Question

If ' $x$ ' is real, then $\frac{x^2-x+c}{x^2+x+2c}$ can take all real values if :

• A. $c\in [0,6]$
• B. $c\in [-6,0]$
• C. $c\in (-\infty ,-6)\cup (0,\infty )$
• D. $c\in (-6,0)$

Answer 1

Answer: (D)

Let $y=\frac{x^2-x+c}{x^2+x+2c};x\in R$ and $y\in R_{\text{. }}$
$\Rightarrow (y-1)x^2+(y+1)x+2yc-c=0$
$\because x\in R$
$\Rightarrow D\ge 0$
$\Rightarrow (y+1{)}^2-4c(y-1)(2y-1)\ge 0$
$\Rightarrow y^2+1+2y-4c\left[2y^2-3y+1\right]\ge 0$
$\Rightarrow (1-8c)y^2+(2+12c)y+1-4c\ge 0$........(1)
Now for all $y\in R(1)$ will be true if
$1-8c>0\Rightarrow c<\frac{1}{8}\text{ and }D\le 0$
$\Rightarrow 4(1+6c{)}^2-4(1-8c)(1-4c)\le 0$
$\Rightarrow 1+36c^2+12c-1-32c^2+12c\le 0$
$\Rightarrow 4c^2+24c\le 0$
$\Rightarrow -6\le c\le 0$
But $c=-6$ and $c=0$ will not satisfy given condition
$\therefore c\in (-6,0)$