The maximum value of `f(x)=sinx(1+cosx)` is

Options

A. `(3sqrt3)/4`
B. `(3sqrt3)/2`
C. `3sqrt3`
D. `sqrt3`

Answer

To find the maximum value of the function \( f(x) = \sin x (1 + \cos x) \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) with respect to \( x \).

\[ f(x) = \sin x (1 + \cos x) \]

Using the product rule, where if \( u = \sin x \) and \( v = 1 + \cos x \), we have:

\[ f'(x) = u'v + uv' \]

Calculating \( u' \) and \( v' \):

- \( u' = \cos x \)
- \( v' = -\sin x \)

Now substituting back into the product rule:

\[ f'(x) = \cos x (1 + \cos x) + \sin x (-\sin x) \]

This simplifies to:

\[ f'(x) = \cos x (1 + \cos x) - \sin^2 x \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

\[ \cos x (1 + \cos x) - \sin^2 x = 0 \]

Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can rewrite the equation:

\[ \cos x (1 + \cos x) - (1 - \cos^2 x) = 0 \]

Expanding and rearranging gives:

\[ \cos x + \cos^2 x - 1 + \cos^2 x = 0 \]

This simplifies to:

\[ 2\cos^2 x + \cos x - 1 = 0 \]

### Step 3: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula \( \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = 1, c = -1 \):

\[ \cos x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ \cos x = \frac{-1 \pm \sqrt{1 + 8}}{4} \]
\[ \cos x = \frac{-1 \pm 3}{4} \]

This gives us two solutions:

1. \( \cos x = \frac{2}{4} = \frac{1}{2} \) (which gives \( x = \frac{\pi}{3} \))
2. \( \cos x = \frac{-4}{4} = -1 \) (which gives \( x = \pi \))

### Step 4: Determine the nature of critical points
To determine if these points are maxima or minima, we can use the second derivative test. We need to find \( f''(x) \).

From our earlier work, we have:

\[ f'(x) = \cos x (1 + \cos x) - \sin^2 x \]

Differentiating again:

\[ f''(x) = -\sin x (1 + \cos x) - \cos^2 x + \sin x \cos x \]

### Step 5: Evaluate the second derivative at critical points
1. For \( x = \frac{\pi}{3} \):
- \( f''\left(\frac{\pi}{3}\right) \) will be evaluated.
- If \( f''\left(\frac{\pi}{3}\right) < 0 \), then \( x = \frac{\pi}{3} \) is a maximum.

2. For \( x = \pi \):
- Similarly evaluate \( f''(\pi) \).

### Step 6: Calculate the maximum value
Finally, we compute the value of \( f(x) \) at the maximum point \( x = \frac{\pi}{3} \):

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

Thus, the maximum value of \( f(x) \) is:

\[ \boxed{\frac{3\sqrt{3}}{4}} \]