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. $f\left(x\right)=\frac{4x^2+3x-5}{4+3x+5x^2},x\in \left(-\infty ,\infty \right)$

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 $1\ne -1$ .
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 $4\ne 5$
The function $f\left(x\right)=\frac{4x^2+3x-5}{4+3x+5x^2},x\in \left(-\infty ,\infty \right)$ is not injective as $f\left(0\right)=f\left(-\frac{27}{41}\right)$ but $0\ne -\frac{27}{41}$
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