Question

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

• A. \(13/6\)
• B. \(13/2\)
• C. \(13/5\)
• D. \(13/7\)

Hint:

Instead of integrating the linear function \(y = mx\) with calculus, save valuable time by treating it as a simple right-angled triangle where \(\text{Base} = b\) and \(\text{Height} = m \cdot b\). The geometric triangle area formula gives: \(\text{Area} = \frac{1}{2} \cdot b \cdot (mb) = \frac{mb^2}{2}\).

Answer 1

The Correct Option is A

Concept: The area under a curve \(y = f(x)\) bounded by the vertical lines \(x = a\) and \(x = b\) above the \(x\)-axis (\(y = 0\)) is calculated using a definite integral: \[ \text{Total Area } (A) = \int_{a}^{b} f(x) \, dx \] When a straight line passing through the origin, \(y = mx\), bisects this bounded space over the same horizontal interval, it splits the total area into two equal parts. The area under the line forms a right-angled trapezoid over the interval \([0, b]\), whose area must equal \(\frac{A}{2}\). Step 1: Calculate the total enclosed area \(A\).
The boundaries are given by the \(y\)-axis (\(x = 0\)), the \(x\)-axis (\(y = 0\)), the vertical line \(x = \frac{3}{2}\), and the downward-opening parabola \(y = 1 + 4x - x^2\). Let us evaluate the definite integral from \(0\) to \(\frac{3}{2}\): \[ A = \int_{0}^{3/2} (1 + 4x - x^2) \, dx \] Integrating term-by-term: \[ A = \left[ x + 2x^2 - \frac{x^3}{3} \right]_{0}^{3/2} \] Substituting the upper limit \(x = \frac{3}{2}\) (the lower limit at \(x = 0\) evaluates cleanly to \(0\)): \[ A = \left( \frac{3}{2} \right) + 2\left( \frac{3}{2} \right)^2 - \frac{1}{3}\left( \frac{3}{2} \right)^3 \] \[ A = \frac{3}{2} + 2\left( \frac{9}{4} \right) - \frac{1}{3}\left( \frac{27}{8} \right) = \frac{3}{2} + \frac{9}{2} - \frac{9}{8} \] Combine the first two fractions: \[ A = 6 - \frac{9}{8} = \frac{48 - 9}{8} = \frac{39}{8} \]

Step 2: Set up the mathematical condition for bisection.
Since the line \(y = mx\) bisects this area, the area bounded under the line \(y = mx\) from \(x = 0\) to \(x = \frac{3}{2}\) must be exactly equal to half of the total area (\(\frac{A}{2}\)): \[ \text{Area under the line } (A_{\text{line}}) = \frac{1}{2} \times \frac{39}{8} = \frac{39}{16} \]

Step 3: Integrate under the line to solve for \(m\).
The region under the line is a simple linear function: \[ A_{\text{line}} = \int_{0}^{3/2} mx \, dx = \left[ \frac{mx^2}{2} \right]_{0}^{3/2} \] Substituting the limits: \[ A_{\text{line}} = \frac{m}{2} \left( \frac{3}{2} \right)^2 = \frac{m}{2} \left( \frac{9}{4} \right) = \frac{9m}{8} \] Equating this area calculation to our value from
Step 2: \[ \frac{9m}{8} = \frac{39}{16} \] Isolating \(m\) by multiplying both sides by \(\frac{8}{9}\): \[ m = \frac{39}{16} \times \frac{8}{9} \] Simplifying the fractions by factoring out common terms (\(8\) with \(16\), and \(3\) with \(39\) and \(9\)): \[ m = \frac{13}{2} \times \frac{1}{3} = \frac{13}{6} \]