Question

Which among the following functions from $R$ to $R$ is not one-one function.

• A. $f\left(x\right)=x^3$
• B. $g\left(x\right)=2x+1$
• C. $h\left(x\right)=x^2+x$
• D. None of the above

Answer 1

The correct option is C $h\left(x\right)=x^2+x$

For $f\left(x\right)=x^3$
Let $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow x_1^3=x_2^3$
$\Rightarrow x_1^3-x_2^3=0$
Let $x_2\ne 0$
$\Rightarrow [{\left(\frac{x_1}{x_2}\right)}^2+\left(\frac{x_1}{x_2}\right)+1](x_1-x_2)=0$
$\Rightarrow x_1=x_2$
$\because {\left(\frac{x_1}{x_2}\right)}^2+\left(\frac{x_1}{x_2}\right)+1>0$
So, $f\left(x\right)=x^3$ is one-one function.

For $g\left(x\right)=2x+1$
$g\left(x_1\right)=g\left(x_2\right)$
$\Rightarrow 2x_1+1=2x_2+1$
$\Rightarrow x_1=x_2$
So, $g\left(x\right)=2x+1$ is one-one function.

For $h\left(x\right)=x^2+x=x(x+1)$
$h\left(x_1\right)=h\left(x_2\right)$
$\Rightarrow x_1(x_1+1)=x_2(x_2+1)$
$\Rightarrow x_1^2-x_2^2+x_1-x_2=0$
$\Rightarrow (x_1-x_2)(x_1+x_2+1)=0$
$\Rightarrow x_1=x_2\text{or}{x}_1=x_2-1$
We are getting two different values for$x_2$
So, $h\left(x\right)=x^2+x$ is not one-one


Alternate solution:
$f\left(x\right)=x^3$

$g\left(x\right)=2x+1$

$h\left(x\right)=x^2+x$

Clearly, from the above graph's, using horizontal line test, we can state that $h\left(x\right)=x^2+x$ is not one-one function.