If the curve `y=a sqrt(x)+bx` passes through `P(1,2)` and lies above the x-axis for `x in [0,9]` and the area bounded by the curve, the x-axis and `x=4` is 8 sq. units the `2a-3b=`

Options

A. 6
B. 9
C. 0
D. 10

Explanation

To solve the problem step by step, we will follow the given conditions and derive the required values for \( a \) and \( b \).

### Step 1: Use the point \( P(1, 2) \)
The curve is given by the equation:
\[ y = a \sqrt{x} + bx \]
Since the curve passes through the point \( P(1, 2) \), we can substitute \( x = 1 \) and \( y = 2 \) into the equation:
\[ 2 = a \sqrt{1} + b(1) \]
This simplifies to:
\[ 2 = a + b \tag{1} \]

### Step 2: Set up the area under the curve
The area \( A \) bounded by the curve, the x-axis, and the line \( x = 4 \) is given as 8 square units. The area under the curve from \( x = 0 \) to \( x = 4 \) can be calculated using integration:
\[ A = \int_0^4 (a \sqrt{x} + bx) \, dx \]
Calculating this integral:
\[ A = \int_0^4 a \sqrt{x} \, dx + \int_0^4 bx \, dx \]
The integral of \( a \sqrt{x} \) is:
\[ \int a \sqrt{x} \, dx = a \cdot \frac{2}{3} x^{3/2} \]
Evaluating from 0 to 4:
\[ \left[ a \cdot \frac{2}{3} x^{3/2} \right]_0^4 = a \cdot \frac{2}{3} (4^{3/2}) = a \cdot \frac{2}{3} \cdot 8 = \frac{16a}{3} \]
The integral of \( bx \) is:
\[ \int bx \, dx = b \cdot \frac{x^2}{2} \]
Evaluating from 0 to 4:
\[ \left[ b \cdot \frac{x^2}{2} \right]_0^4 = b \cdot \frac{4^2}{2} = b \cdot 8 = 8b \]
Thus, the total area is:
\[ A = \frac{16a}{3} + 8b \]
Setting this equal to 8:
\[ \frac{16a}{3} + 8b = 8 \tag{2} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( a + b = 2 \) (from step 1)
2. \( \frac{16a}{3} + 8b = 8 \) (from step 2)

From equation (1), we can express \( b \) in terms of \( a \):
\[ b = 2 - a \]
Substituting this into equation (2):
\[ \frac{16a}{3} + 8(2 - a) = 8 \]
Expanding this:
\[ \frac{16a}{3} + 16 - 8a = 8 \]
Multiplying the entire equation by 3 to eliminate the fraction:
\[ 16a + 48 - 24a = 24 \]
Combining like terms:
\[ -8a + 48 = 24 \]
Solving for \( a \):
\[ -8a = 24 - 48 \]
\[ -8a = -24 \implies a = 3 \]

Now substituting \( a = 3 \) back into equation (1):
\[ 3 + b = 2 \implies b = 2 - 3 = -1 \]

### Step 4: Calculate \( 2a - 3b \)
Now we need to find \( 2a - 3b \):
\[ 2a - 3b = 2(3) - 3(-1) = 6 + 3 = 9 \]

### Final Answer
Thus, the value of \( 2a - 3b \) is:
\[ \boxed{9} \]