Explanation:
, d)
Exp:(b, d)
Given curve, \(\mathrm{y}=\mathrm{x}-\mathrm{x}^{2}\)
Line, \(y=m x\) equals to \(\frac{9}{2}\)
Here, \(m x=x-x^{2}\)
\(\mathrm{x}^{2}=\mathrm{x}(1-\mathrm{m})\)
\(\mathrm{x}=0, \mathrm{x}=1-\mathrm{m}\)
when, \(\mathrm{m}\lt 1,1-\mathrm{m}>0\)
Area, \(\mathrm{A}=\int_{0}^{1-\mathrm{m}} \mathrm{y} \cdot \mathrm{dx}\)
\(\frac{9}{2}=\int_{0}^{1-\mathrm{m}}\left(\mathrm{x}-\mathrm{x}^{2}-\mathrm{mx}\right) \cdot \mathrm{dx}\)
\(\frac{9}{2}=\left[(1-\mathrm{m}) \frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1-\mathrm{m}}\)
\(\frac{9}{2}=\left[(1-\mathrm{m})^{3}\left(\frac{1}{2}-\frac{1}{3}\right)\right]\)
\(\frac{9}{2}=\left[(1-\mathrm{m})^{3} \cdot\left(\frac{1}{6}\right)\right] \Rightarrow \frac{9}{2} \times 6=(1-\mathrm{m})^{3}\)
\(27=(1-\mathrm{m})^{3} \Rightarrow 3^{3}=(1-\mathrm{m})^{3} \Rightarrow 3=1-\mathrm{m} \Rightarrow \mathrm{m}=-2\)
When,
\(\mathrm{m}>1,1-\mathrm{m}\lt 0\)
Area, \(\quad \int_{1-\mathrm{m}}^{0} \mathrm{y} \cdot \mathrm{dx}\)
\(\frac{9}{2}=\int_{1-m}^{0}\left(x-x^{2}-m x\right) \cdot d x\)
\(\frac{9}{2}=\left[(1-m) \cdot \frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{1-m}^{0}\)
\(\frac{9}{2}=\left[-(1-\mathrm{m})^{3}\left(\frac{1}{2}-\frac{1}{3}\right)\right]\)
\(\frac{9 \times 6}{2}=-(1-\mathrm{m})^{3} \Rightarrow(3)^{3}=-(1-\mathrm{m})^{3}\)
\(3=-1+\mathrm{m} \Rightarrow \mathrm{m}=4\)
