Question

Which of the following functions is injective?

• A. $f\left(x\right)=x^{2}+3, \, x\in \left(- \infty , \, \infty\right)$
• B. $f\left(x\right)=\left|x + 1\right|, \, x\in \left[2 , \, \infty\right)$
• C. $f\left(x\right)=\left(x - 4\right)\left(x - 5\right), \, x\in \left(\infty , \, 5\right]$
• D. $f\left(x\right)=\frac{4 x^{2} + 3 x - 5}{4 + 3 x + 5 x^{2}}, \, x\in \, \left(- \infty , \, \infty\right)$

Answer 1

Answer: (B)

The function $f\left(x\right)=x^{2}+3, \, x\in \left(- \infty , \, \infty\right)$ is not injective as $f\left(1\right)=f\left(- 1\right)$ but $1\neq -1$ .
The function $f\left(x\right)=\left(x - 4\right)\left(x - 5\right), \, x\in \left(- \infty , \, 5\right]$ is not injective as $f\left(4\right)=f\left(5\right)$ but $4\neq 5$
The function $f\left(x\right)=\frac{4 x^{2} + 3 x - 5}{4 + 3 x + 5 x^{2}}, \, x\in \, \left(- \infty , \, \infty\right)$ is not injective as $f\left(0\right)=f\left(- \frac{27}{41}\right)$ but $0\neq -\frac{27}{41}$
For the function, $f\left(x\right)=\left|x + 1\right|, \, x\in \left[2 , \, \in fty\right)$
Let $f\left(x\right)=f\left(y\right), \, x,y\in \left[2 , \, \infty\right)\Rightarrow \left|x + 1\right|=\left|y + 1\right|$
$\Rightarrow x+1=y+1$
$\Rightarrow x=y$
So, $f$ is an injective