Question

The function $f:\mathrm{ℝ}\to \mathrm{ℝ}$ given by $f\left(x\right)=x^3-1$ is

• A. a One-One function.
• B. an Onto function.
• C. a Bijection
• D. neither One-One nor Onto

Answer 1

The correct option is C

a Bijection

The explanation for the correct option

Given function, $f\left(x\right)=x^3-1$.

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(x^3-1\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(-1\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{u}\mathbf{\pm }\mathbf{v}\mathbf{)}\mathbf{=}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{\pm }\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ \Rightarrow f\text{'}\left(x\right)=\left(3x^2\right)+0 \\ \Rightarrow f\text{'}\left(x\right)=3x^2 \end{gathered}$$

As $3x^2$ is always non-negative, so $f\text{'}\left(x\right)\ge 0$.

Thus, the given function is always increasing.

Hence, the given function is a One-One function.

As the given function is a polynomial function, thus it is defined for all real values of $x$ and the range is also all real numbers.

Thus, the range is $\mathrm{ℝ}$, which is equal to the given co-domain.

Hence, the given function is an Onto function.

So, the given function is a bijection.

Hence, the correct option is $\mathrm{Option}\left(\mathrm{C}\right)$