Let `f (x)` be a non-negative continuous function such that the area bounded by the curve y `= f (x)`, x-axis and the ordinates `x=pi/4 and x=beta>pi/4` is :
`{betasinbeta+pi/4cosbeta+sqrt2beta}`, then `f(pi/2)` is:

Options

A. `(1-pi/4-sqrt2)`
B. `(1-pi/4+sqrt2)`
C. `(pi/4+sqrt2-1)`
D. `(pi/4-sqrt2-1)`

Answer

To find \( f\left(\frac{\pi}{2}\right) \), we start with the given area bounded by the curve \( y = f(x) \), the x-axis, and the ordinates \( x = \frac{\pi}{4} \) and \( x = \beta \) (where \( \beta > \frac{\pi}{4} \)). The area is given by the expression:

\[ \text{Area} = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \]

### Step 1: Set up the integral for the area
The area can also be expressed using integration:

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

### Step 2: Equate the two expressions for area
We set the two expressions for the area equal to each other:

\[ \int_{\frac{\pi}{4}}^{\beta} f(x) \, dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \]

### Step 3: Differentiate both sides with respect to \( \beta \)
To find \( f(\beta) \), we differentiate both sides with respect to \( \beta \):

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

### Step 4: Apply the product rule and differentiate
Using the product rule on the first term:

\[ \frac{d}{d\beta}(\beta \sin \beta) = \sin \beta + \beta \cos \beta \]

Differentiating the second term:

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

Differentiating the third term:

\[ \frac{d}{d\beta}(\sqrt{2} \beta) = \sqrt{2} \]

### Step 5: Combine the derivatives
Now, we combine the results from the differentiation:

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

### Step 6: Simplify the expression
Combining like terms, we get:

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

### Step 7: Substitute \( \beta = \frac{\pi}{2} \)
Now we substitute \( \beta = \frac{\pi}{2} \) to find \( f\left(\frac{\pi}{2}\right) \):

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

### Step 8: Evaluate the trigonometric functions
We know that:

\[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{and} \quad \cos\left(\frac{\pi}{2}\right) = 0 \]

Substituting these values gives:

\[ f\left(\frac{\pi}{2}\right) = \left(1 - \frac{\pi}{4}\right) \cdot 1 + \frac{\pi}{2} \cdot 0 + \sqrt{2} \]

### Step 9: Final result
Thus, we have:

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

### Conclusion
The final answer is:

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

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