Question

Let,\(f(\theta)=\frac{sin^4\theta+3cos^2\theta}{sin^4\theta+cos^2\theta},\)then range of f(θ) ∈ [a, b]. The sum of infinite G.P., where first term is 64 and common ratio is a/b is equal to:

• A. 32
• B. 64
• C. 96
• D. 108

Answer 1

The Correct Option is C

To solve the given problem, we need to find the sum of the infinite geometric progression (G.P.) where the first term is 64 and the common ratio is \( \frac{a}{b} \). First, however, we need to determine the values of \(a\) and \(b\) by finding the range of the function \( f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta} \).

Step 1: Simplify the function \( f(\theta) \):

Let \( x = \sin^2\theta \). Hence, \( \cos^2\theta = 1-x \).

Substitute these into \( f(\theta) \):

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

We simplify the numerator and denominator:

Numerator: \( x^2 + 3(1-x) = x^2 + 3 - 3x \)

Denominator: \( x^2 + 1 - x = x^2 - x + 1 \)

Thus,

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

Step 2: Analyze the function:

To find the range, we need to find extreme values. We do this by finding the derivatives or analyzing boundary behavior (consider the limits as \( x \rightarrow 0 \) and \( x \rightarrow 1 \)).

Calculate the limits:

• As \( x \rightarrow 0 \), \( f(x) \approx \frac{3}{1} = 3 \).
• As \( x \rightarrow 1 \), \( f(x) \approx \frac{1}{1} = 1 \).

Conclusion on range:

The function \( f(x) \) decreases from 3 to 1 as \( x \) goes from 0 to 1. Thus, a = 1 and b = 3.

Step 3: Calculate the sum of the infinite G.P.:

The first term \( a_1 = 64 \) and the common ratio \( r = \frac{a}{b} = \frac{1}{3} \).

The sum \( S \) of an infinite G.P. is given by:

S = \frac{a_1}{1 - r}

Substitute the values:

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

Final Answer:

The sum of the infinite G.P. is 96. Hence, the correct answer is 96.