JEE Main 2024 — Functions Question with Solution

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:

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

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.