Question

If \( f_n(x) = \frac{1}{2n} \left[ \sin^{2n}x + \cos^{2n}x \right] \), then \( f_1(x) + f_2(x) - f_3(x) = \):

• A. \( 0 \)
• B. \( \frac{5}{12} \)
• C. \( \frac{11}{12} \)
• D. \( \frac{7}{12} \)

Hint:

When dealing with expressions involving trigonometric powers, you can use known average values over a period and identities like: - \( \sin^2x + \cos^2x = 1 \) - \( \sin^4x + \cos^4x = 1 - \frac{1}{2}\sin^22x \) - For integration or average over a period, use average values of periodic functions.

Answer 1

The Correct Option is D

Step 1: Use the given formula \( f_n(x) = \frac{1{2n}(\sin^{2n}x + \cos^{2n}x) \).}
We evaluate each term separately: \[ f_1(x) = \frac{1}{2}(\sin^2x + \cos^2x) = \frac{1}{2}(1) = \frac{1}{2} \] \[ f_2(x) = \frac{1}{4}(\sin^4x + \cos^4x) \] Now use the identity: \( \sin^4x + \cos^4x = 1 - \frac{1}{2}\sin^22x \), and the average value of \( \sin^22x \) over a period is \( \frac{1}{2} \), so: \[ \sin^4x + \cos^4x = 1 - \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} \Rightarrow f_2(x) = \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16} \] \[ f_3(x) = \frac{1}{6}(\sin^6x + \cos^6x) \] Use identity: \[ \sin^6x + \cos^6x = 1 - \frac{3}{4}\sin^2 2x \Rightarrow \text{mean value over a period } = 1 - \frac{3}{4} \cdot \frac{1}{2} = \frac{5}{8} \Rightarrow f_3(x) = \frac{1}{6} \cdot \frac{5}{8} = \frac{5}{48} \]
Step 2: Add and subtract the required expressions.
\[ f_1(x) + f_2(x) - f_3(x) = \frac{1}{2} + \frac{3}{16} - \frac{5}{48} \]
Step 3: Take LCM and simplify.
LCM of 2, 16, and 48 is 48: \[ = \frac{24}{48} + \frac{9}{48} - \frac{5}{48} = \frac{28}{48} = \frac{7}{12} \]