the function by testing specific values of $x$ in the interval $[0, 1]$ (since $\sin^2 \theta$ ranges from 0 to 1):

When $x = 0$ :

$f(0) = \frac{0^2 - 3(0) + 3}{0^2 - 0 + 1} = \frac{3}{1} = 3$

When $x = 1$ :

$f(1) = \frac{1^2 - 3(1) + 3}{1^2 - 1 + 1} = \frac{1 - 3 + 3}{1 - 1 + 1} = \frac{1}{1} = 1$

It appears that $f(x)$ achieves values within $[1, 3]$ . To confirm this, we need to solve the quadratic inequality:

$1 \leq \frac{x^2 - 3x + 3}{x^2 - x + 1} \leq 3$

By solving the inequalities, it can be confirmed that the function indeed ranges from 1 to 3 on the interval [0,1]. Hence, we have:

$\alpha = 1$

$\beta = 3$

The common ratio of the infinite geometric progression is:

$\frac{\alpha}{\beta} = \frac{1}{3}$

Given the first term $a = 64$ , the sum $S$ of the infinite geometric progression can be given as:

$S = \frac{a}{1 - r}$

Substituting the values $a = 64$ and $r = \frac{1}{3}$ , we get:

$S = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \times \frac{3}{2} = 96$

Therefore, the sum of the infinite geometric progression is 96.