The line y=m x bisects the area unclosed by lines x=0, y=0 and...

Given, $y=1+4 x-x^2$$\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=4-2 x=0 \Rightarrow x=2 \text{[point\text{ of } maxima]}\\ & \Rightarrow \quad y_{\max }=1+4 \times 2-4=5 \end{aligned}$when, $x=0, y=1\therefore$ Graph according to question and above information\begin{aligned} & \Rightarrow \text{ If } y=m x \text{ bisects the area bounded } \\ & \Rightarrow \quad A_1=A_2 \\ & \Rightarrow \quad \int_0^{3 / 2}\left(1+4 x-x^2\right) d x=2 \int_0^{3 / 2} m x d x \\ & \Rightarrow \quad\left[x+2 x^2-\frac{x^3}{3}\right]_0^{3 / 2}=\left[m x^2\right]_0^{3 / 2} \\ & \Rightarrow \frac{3}{2}+2 \times \frac{9}{4}-\frac{27}{8} \times \frac{1}{3}=m \times \frac{9}{4} \\ & \therefore \quad m=\frac{13}{6}\end{aligned}$$