Question

Let \( f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x \) & maximum value of f is \( \alpha \). If area bounded between \( g(x) = x^2 \) & \( h(x) = \beta x^3 \) (\( \beta >" 0 \)) is \( \alpha^2 \), then \( 30\beta^3 \) is equal to:

• A. 14
• B. 16
• C. 20
• D. 22

Hint:

For functional equations of the type \( af(x) + bf(g(x)) = h(x) \) where \( g(g(x)) = x \), you can always solve for \( f(x) \) by creating a $2 \times 2$ system of equations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We first solve the functional equation to find \( f(x) \). By substituting \( \left(\frac{\pi}{2} - x\right) \) for \( x \), we get a system of two equations. After finding \( \alpha \), we set up the area integral for the intersection of the two curves.
Step 2: Key Formula or Approach:
1. Solve for \( f(x) \) using substitution.
2. Area \( A = \int |g(x) - h(x)| \, dx = \alpha^2 \).
Step 3: Detailed Explanation:
1. \( f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x \dots (1) \)
2. Replace \( x \) with \( \frac{\pi}{2} - x \): \( f\left(\frac{\pi}{2} - x\right) + 3f(x) = \cos x \dots (2) \)
3. Multiply (2) by 3: \( 3f\left(\frac{\pi}{2} - x\right) + 9f(x) = 3\cos x \)
4. Subtract (1) from this: \( 8f(x) = 3\cos x - \sin x \)
5. \( f(x) = \frac{3\cos x - \sin x}{8} \). Maximum value \( \alpha = \frac{\sqrt{3^2 + 1^2}}{8} = \frac{\sqrt{10}}{8} \).
6. Area between \( x^2 \) and \( \beta x^3 \): Intersection at \( x = 0 \) and \( x = \frac{1}{\beta} \).
7. Area: \[ \int_0^{1/\beta} (x^2 - \beta x^3)\, dx = \left[\frac{x^3}{3} - \frac{\beta x^4}{4}\right]_0^{1/\beta} = \frac{1}{3\beta^3} - \frac{1}{4\beta^3} = \frac{1}{12\beta^3} \] 8. \( \frac{1}{12\beta^3} = \alpha^2 = \frac{10}{64} = \frac{5}{32} \).
9. \( \beta^3 = \frac{32}{60} = \frac{8}{15} \).
10. \( 30\beta^3 = 30 \times \frac{8}{15} = 16 \).
Step 4: Final Answer:
The value of \( 30\beta^3 \) is 16.