Consider f, g and h be three real valued function defined on R. Let `f(x)=sin3x+cosx,g(x)=cos3x+sinx` and `h(x)=f^(2)(x)+g^(2)(x).` Then,
The length of a longest interval in which the function h(x) is increasing, is

Options

A. `pi//8`
B. `pi//4`
C. `pi//6`
D. `pi//2`

Answer

To solve the problem, we need to find the length of the longest interval in which the function \( h(x) = f^2(x) + g^2(x) \) is increasing. We will follow these steps:

### Step 1: Find the derivatives of \( f(x) \) and \( g(x) \)

Given:
- \( f(x) = \sin(3x) + \cos(x) \)
- \( g(x) = \cos(3x) + \sin(x) \)

First, we compute the derivatives:

1. **Derivative of \( f(x) \)**:
\[ f'(x) = \frac{d}{dx}(\sin(3x)) + \frac{d}{dx}(\cos(x)) = 3\cos(3x) - \sin(x) \]

2. **Derivative of \( g(x) \)**:
\[ g'(x) = \frac{d}{dx}(\cos(3x)) + \frac{d}{dx}(\sin(x)) = -3\sin(3x) + \cos(x) \]

### Step 2: Compute \( h'(x) \)

Using the chain rule, the derivative of \( h(x) \) is:
\[ h'(x) = 2f(x)f'(x) + 2g(x)g'(x) \]

Substituting \( f(x) \) and \( g(x) \):
\[ h'(x) = 2(\sin(3x) + \cos(x))(3\cos(3x) - \sin(x)) + 2(\cos(3x) + \sin(x))(-3\sin(3x) + \cos(x)) \]

### Step 3: Simplify \( h'(x) \)

Expanding \( h'(x) \):
\[ h'(x) = 2\left[ (\sin(3x) + \cos(x))(3\cos(3x) - \sin(x)) + (\cos(3x) + \sin(x))(-3\sin(3x) + \cos(x)) \right] \]

Now, we will simplify this expression step by step:
1. Distributing the first term:
\[ = 6\sin(3x)\cos(3x) - 2\sin(3x)\sin(x) + 3\cos(x)\cos(3x) - \cos(x)\sin(x) \]
2. Distributing the second term:
\[ -3\cos(3x)\sin(3x) - 3\sin^2(x) + \cos^2(x) + \sin(x)\cos(3x) \]

Combining like terms, we find:
\[ h'(x) = 2\left[ 3\sin(3x)\cos(3x) - 3\sin^2(x) + \cos^2(x) + \sin(x)\cos(3x) - \sin(3x)\sin(x) \right] \]

### Step 4: Determine when \( h'(x) > 0 \)

To find the intervals where \( h(x) \) is increasing, we need to solve \( h'(x) > 0 \).

Using the property of the cosine function:
\[ h'(x) > 0 \implies \cos(4x) > 0 \]

### Step 5: Find the intervals for \( \cos(4x) > 0 \)

The cosine function is positive in the intervals:
\[ 4x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{2}) \ldots \]

Dividing by 4 gives:
\[ x \in (-\frac{\pi}{8}, \frac{\pi}{8}) \cup (\frac{3\pi}{8}, \frac{5\pi}{8}) \ldots \]

### Step 6: Calculate the length of the longest interval

The length of the interval \( (-\frac{\pi}{8}, \frac{\pi}{8}) \) is:
\[ \text{Length} = \frac{\pi}{8} - (-\frac{\pi}{8}) = \frac{\pi}{8} + \frac{\pi}{8} = \frac{\pi}{4} \]

### Final Answer

The length of the longest interval in which the function \( h(x) \) is increasing is:
\[ \boxed{\frac{\pi}{4}} \]