The maximum ordinate of a point on the graph of the function f(x) = sin x( 1+ cos x ) is

Options

A. `(2+sqrt(3))/(4)`
B. `(3sqrt(3))/(4)`
C. 1
D. non of these

Explanation

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

### Step 1: Find the first derivative of the function
We will use the product rule to differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}[\sin x] \cdot (1 + \cos x) + \sin x \cdot \frac{d}{dx}[1 + \cos x] \]
Calculating the derivatives:
\[ f'(x) = \cos x (1 + \cos x) + \sin x (-\sin x) \]
This simplifies to:
\[ f'(x) = \cos x (1 + \cos x) - \sin^2 x \]
So, we have:
\[ f'(x) = \cos x + \cos^2 x - \sin^2 x \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite this as:
\[ f'(x) = \cos x + \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x + \cos x - 1 \]

### Step 2: Set the first derivative to zero to find critical points
To find the critical points, we set \( f'(x) = 0 \):
\[ 2\cos^2 x + \cos x - 1 = 0 \]
This is a quadratic equation in terms of \( \cos x \). Let \( u = \cos x \):
\[ 2u^2 + u - 1 = 0 \]
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us two solutions:
\[ u = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad u = \frac{-4}{4} = -1 \]
Thus, we have:
\[ \cos x = \frac{1}{2} \quad \text{or} \quad \cos x = -1 \]

### Step 3: Find the corresponding values of \( x \)
1. For \( \cos x = \frac{1}{2} \):
\[ x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
2. For \( \cos x = -1 \):
\[ x = \pi + 2k\pi \quad (k \in \mathbb{Z}) \]

### Step 4: Determine the nature of critical points using the second derivative
Next, we need to find the second derivative \( f''(x) \) to determine if these points are maxima or minima:
\[ f''(x) = \frac{d}{dx}[f'(x)] \]
Calculating \( f''(x) \):
\[ f''(x) = \frac{d}{dx}[2\cos^2 x + \cos x - 1] \]
Using the chain rule:
\[ f''(x) = 2 \cdot 2\cos x (-\sin x) + (-\sin x) = -4\cos x \sin x - \sin x = -\sin x(4\cos x + 1) \]

### Step 5: Evaluate the second derivative at critical points
1. For \( x = \frac{\pi}{3} \):
\[ f''\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right)(4\cos\left(\frac{\pi}{3}\right) + 1) = -\frac{\sqrt{3}}{2}(4 \cdot \frac{1}{2} + 1) = -\frac{\sqrt{3}}{2}(2 + 1) = -\frac{3\sqrt{3}}{2} < 0 \]
This indicates a local maximum at \( x = \frac{\pi}{3} \).

### Step 6: Find the maximum ordinate
Now, we substitute \( x = \frac{\pi}{3} \) back into the original function to find the maximum ordinate:
\[ f\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)(1 + \cos\left(\frac{\pi}{3}\right)) = \frac{\sqrt{3}}{2} \left(1 + \frac{1}{2}\right) = \frac{\sqrt{3}}{2} \cdot \frac{3}{2} = \frac{3\sqrt{3}}{4} \]

### Final Answer
The maximum ordinate of the point on the graph of the function \( f(x) = \sin x (1 + \cos x) \) is:
\[ \frac{3\sqrt{3}}{4} \]