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 \( \frac{13}{6} \).