Question

Let $f:R\to R$ where $f\left(x\right)=\frac{x^2+4x+7}{x^2+x+1},$ then $f\left(x\right)$ is

• A. many-one function
• B. not a function itself
• C. a constant function
• D. one-one function

Answer 1

The correct option is A many-one function

$f\left(x\right)=\frac{x^2+4x+7}{x^2+x+1}$
$\Rightarrow f\left(x\right)=\frac{(x+2{)}^2+3}{{(x+\frac{1}{2})}^2+\frac{3}{4}}$
$\Rightarrow f\left(x\right)=4\left[\frac{(x+2{)}^2+3}{(2x+1{)}^2+3}\right]$

Now $|x+2|=|2x+1|\Rightarrow x=\pm 1$
$\therefore f\left(1\right)=f(-1)\Rightarrow f$ is many one function.