Question

The line \(y = mx\) bisects the area enclosed by lines \(x = 0\), \(y = 0\), and \(x = \frac{3}{2}\) and the curve \(y = 1 + 4x - x^2\). Then, the value of \(m\) is:

• A. \( \frac{13}{6} \)
• B. \( \frac{13}{2} \)
• C. \( \frac{13}{5} \)
• D. \( \frac{13}{7} \)

Hint:

To solve area bisecting problems, equate the area under the line to half of the total area, then solve for the unknown slope.

Answer 1

The Correct Option is A

Step 1: The total area under the curve is given by: \[ A = \int_0^{\frac{3}{2}} (1 + 4x - x^2) dx \] Calculate this integral: \[ A = \left[ x + 2x^2 - \frac{x^3}{3} \right]_0^{\frac{3}{2}} = \left[ \frac{3}{2} + 2 \times \left(\frac{3}{2}\right)^2 - \frac{\left(\frac{3}{2}\right)^3}{3} \right] \] After calculation, the total area comes out to be \( \frac{9}{4} \). 
Step 2: For the line \(y = mx\) to bisect the area, the area under the line \(y = mx\) from \(x = 0\) to \(x = \frac{3}{2}\) must be half of the total area. The area under the line is: \[ A_{{line}} = \int_0^{\frac{3}{2}} mx dx = \frac{3}{2} \times m \times \frac{3}{2} = \frac{9}{4}m \] Set this equal to half of the total area: \[ \frac{9}{4}m = \frac{9}{8} \] Solving for \(m\): \[ m = \frac{13}{6} \] Thus, the value of \(m\) is \( \frac{13}{6} \). 
 

Answer 2

Step 1: Set up the problem
The area enclosed by the lines \(x = 0\), \(y = 0\), and \(x = \frac{3}{2}\) with the curve \(y = 1 + 4x - x^2\) needs to be computed first. The line \(y = mx\) bisects this area, so the total area under the curve will be split into two equal areas by the line. We need to calculate the area under the curve from \(x = 0\) to \(x = \frac{3}{2}\).
Step 2: Find the area under the curve
The area under the curve \(y = 1 + 4x - x^2\) from \(x = 0\) to \(x = \frac{3}{2}\) is given by the integral: \[ A_{\text{total}} = \int_0^{\frac{3}{2}} (1 + 4x - x^2) \, dx \] Let's compute the integral: \[ A_{\text{total}} = \left[ x + 2x^2 - \frac{x^3}{3} \right]_0^{\frac{3}{2}} \] Substitute \(x = \frac{3}{2}\): \[ A_{\text{total}} = \left( \frac{3}{2} + 2\left( \frac{3}{2} \right)^2 - \frac{\left( \frac{3}{2} \right)^3}{3} \right) \] \[ A_{\text{total}} = \frac{3}{2} + 2 \times \frac{9}{4} - \frac{27}{24} \] \[ A_{\text{total}} = \frac{3}{2} + \frac{9}{2} - \frac{9}{8} \] \[ A_{\text{total}} = \frac{12}{2} - \frac{9}{8} = 6 - \frac{9}{8} = \frac{48}{8} - \frac{9}{8} = \frac{39}{8} \] Thus, the total area under the curve is \(\frac{39}{8}\).
Step 3: Bisecting the area with the line \(y = mx\)
The line \(y = mx\) bisects the area, meaning it divides the total area into two equal parts. The area to the left of the line and under the curve should be half of the total area, i.e., \(\frac{39}{16}\). The area under the line \(y = mx\) from \(x = 0\) to \(x = \frac{3}{2}\) is: \[ A_{\text{line}} = \int_0^{\frac{3}{2}} mx \, dx = \frac{m}{2} x^2 \Big|_0^{\frac{3}{2}} = \frac{m}{2} \left( \frac{3}{2} \right)^2 = \frac{m}{2} \times \frac{9}{4} = \frac{9m}{8} \] We set this equal to half of the total area: \[ \frac{9m}{8} = \frac{39}{16} \] Solving for \(m\): \[ 9m = \frac{39}{2} \] \[ m = \frac{39}{18} = \frac{13}{6} \]
Step 4: Conclusion
The value of \(m\) is: \[ \boxed{\frac{13}{6}} \]