Question

If \( \cos^{-1}(2x) + \cos^{-1}(3x) = \frac{\pi}{3} \) and \( 4x^2 = \frac{a}{b} \), then \( a + b = \):

• A. \( 12 \)
• B. \( 11 \)
• C. \( 31 \)
• D. \( 10 \)

Hint:

For trigonometric equations involving inverse trigonometric functions, apply known identities and simplify step-by-step. In particular, be mindful of squaring both sides to eliminate square roots.

Answer 1

The Correct Option is D

Step 1: We are given the equation \( \cos^{-1}(2x) + \cos^{-1}(3x) = \frac{\pi}{3} \). To simplify this, use the identity for the sum of inverse cosines: \[ \cos^{-1}(A) + \cos^{-1}(B) = \cos^{-1}(AB - \sqrt{(1-A^2)(1-B^2)}) \] Substituting \( A = 2x \) and \( B = 3x \): \[ \cos^{-1}(2x) + \cos^{-1}(3x) = \cos^{-1}\left( 2x \cdot 3x - \sqrt{(1 - (2x)^2)(1 - (3x)^2)} \right) \] We are told that this is equal to \( \frac{\pi}{3} \), so: \[ \cos^{-1}\left( 6x^2 - \sqrt{(1 - 4x^2)(1 - 9x^2)} \right) = \frac{\pi}{3} \] Taking the cosine of both sides, we get: \[ 6x^2 - \sqrt{(1 - 4x^2)(1 - 9x^2)} = \frac{1}{2} \] Step 2: Now, simplify the equation. First isolate the square root term: \[ \sqrt{(1 - 4x^2)(1 - 9x^2)} = 6x^2 - \frac{1}{2} \] Square both sides: \[ (1 - 4x^2)(1 - 9x^2) = \left( 6x^2 - \frac{1}{2} \right)^2 \] Expanding both sides: \[ (1 - 4x^2)(1 - 9x^2) = 1 - 9x^2 - 4x^2 + 36x^4 = 1 - 13x^2 + 36x^4 \] \[ \left( 6x^2 - \frac{1}{2} \right)^2 = 36x^4 - 6x^2 + \frac{1}{4} \] Equating both sides: \[ 1 - 13x^2 + 36x^4 = 36x^4 - 6x^2 + \frac{1}{4} \] Simplifying: \[ 1 - 13x^2 = -6x^2 + \frac{1}{4} \] \[ 1 - 13x^2 + 6x^2 = \frac{1}{4} \] \[ 1 - 7x^2 = \frac{1}{4} \] \[ 7x^2 = \frac{3}{4} \] \[ x^2 = \frac{3}{28} \] Step 3: Substitute this value of \( x^2 \) into the equation \( 4x^2 = \frac{a}{b} \): \[ 4x^2 = 4 \cdot \frac{3}{28} = \frac{12}{28} = \frac{3}{7} \] Thus, \( \frac{a}{b} = \frac{3}{7} \), so \( a = 3 \) and \( b = 7 \). 
Step 4: Finally, calculate \( a + b \): \[ a + b = 3 + 7 = 10 \]