Let `f_K(x)""=1/k(s in^k x+cos^k x)` where `x in R` and `kgeq1` . Then `f_4(x)-f_6(x)` equals (1) `1/6` (2) `1/3` (3) `1/4` (4) `1/(12)`

Options

A. `1/6`
B. `1/3`
C. `1/4`
D. `1/12`

Explanation

To solve the problem, we need to find \( f_4(x) - f_6(x) \) where:

\[ f_k(x) = \frac{1}{k} (\sin^k x + \cos^k x) \]

### Step 1: Write down the expressions for \( f_4(x) \) and \( f_6(x) \)

Using the definition of \( f_k(x) \):

\[ f_4(x) = \frac{1}{4} (\sin^4 x + \cos^4 x) \]

\[ f_6(x) = \frac{1}{6} (\sin^6 x + \cos^6 x) \]

### Step 2: Set up the expression for \( f_4(x) - f_6(x) \)

Now, we can express \( f_4(x) - f_6(x) \):

\[ f_4(x) - f_6(x) = \frac{1}{4} (\sin^4 x + \cos^4 x) - \frac{1}{6} (\sin^6 x + \cos^6 x) \]

### Step 3: Find a common denominator

The common denominator for \( 4 \) and \( 6 \) is \( 12 \). We rewrite the expression with this common denominator:

\[ f_4(x) - f_6(x) = \frac{3}{12} (\sin^4 x + \cos^4 x) - \frac{2}{12} (\sin^6 x + \cos^6 x) \]

### Step 4: Combine the fractions

Now we can combine the fractions:

\[ f_4(x) - f_6(x) = \frac{3(\sin^4 x + \cos^4 x) - 2(\sin^6 x + \cos^6 x)}{12} \]

### Step 5: Simplify the numerator

Using the identity \( a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \) and \( a^6 + b^6 = (a^2 + b^2)(a^4 - a^2b^2 + b^4) \):

Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then:

\[ \sin^4 x + \cos^4 x = (a^2 + b^2)^2 - 2a^2b^2 = 1 - 2a^2b^2 \]

\[ \sin^6 x + \cos^6 x = (a + b)(a^4 - a^2b^2 + b^4) = (1)(1 - 3a^2b^2) = 1 - 3a^2b^2 \]

Substituting these into our expression:

\[ f_4(x) - f_6(x) = \frac{3(1 - 2a^2b^2) - 2(1 - 3a^2b^2)}{12} \]

### Step 6: Simplify further

Now simplify the numerator:

\[ = \frac{3 - 6a^2b^2 - 2 + 6a^2b^2}{12} \]

\[ = \frac{1}{12} \]

### Final Answer

Thus, we find that:

\[ f_4(x) - f_6(x) = \frac{1}{12} \]

The correct answer is option (4) \( \frac{1}{12} \).

---