Question

If the line $y=x$ cuts the curve $y=2x^3+6x^2+x-4$ at three points $A,B$ and $C$. Then the value of $|OA.OB.OC|$ with $O$ being the origin is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $4\sqrt{2}$
• D. $8\sqrt{2}$

Answer 1

The correct option is C $4\sqrt{2}$

$y=x...\left(1\right)$
$y=2x^3+6x^2+x-4...\left(2\right)$
Equating eqn $\left(1\right)$ and $\left(2\right)$, we get
$x^3+3x^2-2=0...\left(3\right)$
Coordinates of point
$A\equiv (x_1,x_1)$ $(\because x_i=y_i)$
$B\equiv (x_2,x_2)$
$C\equiv (x_3,x_3)$

Here
$OA=\sqrt{(x_1-0{)}^2+(x_1-0{)}^2}=\sqrt{2}{x}_1$
$OB=\sqrt{2}{x}_2$
$OC=\sqrt{2}{x}_3$

$|OA.OB.OC|=|\sqrt{2}{x}_1\times \sqrt{2}{x}_2\times \sqrt{2}{x}_3|=2\sqrt{2}{x}_1.x_2.x_3$
Here $x_1.x_2.x_3=-\frac{d}{a}=-\frac{(-2)}{1}=2$ from eqn $\left(3\right)$

Putting this value, we get
$|OA.OB.OC|=2\sqrt{2}\times 2=4\sqrt{2}$