Question

Let $f:R\to R$ be a function defined by $f\left(x\right)=\frac{x^2-3x+4}{x^2+3x+4}$ then $f$ is

• A. One - one but not onto
• B. Onto but not one - one
• C. Onto as well as one - one
• D. Neither onto nor one - one

Answer 1

The correct option is D Neither onto nor one - one

Discriminant of $x^2-3x+4\to (3{)}^2-4(1\left)\right(4)=-7$

So, $x^2-3x+4>0$ for $x\in R$

Similarly, Discriminant of $x^2+3x+4\to (3{)}^2-4(1\left)\right(4)=-7$

So, $x^2+3x+4>0$ for $x\in R$

Thus,$f\left(x\right)\ge 0$ for $x\in R$ and Hence $f\left(x\right)$ is not onto .

Now, $f\left(x\right)=\frac{x^2-3x+4}{x^2+3x+4}=1-\frac{6x}{x^2+3x+4}$

$f^{\prime }\left(x\right)=-6\left\{\frac{(x^2+3x+4)-x(2x+3)}{(x^2+3x+4{)}^2}\right\}=\frac{6(x-2)(x+2)}{(x^2+3x+4{)}^2}$ .

Since $f^{\prime }\left(x\right)$ takes both positive as well as a negative value, we can claim that f(x) is not a one-one function.