Let `f (x= sin ^(6) ((x )/(4)) + cos ^(6) ((x)/(4)). If f ^(n) (x)` denotes `n ^(th)` derivative of f evaluated at x. Then which of the following hold ?

Options

A. `f ^(2014) (0) =-3/8`
B. `f ^(2015) (0) =3/8`
C. `f ^(2010) ((pi)/(2))=0`
D. `f ^(2011) ((pi)/(2)) =3/8`

Answer

To solve the problem, we need to analyze the function given and find its derivatives. Let's go through the steps systematically.

### Step 1: Define the function
We are given the function:
\[ f(x) = \sin^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right) \]

### Step 2: Simplify the function
Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we can express \( f(x) \) as:
Let \( a = \sin^2\left(\frac{x}{4}\right) \) and \( b = \cos^2\left(\frac{x}{4}\right) \). Then,
\[ f(x) = a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Since \( a + b = 1 \), we have:
\[ f(x) = 1 \cdot \left( a^2 - ab + b^2 \right) = a^2 - ab + b^2 \]
Now, substituting back:
\[ f(x) = \sin^4\left(\frac{x}{4}\right) - \sin^2\left(\frac{x}{4}\right)\cos^2\left(\frac{x}{4}\right) + \cos^4\left(\frac{x}{4}\right) \]

### Step 3: Further simplify using trigonometric identities
Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[ f(x) = \sin^4\left(\frac{x}{4}\right) + \cos^4\left(\frac{x}{4}\right) - \sin^2\left(\frac{x}{4}\right)\cos^2\left(\frac{x}{4}\right) \]
We can express \( \sin^4\theta + \cos^4\theta \) as:
\[ \sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1 - 2\sin^2\theta\cos^2\theta \]
Thus,
\[ f(x) = 1 - 3\sin^2\left(\frac{x}{4}\right)\cos^2\left(\frac{x}{4}\right) \]

### Step 4: Use double angle identity
Using the double angle identity \( \sin(2\theta) = 2\sin\theta\cos\theta \):
\[ \sin^2\left(\frac{x}{4}\right)\cos^2\left(\frac{x}{4}\right) = \frac{1}{4}\sin^2\left(\frac{x}{2}\right) \]
So,
\[ f(x) = 1 - \frac{3}{4}\cdot\frac{1}{4}\sin^2\left(\frac{x}{2}\right) = 1 - \frac{3}{16}\sin^2\left(\frac{x}{2}\right) \]

### Step 5: Find the nth derivative
Now we need to find the nth derivative of \( f(x) \). The function is periodic due to the sine function. The derivatives will cycle through a pattern based on the periodicity of sine and cosine functions.

1. **First derivative**:
\[ f'(x) = -\frac{3}{16}\cdot\frac{1}{2}\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = -\frac{3}{32}\sin(x) \]
2. **Second derivative**:
\[ f''(x) = -\frac{3}{32}\cos(x) \]
3. **Third derivative**:
\[ f'''(x) = \frac{3}{32}\sin(x) \]
4. **Fourth derivative**:
\[ f^{(4)}(x) = \frac{3}{32}\cos(x) \]

### Step 6: Identify the pattern
From the derivatives, we can see that:
- \( f^{(n)}(x) \) will cycle every four derivatives:
- \( f^{(0)}(x) = f(x) \)
- \( f^{(1)}(x) = -\frac{3}{32}\sin(x) \)
- \( f^{(2)}(x) = -\frac{3}{32}\cos(x) \)
- \( f^{(3)}(x) = \frac{3}{32}\sin(x) \)
- \( f^{(4)}(x) = \frac{3}{32}\cos(x) \)

### Conclusion
Thus, we can conclude that:
- For \( n \equiv 0 \mod 4 \): \( f^{(n)}(x) = f(x) \)
- For \( n \equiv 1 \mod 4 \): \( f^{(n)}(x) = -\frac{3}{32}\sin(x) \)
- For \( n \equiv 2 \mod 4 \): \( f^{(n)}(x) = -\frac{3}{32}\cos(x) \)
- For \( n \equiv 3 \mod 4 \): \( f^{(n)}(x) = \frac{3}{32}\sin(x) \)