Question 511 Mark
Let $f: R \rightarrow R$ be defined by $f(x)=x+|x|$. Then $f(x)$ is
Answer
View full question & answer→(d) : Given, $f(x)=x+|x|$
Now, $f(-2)=-2+|-2|=-2+2=0$
and $f(-3)=-3+|-3|=-3+3=0$
Hence, $f$ is not one-one
Also, $f(x)=\left\{\begin{array}{ll}x+x & \text { if } x \geq 0 \\ x-x & \text { if } x<0\end{array} \Rightarrow f(x)=\left\{\begin{array}{ll}2 x, & x \geq 0 \\ 0, & x<0\end{array}\right.\right.$
Thus, $f(x)=2 x \geq 0$ for all $x \geq 0$ and $f(x)=0$ for $x<0$. This means that $f(x)$ cannot be negative for any $x \in R$. So, $f$ is not onto. Note that $R_f=[0, \infty)$, which is a proper subset of $R$.
Now, $f(-2)=-2+|-2|=-2+2=0$
and $f(-3)=-3+|-3|=-3+3=0$
Hence, $f$ is not one-one
Also, $f(x)=\left\{\begin{array}{ll}x+x & \text { if } x \geq 0 \\ x-x & \text { if } x<0\end{array} \Rightarrow f(x)=\left\{\begin{array}{ll}2 x, & x \geq 0 \\ 0, & x<0\end{array}\right.\right.$
Thus, $f(x)=2 x \geq 0$ for all $x \geq 0$ and $f(x)=0$ for $x<0$. This means that $f(x)$ cannot be negative for any $x \in R$. So, $f$ is not onto. Note that $R_f=[0, \infty)$, which is a proper subset of $R$.