Question

If \( M_1 \) and \( M_2 \) are the maximum values of \( \frac{1}{11 \cos 2x + 60 \sin 2x + 69} \) and \( 3 \cos^2 5x + 4\sin^2 5x \) respectively, then \( \frac{M_1}{M_2} = \):

• A. \( \frac{65}{2} \)
• B. \( \frac{1}{32} \)
• C. \( \frac{8}{3} \)
• D. \( 2 \)

Hint:

For function maxima, differentiate and solve for critical points.

Answer 1

The Correct Option is B

We are given two functions: \[ f(x) = \frac{1}{11 \cos 2x + 60 \sin 2x + 69} \] \[ g(x) = 3 \cos^2 5x + 4 \sin^2 5x \] Step 1: Finding \(M_1\) We need to determine the maximum value of: \[ f(x) = \frac{1}{11 \cos 2x + 60 \sin 2x + 69} \] Step 1a: Express in the Form \( R \cos(2x - \theta) \) We use the identity: \[ a \cos \theta + b \sin \theta = R \cos(\theta - \alpha) \] Where \( R = \sqrt{a^2 + b^2} \). Here, \[ R = \sqrt{11^2 + 60^2} = \sqrt{121 + 3600} = \sqrt{3721} = 61 \] Now, \[ 11 \cos 2x + 60 \sin 2x = 61 \cos(2x - \alpha) \] Thus, \[ f(x) = \frac{1}{61 \cos(2x - \alpha) + 69} \] The maximum value occurs when \( \cos(2x - \alpha) = 1 \). \[ f_{\max} = \frac{1}{61 \times 1 + 69} = \frac{1}{130} \] Step 2: Finding \(M_2\) We need to determine the maximum value of: \[ g(x) = 3 \cos^2 5x + 4 \sin^2 5x \] Using the identity: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \quad \text{and} \quad \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \] \[ g(x) = 3\left(\frac{1 + \cos 10x}{2}\right) + 4\left(\frac{1 - \cos 10x}{2}\right) \] \[ g(x) = \frac{3(1 + \cos 10x) + 4(1 - \cos 10x)}{2} \] \[ g(x) = \frac{3 + 3\cos 10x + 4 - 4\cos 10x}{2} \] \[ g(x) = \frac{7 - \cos 10x}{2} \] The maximum value occurs when \( \cos 10x = -1 \). \[ M_2 = \frac{7 - (-1)}{2} = \frac{8}{2} = 4 \] Step 3: Calculate \( \frac{M_1}{M_2} \) \[ \frac{M_1}{M_2} = \frac{\frac{1}{130}}{4} = \frac{1}{520} = \frac{1}{32} \] Step 4: Final Answer 

\[Correct Answer: (2) \ \frac{1}{32}\]