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.
$$\frac{{13}}{7}$$
Answer :
$$\frac{{13}}{6}$$
Solution :
$$y = 1 + 4x - {x^2} = 5 - {\left( {x - 2} \right)^2}$$
We have $$\int\limits_0^{\frac{3}{2}} {\left( {1 + 4x - {x^2}} \right)} dx$$
$$\eqalign{ & = 2\int\limits_0^{\frac{3}{2}} {mx\,dx} \cr & = \frac{3}{2} + 2\left( {\frac{9}{4}} \right) - \frac{1}{3}\left( {\frac{{27}}{8}} \right) \cr & = m.\frac{9}{4} \cr} $$
On solving we get $$m = \frac{{13}}{6}$$
$$y = 1 + 4x - {x^2} = 5 - {\left( {x - 2} \right)^2}$$
We have $$\int\limits_0^{\frac{3}{2}} {\left( {1 + 4x - {x^2}} \right)} dx$$
$$\eqalign{ & = 2\int\limits_0^{\frac{3}{2}} {mx\,dx} \cr & = \frac{3}{2} + 2\left( {\frac{9}{4}} \right) - \frac{1}{3}\left( {\frac{{27}}{8}} \right) \cr & = m.\frac{9}{4} \cr} $$
On solving we get $$m = \frac{{13}}{6}$$