Let f(x) be a continuous and positive function, such that the area bounded by`y=f(x),` x - axis and the lines `x^(2)=2ax` is `6a^(2)+sina(AA a gt 0)` sq. units. If `f(pi)=kpi`, then the value of k is equal to

Answer

To solve the problem, we need to find the value of \( k \) given that \( f(\pi) = k\pi \) and the area bounded by \( y = f(x) \), the x-axis, and the lines defined by \( x^2 = 2ax \) is \( 6a^2 + \sin a \) square units.

### Step-by-Step Solution:

1. **Identify the bounds of integration**:
The equation \( x^2 = 2ax \) can be rearranged to \( x(x - 2a) = 0 \), giving us the roots \( x = 0 \) and \( x = 2a \). Thus, the area we need to consider is bounded between \( x = 0 \) and \( x = 2a \).

2. **Set up the area integral**:
The area \( A \) under the curve \( y = f(x) \) from \( x = 0 \) to \( x = 2a \) can be expressed as:
\[ A = \int_0^{2a} f(x) \, dx \]
According to the problem, this area is equal to \( 6a^2 + \sin a \):
\[ \int_0^{2a} f(x) \, dx = 6a^2 + \sin a \]

3. **Differentiate both sides with respect to \( a \)**:
We will apply Leibniz's rule for differentiation under the integral sign:
\[ \frac{d}{da} \left( \int_0^{2a} f(x) \, dx \right) = f(2a) \cdot \frac{d(2a)}{da} = 2f(2a) \]
Now differentiate the right-hand side:
\[ \frac{d}{da} (6a^2 + \sin a) = 12a + \cos a \]
Setting these equal gives:
\[ 2f(2a) = 12a + \cos a \]
Therefore, we can express \( f(2a) \) as:
\[ f(2a) = 6a + \frac{1}{2} \cos a \]

4. **Substituting \( a \) to find \( f(\pi) \)**:
To find \( f(\pi) \), we need to express it in terms of \( a \). Set \( 2a = \pi \), which gives \( a = \frac{\pi}{2} \). Substitute \( a = \frac{\pi}{2} \) into the expression for \( f(2a) \):
\[ f(\pi) = 6\left(\frac{\pi}{2}\right) + \frac{1}{2} \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), we have:
\[ f(\pi) = 6 \cdot \frac{\pi}{2} + 0 = 3\pi \]

5. **Relate \( f(\pi) \) to \( k \)**:
We know from the problem statement that \( f(\pi) = k\pi \). Thus, we equate:
\[ 3\pi = k\pi \]
Dividing both sides by \( \pi \) (since \( \pi \neq 0 \)):
\[ k = 3 \]

### Final Answer:
The value of \( k \) is \( 3 \).