Let `f(x)=sinx+2cos^2x`, x `in [pi/6,(2pi)/3]`, then maximum value of f(x) is

Options

A. 1
B. `(3)/(2)`
C. 2
D. `(5)/(2)`

Answer

To find the maximum value of the function \( f(x) = \sin x + 2 \cos^2 x \) for \( x \) in the interval \( \left[\frac{\pi}{6}, \frac{2\pi}{3}\right] \), we can follow these steps:

### Step 1: Rewrite the function
We know that \( \cos^2 x = 1 - \sin^2 x \). Therefore, we can rewrite \( f(x) \) as:
\[ f(x) = \sin x + 2(1 - \sin^2 x) \]
This simplifies to:
\[ f(x) = \sin x + 2 - 2\sin^2 x \]
Thus, we have:
\[ f(x) = 2 - 2\sin^2 x + \sin x \]

### Step 2: Substitute \( \sin x \) with \( t \)
Let \( t = \sin x \). The range of \( t \) for \( x \) in \( \left[\frac{\pi}{6}, \frac{2\pi}{3}\right] \) is:
\[ t \in \left[\frac{1}{2}, 1\right] \]
Now, we can express \( f(t) \) as:
\[ f(t) = 2 - 2t^2 + t \]

### Step 3: Differentiate \( f(t) \)
To find the maximum value, we need to differentiate \( f(t) \) with respect to \( t \):
\[ f'(t) = -4t + 1 \]
Setting the derivative equal to zero to find critical points:
\[ -4t + 1 = 0 \implies t = \frac{1}{4} \]
However, \( t = \frac{1}{4} \) is outside the range \( \left[\frac{1}{2}, 1\right] \).

### Step 4: Evaluate \( f(t) \) at the endpoints
Since the critical point is not in the interval, we evaluate \( f(t) \) at the endpoints \( t = \frac{1}{2} \) and \( t = 1 \).

1. For \( t = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = 2 - 2\left(\frac{1}{2}\right)^2 + \frac{1}{2} = 2 - 2 \cdot \frac{1}{4} + \frac{1}{2} = 2 - \frac{1}{2} + \frac{1}{2} = 2 \]

2. For \( t = 1 \):
\[ f(1) = 2 - 2(1)^2 + 1 = 2 - 2 + 1 = 1 \]

### Step 5: Determine the maximum value
Comparing the values:
- \( f\left(\frac{1}{2}\right) = 2 \)
- \( f(1) = 1 \)

Thus, the maximum value of \( f(x) \) in the interval \( \left[\frac{\pi}{6}, \frac{2\pi}{3}\right] \) is:
\[ \boxed{2} \]