Question

If the function: $R\to R$, is defined by $f\left(x\right)=\left|x\right|\left(x-\sin x\right)$, then which of the following statements is TRUE?

• A. Function is one-one, but not onto
• B. Function is onto, but not one-one
• C. Function is both one-one and onto
• D. Function is neither one-one nor onto

Answer 1

The correct option is C

Function is both one-one and onto

Explanation for the correct option

Step 1: Identify the nature of the given function

Given function, $f\left(x\right)=\left|x\right|\left(x-\sin x\right)$

Identifying if the function is odd, by applying the condition $f(-x)=-f\left(x\right)$

$$\begin{gathered} \Rightarrow f(-x)=-\left(\left|x\right|\left(x-\sin x\right)\right) \\ \Rightarrow -f\left(x\right)=-\left(\left|x\right|\left(x-\sin x\right)\right) \end{gathered}$$

So, $f(-x)=-f\left(x\right)$ for the given function. This also demonstrates that the function is continuous.

Thus, $f\left(x\right)=\left|x\right|\left(x-\sin x\right)$ is an odd, non-periodic continuous function.

Step 2: Evaluate the domain and range of the function

The function can be written as $f\left(x\right)=\left\{\begin{array}{l}{x}^2-x\sin x,x\ge 0\\ -x^2+x\sin x,x<0\end{array}\right.\left[\because \left|x\right|=\left\{\begin{array}{l}+x;\forall x\ge 0\\ -x;\forall x<0\end{array}\right.\right]$

So,

$$\begin{gathered} f\left(x^{+}\right)=\underset{x\to \infty }{\lim }\left(x^2\right)\left(1-\frac{\sin x}{x}\right) \\ f\left(x^{+}\right)=\infty \end{gathered}$$

and,

$$\begin{gathered} f\left(x^{-}\right)=\underset{x\to \infty }{\lim }\left(-x^2\right)\left(1-\frac{\sin x}{x}\right) \\ f\left(x^{-}\right)=-\infty \end{gathered}$$

Thus, the range of $f\left(x\right)=R$ and $f\left(x\right)$ is an onto function.

Differentiate the given function, $f\left(x\right)=\left\{\begin{array}{l}{x}^2-x\sin x,x\ge 0\\ -x^2+x\sin x,x<0\end{array}\right.$, we get,

$\Rightarrow f\text{'}\left(x\right)=\left\{\begin{array}{l}2x-\sin x-x\cos x,x\ge 0\\ -2x+\sin x+x\cos x,x<0\end{array}\right.$

$\Rightarrow f\text{'}\left(x\right)=\left(x-\sin x\right)+\left(x\left(1-\cos x\right)\right)$, where the result will always be positive or zero.

So, $f\text{'}\left(x\right)>0\forall x\in \left(-\infty ,\infty \right)$

Thus, $f\left(x\right)$ is a one-one function.

Therefore, option (C) i.e. Function is both one-one and onto, is the correct answer.