Question

Which among the following functions is not an injective function?

• A. $f\left(x\right)=|x+1|,x\in [-1,\infty )$
• B. $g\left(x\right)=x+\frac{1}{x},x\in (0,\infty )$
• C. $h\left(x\right)=x^2+4x-5,x\in (0,\infty )$
• D. $k\left(x\right)=e^{-x},x\in [0,\infty )$

Answer 1

The correct option is B $g\left(x\right)=x+\frac{1}{x},x\in (0,\infty )$

$f\left(x\right)=|x+1|,x\in [-1,\infty )$




$h\left(x\right)=x^2+4x-5,x\in (0,\infty )$



$k\left(x\right)=e^{-x},x\in [0,\infty )$


Clearly, from above graphs $f\left(x\right),h\left(x\right),k\left(x\right)$ using horizontal line test in the given intervals of each function, we can say that
they are injective functions in the given interval.

For $g\left(x\right)=x+\frac{1}{x},x\in (0,\infty )$
Let $g\left(x_1\right)=g\left(x_2\right)$
$\Rightarrow x_1+\frac{1}{x_1}=x_2+\frac{1}{x_2}$
$\Rightarrow x_1^2{x}_2+x_2=x_1{x}_2^2+x_1$
$\Rightarrow x_1{x}_2(x_1-x_2)-1(x_1-x_2)=0$
$\Rightarrow (x_1-x_2)(x_1{x}_2-1)=0$
Let $x_1{x}_2=1$
This relation is satisfied by infinite number of pairs $(x_1,x_2)$ where $x_1\ne x_2.$
Example: $(2,\frac{1}{2}),(3,\frac{1}{3}),(4,\frac{1}{4}),...$
$\therefore g\left(x\right)$ is many-one function.