Question

For which of the following value of $m$, is the area of the region bounded by the curve $y=x-x^2$ and the line $y=mx$ equals to $\frac{9}{2}$?

• A. $-4$
• B. $-2$
• C. $2$
• D. $4$

Answer 1

The correct option is D $4$

Given curve $y=x-x^2$ and the line $y=mx$
$\Rightarrow mx=x-x^2\Rightarrow x=0,1-m$
$m$ can be $0,<0$ or $>0$
Case 1: If $m=0\Rightarrow$ limits are $0\text{to}1$
Area $=\underset{0}{\overset{1}{\int }}\left[\right(x-x^2)-mx]dx$
$=\underset{0}{\overset{1}{\int }}\left[\right(x-x^2]dx$
$={[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1$
$=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\ne \frac{9}{2}$
Case 2: $m<1\Rightarrow$ limits are $0\text{to}1-m$
Area $=\underset{0}{\overset{1-m}{\int }}\left[\right(x-x^2)-mx]dx=\frac{9}{2}$
$\Rightarrow \underset{0}{\overset{1-m}{\int }}\left[\right(1-m)x-x^2]dx=\frac{9}{2}$
$\Rightarrow {\left[\right(1-m)\frac{x^2}{2}-\frac{x^3}{3}]}_0^{1-m}=\frac{9}{2}$
$\Rightarrow m=-2$
Case 3: $m>1\Rightarrow$ limits are $1-m\text{to}0$
Area $=\underset{1-m}{\overset{0}{\int }}\left[\right(x-x^2)-mx]dx=\frac{9}{2}$
$\Rightarrow \underset{1-m}{\overset{0}{\int }}\left[\right(1-m)x-x^2]dx=\frac{9}{2}$
$\Rightarrow {\left[\right(1-m)\frac{x^2}{2}-\frac{x^3}{3}]}_{1-m}^0=\frac{9}{2}$
$\Rightarrow m=4$