If f(x) is a non - negative function such that the area bounded by `y=f(x),` x - axis and the lines x = 0 and `x=alpha` is `4alpha sin alpha+2` sq. Units
(`AA alpha in [0, pi]`), then the value of `f((pi)/(2))` is equal to

Options

A. 0
B. 1
C. `2pi`
D. `8pi`

Answer

To solve the problem, we need to find the value of \( f\left(\frac{\pi}{2}\right) \) given that the area under the curve \( y = f(x) \) from \( x = 0 \) to \( x = \alpha \) is equal to \( 4\alpha \sin \alpha + 2 \).

### Step-by-Step Solution:

1. **Set up the equation for the area**:
The area under the curve from \( x = 0 \) to \( x = \alpha \) can be expressed as:
\[ \int_0^\alpha f(x) \, dx = 4\alpha \sin \alpha + 2 \]

2. **Differentiate both sides with respect to \( \alpha \)**:
By applying the Fundamental Theorem of Calculus and the Leibniz rule for differentiation under the integral sign, we differentiate the left side:
\[ \frac{d}{d\alpha} \left( \int_0^\alpha f(x) \, dx \right) = f(\alpha) \]
For the right side, we differentiate \( 4\alpha \sin \alpha + 2 \):
\[ \frac{d}{d\alpha} (4\alpha \sin \alpha + 2) = 4\sin \alpha + 4\alpha \cos \alpha \]

3. **Set the derivatives equal**:
Now we equate the two results:
\[ f(\alpha) = 4\sin \alpha + 4\alpha \cos \alpha \]

4. **Substitute \( \alpha = \frac{\pi}{2} \)**:
We need to find \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = 4\sin\left(\frac{\pi}{2}\right) + 4\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) \]

5. **Evaluate the trigonometric functions**:
Knowing that \( \sin\left(\frac{\pi}{2}\right) = 1 \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ f\left(\frac{\pi}{2}\right) = 4 \cdot 1 + 4 \cdot \frac{\pi}{2} \cdot 0 \]
This simplifies to:
\[ f\left(\frac{\pi}{2}\right) = 4 + 0 = 4 \]

### Final Answer:
Thus, the value of \( f\left(\frac{\pi}{2}\right) \) is \( \boxed{4} \).