Question

If the line $y=mx+2$ cuts the parabola $2y=x^2$ at points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ where $\left(x_1<x_2\right)$, then the value of $m$ for which ${\int }_{x_1}^{x_2}\left(mx+2-\frac{x^2}{2}\right)dx$ is minimum, is

• A. $\sqrt{2}$
• B. $\frac{8}{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. 0

Answer 1

Answer: (D)

Eliminating $y$ from two equations,
We get $2mx+4=x^2\Rightarrow x^2-2m-4=0$
$x_1$ and $x_2$ are roots of this quadratic equation

$\therefore x_1+x_2=2m$ and $x_1{x}_2=-4$
Now, $\underset{x_1}{\overset{x_2}{\int }}\left(mx+2-\frac{x^2}{2}\right)dx$
$=\frac{m}{2}\left(x_2^2-x_1^2\right)+2\left(x_2-x_1\right)-\frac{1}{6}\left(x_2^3-x_1^3\right)$
$=\left(x_2-x_1\right)\left[\frac{m}{2}\left(x_2+x_1\right)+2-\frac{1}{6}\left(x_1^2+x_1{x}_2+x_2^2\right)\right]$
$=\sqrt{4m^2+16}\left[\frac{m^2+4}{3}\right]$
Clearly above is minumum if $m=0$