The maximum value of sin x. cos x is :

Options

A. `1/4`
B. `1/2`
C. `sqrt2`
D. `2sqrt2`

Answer

To find the maximum value of \( f(x) = \sin x \cdot \cos x \), we can follow these steps:

### Step 1: Rewrite the function
We can use the double angle identity for sine:
\[ f(x) = \sin x \cdot \cos x = \frac{1}{2} \sin(2x) \]
This transformation helps us to analyze the function more easily.

### Step 2: Find the derivative
To find the maximum value, we need to find the critical points. We differentiate \( f(x) \):
\[ f'(x) = \frac{1}{2} \cdot 2 \cos(2x) = \cos(2x) \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ \cos(2x) = 0 \]

### Step 4: Solve for \( x \)
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ 2x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Thus,
\[ x = \frac{\pi}{4} + \frac{n\pi}{2} \]
The first critical point in the interval \( [0, 2\pi] \) is \( x = \frac{\pi}{4} \).

### Step 5: Determine if it's a maximum or minimum
To confirm whether this point is a maximum, we can use the second derivative test. We find the second derivative:
\[ f''(x) = -2 \sin(2x) \]
Now, we evaluate the second derivative at \( x = \frac{\pi}{4} \):
\[ f''\left(\frac{\pi}{4}\right) = -2 \sin\left(2 \cdot \frac{\pi}{4}\right) = -2 \sin\left(\frac{\pi}{2}\right) = -2 \cdot 1 = -2 \]
Since \( f''\left(\frac{\pi}{4}\right) < 0 \), this indicates that \( x = \frac{\pi}{4} \) is indeed a maximum.

### Step 6: Calculate the maximum value
Now we can find the maximum value of \( f(x) \):
\[ f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) \cdot \cos\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right) \cdot \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2} \]

### Conclusion
The maximum value of \( \sin x \cdot \cos x \) is:
\[ \boxed{\frac{1}{2}} \]