Question

The area of the region enclosed between parabola $y^2=x$ and the line $y=mx$ is $148$. The value of $m$ is:

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

The graph is shown in the image.

The intersection points of $y^2=x$ and $y=mx$ are $(0,0)$ and $(\frac{1}{m^2},\frac{1}{m})$

Given: area enclosed by $f\left(x\right)$ ,$y=mx$ is $\frac{1}{48}$

Now $A={\int }_0^{\frac{1}{m^2}}[\sqrt{x}]-{\int }_0^{\frac{1}{m^2}}[mx]$

$\Rightarrow \frac{1}{48}=[\frac{2x^{\frac{3}{2}}}{3}]{|}_0^{\frac{1}{m^2}}-[\frac{m{x}^2}{2}]{|}_0^{\frac{1}{m^2}}$

$\Rightarrow \frac{1}{48}=\frac{2}{3m^2}-\frac{1}{2m^3}$

$\Rightarrow \frac{1}{48}=\frac{1}{6m^3}$

$\Rightarrow 8=m^3$

$\Rightarrow m=2$