If a curve `y =a sqrtx+bx` passes through point `(1,2)` and the area bounded by curve, line `x=4` and x-axis is 8, then : (a) `a=3` (b) `b=3` (c) `a=-1` (d) `b=-1`

Options

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

Answer

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

### Step 1: Use the point (1, 2) to create an 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:
\[ 2 = a + b \tag{1} \]

### Step 2: Set up the area equation
The area under the curve from \(x = 0\) to \(x = 4\) is given by:
\[ \text{Area} = \int_0^4 (a \sqrt{x} + bx) \, dx = 8 \]
We will calculate this integral.

### Step 3: Calculate the integral
First, we find the integral:
\[ \int (a \sqrt{x} + bx) \, dx = \int a x^{1/2} \, dx + \int b x \, dx \]
Calculating each part:
\[ \int a x^{1/2} \, dx = \frac{2a}{3} x^{3/2} \quad \text{and} \quad \int b x \, dx = \frac{b}{2} x^2 \]
Thus,
\[ \int_0^4 (a \sqrt{x} + bx) \, dx = \left[\frac{2a}{3} x^{3/2} + \frac{b}{2} x^2\right]_0^4 \]
Calculating at the limits:
\[ = \left(\frac{2a}{3} (4)^{3/2} + \frac{b}{2} (4)^2\right) - 0 \]
Calculating \(4^{3/2} = 8\) and \(4^2 = 16\):
\[ = \frac{2a}{3} \cdot 8 + \frac{b}{2} \cdot 16 = \frac{16a}{3} + 8b \]
Setting this equal to the area:
\[ \frac{16a}{3} + 8b = 8 \tag{2} \]

### Step 4: Solve the system of 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 \(a\):
\[ b = 2 - a \tag{3} \]

Substituting equation (3) into equation (2):
\[ \frac{16a}{3} + 8(2 - a) = 8 \]
Expanding this gives:
\[ \frac{16a}{3} + 16 - 8a = 8 \]
Multiplying the entire equation by 3 to eliminate the fraction:
\[ 16a + 48 - 24a = 24 \]
Combining like terms:
\[ -8a + 48 = 24 \]
Solving for \(a\):
\[ -8a = 24 - 48 \]
\[ -8a = -24 \implies a = 3 \]

### Step 5: Find \(b\)
Using the value of \(a\) in equation (3):
\[ b = 2 - 3 = -1 \]

### Conclusion
Thus, we find:
\[ a = 3, \quad b = -1 \]

### Final Answer
The correct options are:
(a) \(a = 3\) and (d) \(b = -1\).