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 $\int_{x_{1}}^{x_{2}}\left(m x+2-\frac{x^{2}}{2}\right) dx$ 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 $2 m x+4=x^{2} \Rightarrow x^{2}-2 m-4=0$
$x_{1}$ and $x_{2}$ are roots of this quadratic equation

$\therefore x_{1}+x_{2}=2 m$ and $x_{1} x_{2}=-4$
Now, $\int\limits_{x_{1}}^{x_{2}}\left(m x+2-\frac{x^{2}}{2}\right) d x$
$=\frac{m}{2}\left(x_{2}^{2}-x_{1}^{2}\right)+2\left(x_{2}-x_{1}\right)-\frac{1}{6}\left(x_{2}^{3}-x_{1}^{3}\right)$
$=\left(x_{2}-x_{1}\right)\left[\frac{m}{2}\left(x_{2}+x_{1}\right)+2-\frac{1}{6}\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)\right]$
$=\sqrt{4 m^{2}+16}\left[\frac{m^{2}+4}{3}\right]$
Clearly above is minumum if $m=0$