Let f(x) be a non-negative continuous function such that the area bounded by the curve y=f(x), the x-axis, and the ordinates `x=(pi)/(4) and x=betagt(pi)/(4)" is "beta sin beta +(pi)/(4)cos beta +sqrt(2)beta.` Then `f'((pi)/(2))` is

Options

A. `((pi)/(2)-sqrt(2)-1)`
B. `((pi)/(4)+sqrt(2)-1)`
C. `-(pi)/(2)`
D. `(1-(pi)/(2)-sqrt(2))`

Explanation

To solve the problem, we need to find \( f'(\frac{\pi}{2}) \) given the area bounded by the curve \( y = f(x) \), the x-axis, and the ordinates \( x = \frac{\pi}{4} \) and \( x = \beta \) (where \( \beta > \frac{\pi}{4} \)) is given by:

\[ A = \beta \sin(\beta) + \frac{\pi}{4} \cos(\beta) + \sqrt{2} \beta \]

### Step 1: Set up the area expression
The area \( A \) can also be expressed using the integral of the function \( f(x) \):

\[ A = \int_{\frac{\pi}{4}}^{\beta} f(x) \, dx \]

### Step 2: Differentiate both sides with respect to \( \beta \)
Using the Leibniz rule for differentiation under the integral sign, we differentiate both sides:

\[ \frac{dA}{d\beta} = f(\beta) \]

On the right-hand side, we differentiate the expression for \( A \):

\[ \frac{d}{d\beta} \left( \beta \sin(\beta) + \frac{\pi}{4} \cos(\beta) + \sqrt{2} \beta \right) \]

### Step 3: Apply the product and chain rule
Now we differentiate each term:

1. For \( \beta \sin(\beta) \):
\[ \frac{d}{d\beta}(\beta \sin(\beta)) = \sin(\beta) + \beta \cos(\beta) \]

2. For \( \frac{\pi}{4} \cos(\beta) \):
\[ \frac{d}{d\beta}\left(\frac{\pi}{4} \cos(\beta)\right) = -\frac{\pi}{4} \sin(\beta) \]

3. For \( \sqrt{2} \beta \):
\[ \frac{d}{d\beta}(\sqrt{2} \beta) = \sqrt{2} \]

Combining these results, we have:

\[ f(\beta) = \sin(\beta) + \beta \cos(\beta) - \frac{\pi}{4} \sin(\beta) + \sqrt{2} \]

### Step 4: Simplify the expression for \( f(\beta) \)
Combining like terms gives us:

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

### Step 5: Differentiate to find \( f'(\beta) \)
Now, we differentiate \( f(\beta) \) to find \( f'(\beta) \):

\[ f'(\beta) = \left(1 - \frac{\pi}{4}\right) \cos(\beta) + \cos(\beta) - \beta \sin(\beta) \]

Combining the cosine terms:

\[ f'(\beta) = \left(1 - \frac{\pi}{4} + 1\right) \cos(\beta) - \beta \sin(\beta) \]

This simplifies to:

\[ f'(\beta) = \left(2 - \frac{\pi}{4}\right) \cos(\beta) - \beta \sin(\beta) \]

### Step 6: Evaluate \( f'(\frac{\pi}{2}) \)
Now we substitute \( \beta = \frac{\pi}{2} \):

\[ f'\left(\frac{\pi}{2}\right) = \left(2 - \frac{\pi}{4}\right) \cos\left(\frac{\pi}{2}\right) - \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) \]

Since \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \):

\[ f'\left(\frac{\pi}{2}\right) = 0 - \frac{\pi}{2} = -\frac{\pi}{2} \]

### Final Answer
Thus, the value of \( f'\left(\frac{\pi}{2}\right) \) is:

\[ \boxed{-\frac{\pi}{2}} \]