Question

The area of the region bounded by the curve $y=x-x^2$ and the line $y=mx$ equals $\frac{9}{2}\text{sq.units}$. Then the possible values of $m$ is/are:

• A. $-4$
• B. $-2$
• C. $2$
• D. $4$

Answer 1

Answer: (B), (D)

$4$


The two curves meet at
$mx=x-x^2$
$\Rightarrow x^2=x(1-m),\therefore x=0,1-m$
$A=\underset{0}{\overset{1-m}{\int }}(y_1-y_2)dx=\underset{0}{\overset{1-m}{\int }}(x-x^2-mx)dx$
Clearly $m<1$ or $m>1,$ but $m\ne 1$
Now, if $m<1$
$A={\left[\right(1-m)\frac{x^2}{2}-\frac{x^3}{3}]}_0^{1-m}=\frac{9}{2}$
$\Rightarrow (1-m{)}^3=27\therefore m=-2$
But if $m>1$ then $(1-m)$ is $-ve,$ then
$A={\left[\right(1-m)\frac{x^2}{2}-\frac{x^3}{3}]}_{1-m}^0=\frac{9}{2}$
$\Rightarrow (1-m{)}^3=-27$
$\Rightarrow 1-m=-3\therefore m=4$