Question

Let $A=\left\{x\in R:-1\le x\le 1\right\}=B$. Then, the mapping $f:A\to B\mathrm{given}\mathrm{by}f\left(x\right)=x\left|x\right|$ is
(a) injective but not surjective
(b) surjective but not injective
(c) bijective
(d) none of these

Answer 1

Injectivity:
Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow x\left|x\right|=y\left|y\right| \\ \Rightarrow x\left(x\right)=y\left(y\right) \\ \Rightarrow x^2=y^2 \\ \Rightarrow x=y \end{gathered}$$

Case-2: Let x and y be two negative numbers, such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow x\left|x\right|=y\left|y\right| \\ \Rightarrow x\left(-x\right)=y\left(-y\right) \\ \Rightarrow -x^2=-y^2 \\ \Rightarrow x^2=y^2 \\ \Rightarrow x=y \end{gathered}$$

Case-3: Let x be positive and y be negative.
$$\begin{gathered} \text{Then,}x\ne y \\ \Rightarrow f\left(x\right)=x\left|x\right|\text{is positive and} \\ f\left(y\right)=y\left|y\right|\text{is negative} \\ \Rightarrow f\left(x\right)\ne f\left(y\right) \\ \text{So, x}\ne y \\ \Rightarrow f\left(x\right)\ne f\left(y\right) \end{gathered}$$
From the 3 cases, we can conclude that f is one-one.

Surjectivity:
Let y be an element in the co-domain, such that y = f (x)
$$\begin{gathered} \text{Case-1: Let}\mathit{\text{y}}\text{>0. Then, 0<}\mathit{\text{y}}\text{≤1} \\ \Rightarrow y=f\left(x\right)=x\left|x\right|>0 \\ \Rightarrow x>0 \\ \Rightarrow \left|x\right|=x \\ f\left(x\right)=y \\ \Rightarrow x\left|x\right|=y \\ \Rightarrow x\left(x\right)=y \\ \Rightarrow x^2=y \\ \Rightarrow x=\sqrt{y}\in A\mathit{}\left(\text{We do not get ± because}\mathit{\text{x}}\text{>0}\right) \\ \text{Case-2: Let}\mathit{\text{y}}\text{<0. Then, -1≤}\mathit{\text{y}}\text{<0} \\ \Rightarrow y=f\left(x\right)=x\left|x\right|<0 \\ \Rightarrow x<0 \\ \Rightarrow \left|x\right|=-x \\ f\left(x\right)=y \\ \Rightarrow x\left|x\right|=y \\ \Rightarrow x\left(-x\right)=y \\ \Rightarrow -x^2=y \\ \Rightarrow x^2=-y \\ \Rightarrow x=-\sqrt{-y}\in A\mathit{}\left(\text{We do not get ± because}\mathit{\text{x}}\text{>0}\right) \end{gathered}$$
$\Rightarrow$f is onto.
$\Rightarrow$f is a bijection.
So, the answer is (c).