Statement I `phi(x)=int_(0)^(x)(3 sin t+4 cos t)dt,[(pi)/(6),(pi)/(3)]phi(x)-`
attains its maximum value at `x=(pi)/(3).`
Statement II `phi(x)=int_(0)^(x)(3sint+4cost)dt,phi(x)` is
increasing function in `[(pi)/(6),(pi)/(3)]`

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of statement I.
B. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.
C. Statement I is true, Statement II is false
D. Statement I is false, Statement II is true

Answer

To solve the problem, we need to analyze the function \( \phi(x) \) defined as:

\[ \phi(x) = \int_{0}^{x} (3 \sin t + 4 \cos t) \, dt \]

### Step 1: Evaluate the Integral
First, we need to compute the integral \( \phi(x) \).

\[ \phi(x) = \int_{0}^{x} (3 \sin t + 4 \cos t) \, dt \]

Using the properties of integrals, we can separate the integral:

\[ \phi(x) = \int_{0}^{x} 3 \sin t \, dt + \int_{0}^{x} 4 \cos t \, dt \]

Now, we compute each integral separately:

1. The integral of \( 3 \sin t \):

\[ \int 3 \sin t \, dt = -3 \cos t \]

Evaluating from \( 0 \) to \( x \):

\[ \left[-3 \cos t \right]_{0}^{x} = -3 \cos x + 3 \cos 0 = -3 \cos x + 3 \]

2. The integral of \( 4 \cos t \):

\[ \int 4 \cos t \, dt = 4 \sin t \]

Evaluating from \( 0 \) to \( x \):

\[ \left[4 \sin t \right]_{0}^{x} = 4 \sin x - 0 = 4 \sin x \]

Combining both results, we have:

\[ \phi(x) = (-3 \cos x + 3) + 4 \sin x = 3 - 3 \cos x + 4 \sin x \]

### Step 2: Differentiate \( \phi(x) \)
Next, we differentiate \( \phi(x) \) to analyze its monotonicity:

\[ \phi'(x) = \frac{d}{dx}(3 - 3 \cos x + 4 \sin x) \]

Using the derivatives of sine and cosine:

\[ \phi'(x) = 0 + 3 \sin x + 4 \cos x \]

### Step 3: Analyze the Sign of \( \phi'(x) \)
We need to check the sign of \( \phi'(x) \) in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{3}\right] \).

1. At \( x = \frac{\pi}{6} \):

\[ \phi'\left(\frac{\pi}{6}\right) = 3 \sin\left(\frac{\pi}{6}\right) + 4 \cos\left(\frac{\pi}{6}\right) = 3 \cdot \frac{1}{2} + 4 \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} + 2\sqrt{3} \]

2. At \( x = \frac{\pi}{3} \):

\[ \phi'\left(\frac{\pi}{3}\right) = 3 \sin\left(\frac{\pi}{3}\right) + 4 \cos\left(\frac{\pi}{3}\right) = 3 \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2} = \frac{3\sqrt{3}}{2} + 2 \]

Both values are positive since \( \sin x \) and \( \cos x \) are positive in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{3}\right] \).

### Conclusion
Since \( \phi'(x) > 0 \) in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{3}\right] \), \( \phi(x) \) is increasing in this interval. Therefore, \( \phi(x) \) attains its maximum value at \( x = \frac{\pi}{3} \).

### Final Statements
- **Statement I**: True, \( \phi(x) \) attains its maximum value at \( x = \frac{\pi}{3} \).
- **Statement II**: True, \( \phi(x) \) is an increasing function in the interval \( \left[\frac{\pi}{6}, \frac{\pi}{3}\right] \).