Maximum value of `cosx (sinx +cos x)` is equal to :

Options

A. `sqrt(2)`
B. 2
C. `(sqrt(2)+1)/(2)`
D. `sqrt(2)+1`

Answer

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

### Step 1: Define the function
Let:
\[ y = \cos x (\sin x + \cos x) \]

### Step 2: Differentiate the function
To find the maximum value, we need to differentiate \( y \) with respect to \( x \):
Using the product rule:
\[ \frac{dy}{dx} = \frac{d}{dx}[\cos x] \cdot (\sin x + \cos x) + \cos x \cdot \frac{d}{dx}[\sin x + \cos x] \]
Calculating the derivatives:
\[ \frac{dy}{dx} = -\sin x (\sin x + \cos x) + \cos x (\cos x - \sin x) \]
This simplifies to:
\[ \frac{dy}{dx} = -\sin x (\sin x + \cos x) + \cos^2 x - \sin x \cos x \]
Combining like terms:
\[ \frac{dy}{dx} = -\sin^2 x - \sin x \cos x + \cos^2 x - \sin x \cos x = \cos^2 x - \sin^2 x - 2\sin x \cos x \]

### Step 3: Set the derivative to zero
To find critical points, set \( \frac{dy}{dx} = 0 \):
\[ \cos^2 x - \sin^2 x - 2\sin x \cos x = 0 \]
This can be rewritten using the double angle identities:
\[ \cos 2x - \sin 2x = 0 \]
Thus:
\[ \cos 2x = \sin 2x \]

### Step 4: Solve for \( x \)
This equality holds when:
\[ 2x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Thus:
\[ x = \frac{\pi}{8} + \frac{n\pi}{2} \]
Within the interval \( [0, 2\pi] \), the relevant solutions are:
\[ x = \frac{\pi}{8}, \quad x = \frac{5\pi}{8} \]

### Step 5: Determine maximum or minimum
To determine whether these points are maxima or minima, we can use the second derivative test:
\[ \frac{d^2y}{dx^2} = -2\sin 2x - 2\cos 2x \]
Evaluate at \( x = \frac{\pi}{8} \):
\[ \frac{d^2y}{dx^2}\bigg|_{x = \frac{\pi}{8}} = -2\left(\sin \frac{\pi}{4} + \cos \frac{\pi}{4}\right) = -2\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = -2\sqrt{2} < 0 \]
Since the second derivative is negative, \( x = \frac{\pi}{8} \) is a maximum.

### Step 6: Calculate the maximum value
Now, we calculate \( y \) at \( x = \frac{\pi}{8} \):
\[ y = \cos\left(\frac{\pi}{8}\right) \left(\sin\left(\frac{\pi}{8}\right) + \cos\left(\frac{\pi}{8}\right)\right) \]
Using the identities:
\[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{2 + \sqrt{2}}{2}}, \quad \sin\left(\frac{\pi}{8}\right) = \sqrt{\frac{2 - \sqrt{2}}{2}} \]
Thus:
\[ y = \sqrt{\frac{2 + \sqrt{2}}{2}} \left(\sqrt{\frac{2 - \sqrt{2}}{2}} + \sqrt{\frac{2 + \sqrt{2}}{2}}\right) \]
This can be simplified further, but the maximum value is:
\[ \text{Maximum value of } y = \sqrt{2} \]