Question

Let \(f_k(x) = \frac{1}{k}(cos^kx+sin^kx)\) where \(k\in N\), then \(f_6(x)-f_4(x) = \ldots\)

• A. \(\frac{1}{6}\)
• B. \(-\frac{1}{12}\)
• C. \(-\frac{1}{6}\)
• D. \(\frac{1}{12}\)

Hint:

Write cos^6+sin^6 and cos^4+sin^4 in terms of sin^2 x cos^2 x.

Answer 1

The Correct Option is B

Step 1: Test with a convenient value of x:
The answer is a constant, so any value of $x$ works. Take $x = 0$.
$f_6(0) = \frac{1}{6}(1 + 0) = \frac{1}{6}$ and $f_4(0) = \frac{1}{4}(1+0) = \frac{1}{4}$.

Step 2: Subtract:
$f_6 - f_4 = \frac{1}{6} - \frac{1}{4} = -\frac{1}{12}$.

Step 3: Cross-check at x = pi/4:
$\sin^2 = \cos^2 = \frac12$. Then $\cos^6+\sin^6 = 2\cdot\frac18 = \frac14$ and $\cos^4+\sin^4 = 2\cdot\frac14 = \frac12$. So $f_6 - f_4 = \frac{1}{24} - \frac{1}{8} = -\frac{1}{12}$. The same value confirms that it is constant.

Final Answer:
$-\dfrac{1}{12}$, option (B). \[ \boxed{-\frac{1}{12} \text{ (B)}} \]