Question

The function $f:\left[0,3\right]\to \left[1,29\right]$ defined by $f\left(x\right)=2x^3-15x^2+36x+1$ is

• A. One-One and Onto
• B. Onto but not One-One
• C. One-One but not 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, $f\left(x\right)=2x^3-15x^2+36x+1$.

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(2x^3-15x^2+36x+1\right) \\ f\text{'}\left(x\right)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(-15x^2\right)+\frac{d}{dx}\left(36x\right)+\frac{d}{dx}\left(1\right)\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(u\pm v\right)\mathbf{=}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{\pm }\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\right] \\ f\text{'}\left(x\right)=\left(2\times 3x^2\right)-\left(15\times 2x\right)+36+0 \\ f\text{'}\left(x\right)=6x^2-30x+36 \\ f\text{'}\left(x\right)=6\left(x^2-5x+6\right) \\ f\text{'}\left(x\right)=6\left(x^2-3x-2x+6\right) \\ f\text{'}\left(x\right)=6\left[x\left(x-3\right)-2\left(x-3\right)\right] \\ f\text{'}\left(x\right)=6\left(x-3\right)\left(x-2\right) \end{gathered}$$

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

$$\begin{gathered} \Rightarrow 6\left(x-3\right)\left(x-2\right)>0 \\ \Rightarrow \left(x-3\right)\left(x-2\right)>0 \\ \Rightarrow x\in \left(-\infty ,2\right)\cup \left(3,\infty \right) \end{gathered}$$

The given function is defined in the domain $\left[0,3\right]$.

Thus, the function is increasing in nature in $[0,2)$ and decreasing in nature in $\left[2,3\right]$.

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

Now, for $x=0$, $f\left(0\right)=2{\left(0\right)}^3-15{\left(0\right)}^2+36\left(0\right)+1$.

$$\begin{gathered} \Rightarrow f\left(0\right)=0-0+0+1 \\ \Rightarrow f\left(0\right)=1 \end{gathered}$$

Now, for $x=2$, $f\left(2\right)=2{\left(2\right)}^3-15{\left(2\right)}^2+36\left(2\right)+1$.

$$\begin{gathered} \Rightarrow f\left(2\right)=16-60+72+1 \\ \Rightarrow f\left(2\right)=29 \end{gathered}$$

Now, for $x=3$, $f\left(3\right)=2{\left(3\right)}^3-15{\left(3\right)}^2+36\left(3\right)+1$.

$$\begin{gathered} \Rightarrow f\left(3\right)=54-135+108+1 \\ \Rightarrow f\left(3\right)=28 \end{gathered}$$

Thus, the range is $\left[0,29\right]$, which is equal to the given co-domain.

Hence, the given function is an Onto function.

Therefore, the function $f\left(x\right)=2x^3-15x^2+36x+1$ is Onto but not One-One.