Question

If $a{x}^2+bx+c=0$ has imaginary roots and $a-b+c>0,$ then the set of points $(x,y)$ satisfying the equation $|a(x^2+\frac{y}{z})+(b+1)x+c|=|a{x}^2+bx+c|+|x+y|$ consists of the region in the $xy-$plane which is

• A. on or above the bisector of I and III quadrant
• B. on or above the bisector of II and IV quadrant
• C. on or below the bisector of I and III quadrant
• D. on or bellow the bisector of II and IV quadrant

Answer 1

The correct option is B on or above the bisector of II and IV quadrant

Let $f\left(x\right)=a{x}^2+bx+c=0$

Given that $f\left(x\right)$ has imaginary roots

$\therefore f\left(x\right)$ is either $>0$ (or) $<0$always for all real x

$f(-1)=a-b+c$

Given that $a-b+c>0$

$\therefore f(-1)>0$

$\therefore f\left(x\right)=a{x}^2+bx+c>0\forall xϵR$

Given

$|a(x^2+\frac{y}{a})+x(b+1)+c|=|a{x}^2+bx+c|+|x+y|\to 1$

$|(a{x}^2+bx+c)+(x+y)|=a{x}^2+bx+c+|x+y|(\because a{x}^2+bx+c>0)$

If $k>0$ and $|m+k|=\left|m\right|+k$ then 'm' must be $\ge 0$

$\therefore 1$ valids if $x+y\ge 0$

$1$ is consists of the region $x+y\ge 0$ in xy-plane.

$x+y>0$ is region above (or) on bisector of and $II$ and $IV$ quadrant.