Question

$f:R\to R,f\left(x\right)=\frac{x^2+2x+c}{x^2+4x+c}$ is onto only if

• A. $c<2$
• B. $c\ge 0$
• C. $c<0$
• D. $|c|\le 1$

Answer 1

Answer: (A)

$c<2$

$f:R\to R,f\left(x\right)=\frac{x^2+2x+c}{x^2+4x+c}\Rightarrow y=\frac{x^2+2x+c}{x^2+4x+c}\Rightarrow x^{2}y+4xy+cy=x^2+2x+c\Rightarrow x^2(y-1)+2x(2y-1)+(cy-c)=0\because xϵR\therefore 4{(2y-1)}^2-4c{(y-1)}^2\ge 0{(2y-1)}^2-c{(y-1)}^2\ge 04y^2+1-4y-c{y}^2+2cy-1\ge 0y^2(4-c)-y(4-2c)\ge 0\because yϵR{(4-2c)}^2-4(4-c)\times 0<0{(4-2c)}^2<0{(2c-4)}^2<02c-4<0c<2$