If a curve `y = asqrt(x)+bx` passes through the point (1,2) and the area bounded by the curve, line x = 4 and X-axis is 8 sq units, then

Options

A. a = 3, b = - 1
B. a = 3, b = 1
C. a = -3, b = 1
D. a = -3, b = -1

Explanation

To solve the problem step by step, we need to find the values of \( a \) and \( b \) for the curve \( y = a\sqrt{x} + bx \) given that it passes through the point \( (1, 2) \) and the area bounded by the curve, the line \( x = 4 \), and the x-axis is \( 8 \) square units.

### Step 1: Use the point (1, 2) to form the first equation.

Since the curve passes through the point \( (1, 2) \), we can substitute \( x = 1 \) and \( y = 2 \) into the equation of the curve:

\[ 2 = a\sqrt{1} + b(1) \]

This simplifies to:

\[ 2 = a + b \quad \text{(Equation 1)} \]

### Step 2: Set up the integral for the area.

The area \( A \) under the curve from \( x = 0 \) to \( x = 4 \) is given by the integral:

\[ A = \int_{0}^{4} (a\sqrt{x} + bx) \, dx \]

We know this area is equal to \( 8 \):

\[ \int_{0}^{4} (a\sqrt{x} + bx) \, dx = 8 \]

### Step 3: Calculate the integral.

We can split the integral:

\[ A = \int_{0}^{4} a\sqrt{x} \, dx + \int_{0}^{4} bx \, dx \]

Calculating each part:

1. For \( \int_{0}^{4} a\sqrt{x} \, dx \):

\[ \int a\sqrt{x} \, dx = a \cdot \frac{2}{3} x^{3/2} \bigg|_{0}^{4} = a \cdot \frac{2}{3} (4^{3/2} - 0) = a \cdot \frac{2}{3} \cdot 8 = \frac{16a}{3} \]

2. For \( \int_{0}^{4} bx \, dx \):

\[ \int bx \, dx = b \cdot \frac{x^2}{2} \bigg|_{0}^{4} = b \cdot \frac{16}{2} = 8b \]

Combining these results, we have:

\[ A = \frac{16a}{3} + 8b \]

Setting this equal to \( 8 \):

\[ \frac{16a}{3} + 8b = 8 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations.

Now we have two equations:

1. \( a + b = 2 \) (Equation 1)
2. \( \frac{16a}{3} + 8b = 8 \) (Equation 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 and simplifying:

\[ \frac{16a}{3} + 16 - 8a = 8 \]

Multiplying through by \( 3 \) to eliminate the fraction:

\[ 16a + 48 - 24a = 24 \]

Combining like terms:

\[ -8a + 48 = 24 \]

Solving for \( a \):

\[ -8a = 24 - 48 \]
\[ -8a = -24 \]
\[ a = 3 \]

Now substituting back to find \( b \):

\[ b = 2 - a = 2 - 3 = -1 \]

### Final Result:

Thus, the values of \( a \) and \( b \) are:

\[ a = 3, \quad b = -1 \]