Question

The function $f:\mathrm{ℝ}\to \mathrm{ℝ}$ defined by $f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$ is

• A. One-One but not Onto
• B. Onto but not One-One
• C. both One-One and Onto
• D. neither One-One nor Onto

Answer 1

The correct option is B

Onto but not One-One

The explanation for the correct option

Given function, $f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$.

$\Rightarrow f\left(x\right)=x^3-6x^2+11x-6$

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(x^3-6x^2+11x-6\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(-6x^2\right)+\frac{d}{dx}\left(11x\right)+\frac{d}{dx}\left(-6\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{u}\mathbf{\pm }\mathbf{v}\mathbf{)}\mathbf{=}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{\pm }\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ \Rightarrow f\text{'}\left(x\right)=\left(3x^2\right)-\left(6\times 2x\right)+11+0 \\ \Rightarrow f\text{'}\left(x\right)=3x^2-12x+11 \end{gathered}$$

Factorize the given quadratic expression using the quadratic formula.

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\left(x-\frac{-\left(-12\right)+\sqrt{{\left(-12\right)}^2-4\left(3\right)\left(11\right)}}{2\left(3\right)}\right)\left(x-\frac{-\left(-12\right)-\sqrt{{\left(-12\right)}^2-4\left(3\right)\left(11\right)}}{2\left(3\right)}\right) \\ \Rightarrow f\text{'}\left(x\right)=\left(x-\frac{6+\sqrt{3}}{3}\right)\left(x-\frac{6-\sqrt{3}}{3}\right) \end{gathered}$$

Now, for increasing intervals $\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{>}\mathbf{0}$.

$$\begin{gathered} \Rightarrow \left(x-\frac{6+\sqrt{3}}{3}\right)\left(x-\frac{6-\sqrt{3}}{3}\right)>0 \\ \Rightarrow x\in \left(-\infty ,\frac{6-\sqrt{3}}{3}\right)\cup \left(\frac{6+\sqrt{3}}{3},\infty \right) \end{gathered}$$

Thus, the given function is increasing in nature in $\left(-\infty ,\frac{6-\sqrt{3}}{3}\right)\cup \left(\frac{6+\sqrt{3}}{3},\infty \right)$ and decreasing in nature elsewhere.

Hence, the given function is a Many-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.

So, the given function is an Onto function.

Hence, option B is correct .