Mathematics - Area Under Curves Question with Solution | TestHub

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