Question

If ${}^{\prime }{x}^{\prime }$ is real, then the function $\left(\frac{x^2-x+c}{x^2+x+2c}\right)$ can take all real value if the value of c lies in

• 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: (B)

$c\in [-6,0]$

Let $y=\frac{x^2-x+c}{x^2+x+2c}$

$\Rightarrow y{x}^2+yx+2cy=x^2-x+c$

$\Rightarrow y{x}^2+yx+2cy-x^2+x-c=0$

$\Rightarrow (y-1)x^2+(y+1)x+2cy-c=0$ .......$\left(1\right)$

Since $x$ is real,

$\therefore$ Discriminant of equation$\left(1\right)\ge 0$

$\Rightarrow {(y+1)}^2-4\times (y-1)(2cy-c)\ge 0$

$\Rightarrow (y^2+2y+1)-4(2c{y}^2-cy-2cy+c)\ge 0$

$\Rightarrow (1-8c)y^2+2(1+6c)y+(1-4c)\ge 0$

$\Rightarrow 1-8c>0$ and $4{(1+6c)}^2-4(1-8c)(1-4c)\le 0$

$\Rightarrow 1-8c>0$ and ${(1+6c)}^2-(1-8c)(1-4c)\le 0$

$\Rightarrow -8c>-1$ and $1+36c^2+12c-1+12c-32c^2\le 0$

$\Rightarrow 8c<1$ and $4c^2+24c\le 0$

$\Rightarrow c<\frac{1}{8}$ and $4c(c+6)\le 0$

$\Rightarrow c<\frac{1}{8}$ and $c(c+6)\le 0$

$\Rightarrow c<\frac{1}{8}$ and $-6\le c\le 0$

$\therefore c\in (-\infty ,\frac{1}{8})\cap [-6,0]$

$\therefore c\in [-6,0]$