Find the points of absolute maximum and minimum of each of the following :
`y=sqrt5(sinx+(1)/(2)cos 2x),0lexle(pi)/(2).`

Answer

To find the points of absolute maximum and minimum of the function \( y = \sqrt{5} \left( \sin x + \frac{1}{2} \cos 2x \right) \) for \( 0 \leq x \leq \frac{\pi}{2} \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( y \) with respect to \( x \):

\[ y = \sqrt{5} \left( \sin x + \frac{1}{2} \cos 2x \right) \]

Using the chain rule and product rule, we differentiate:

\[ \frac{dy}{dx} = \sqrt{5} \left( \cos x - \frac{1}{2} \cdot 2 \sin 2x \right) \]
\[ = \sqrt{5} \left( \cos x - \sin 2x \right) \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:

\[ \sqrt{5} \left( \cos x - \sin 2x \right) = 0 \]

This simplifies to:

\[ \cos x - \sin 2x = 0 \]

Using the double angle identity \( \sin 2x = 2 \sin x \cos x \), we rewrite the equation:

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

### Step 3: Solve for \( x \)
From the equation above, we have two cases:

1. \( \cos x = 0 \)
2. \( 1 - 2 \sin x = 0 \)

For \( \cos x = 0 \):
- This occurs at \( x = \frac{\pi}{2} \) within the interval \( [0, \frac{\pi}{2}] \).

For \( 1 - 2 \sin x = 0 \):
- This gives \( \sin x = \frac{1}{2} \), which leads to \( x = \frac{\pi}{6} \).

### Step 4: Evaluate the function at critical points and endpoints
We need to evaluate the function \( y \) at the critical points and the endpoints of the interval \( [0, \frac{\pi}{2}] \):

1. At \( x = 0 \):
\[ y(0) = \sqrt{5} \left( \sin 0 + \frac{1}{2} \cos 0 \right) = \sqrt{5} \left( 0 + \frac{1}{2} \cdot 1 \right) = \frac{\sqrt{5}}{2} \]

2. At \( x = \frac{\pi}{6} \):
\[ y\left(\frac{\pi}{6}\right) = \sqrt{5} \left( \sin \frac{\pi}{6} + \frac{1}{2} \cos \frac{\pi}{3} \right) = \sqrt{5} \left( \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \right) = \sqrt{5} \left( \frac{1}{2} + \frac{1}{4} \right) = \sqrt{5} \cdot \frac{3}{4} = \frac{3\sqrt{5}}{4} \]

3. At \( x = \frac{\pi}{2} \):
\[ y\left(\frac{\pi}{2}\right) = \sqrt{5} \left( \sin \frac{\pi}{2} + \frac{1}{2} \cos \pi \right) = \sqrt{5} \left( 1 + \frac{1}{2} \cdot (-1) \right) = \sqrt{5} \left( 1 - \frac{1}{2} \right) = \sqrt{5} \cdot \frac{1}{2} = \frac{\sqrt{5}}{2} \]

### Step 5: Determine absolute maximum and minimum
Now we compare the values obtained:

- \( y(0) = \frac{\sqrt{5}}{2} \)
- \( y\left(\frac{\pi}{6}\right) = \frac{3\sqrt{5}}{4} \)
- \( y\left(\frac{\pi}{2}\right) = \frac{\sqrt{5}}{2} \)

The maximum value is \( \frac{3\sqrt{5}}{4} \) at \( x = \frac{\pi}{6} \) and the minimum value is \( \frac{\sqrt{5}}{2} \) at \( x = 0 \) and \( x = \frac{\pi}{2} \).

### Final Result
- **Absolute Maximum**: \( \frac{3\sqrt{5}}{4} \) at \( x = \frac{\pi}{6} \)
- **Absolute Minimum**: \( \frac{\sqrt{5}}{2} \) at \( x = 0 \) and \( x = \frac{\pi}{2} \)