# Let A={x ∈ R : −1≤x≤1}=B. Then, the mapping

Question:

Let $A=\{x \in R:-1 \leq x \leq 1\}=B$. Then, the mapping $f: A \rightarrow B$ given by $f(x)=x|x|$ is

(a) injective but not surjective
(b) surjective but not injective
(c) bijective
(d) none of these

Solution:

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

$f(x)=f(y)$

$\Rightarrow x|x|=y|y|$

$\Rightarrow x(x)=y(y)$

$\Rightarrow x^{2}=y^{2}$

$\Rightarrow x=y$

Case-2: Let x and y be two negative numbers, such that

$f(x)=f(y)$

$\Rightarrow x|x|=y|y|$

$\Rightarrow x(-x)=y(-y)$

$\Rightarrow-x^{2}=-y^{2}$

$\Rightarrow x^{2}=y^{2}$

$\Rightarrow x=y$

Case-3: Let be positive and y be negative.

Then, $x \neq y$

$\Rightarrow f(x)=x|x|$ is positive and

$f(y)=y|y|$ is negative

$\Rightarrow f(x) \neq f(y)$

So, $x \neq y$

$\Rightarrow f(x) \neq f(y)$

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)

Case-1: Let $y>0 .$ Then, $0$\Rightarrow y=f(x)=x|x|>0\Rightarrow x>0\Rightarrow|x|=xf(x)=y\Rightarrow x|x|=y\Rightarrow x(x)=y\Rightarrow x^{2}=y\Rightarrow x=\sqrt{y} \in A($We do not get$\pm$because$x>0)$Case-2: Let$y<0$. Then,$-1 \leq y<0\Rightarrow y=f(x)=x|x|<0\Rightarrow x<0\Rightarrow|x|=-xf(x)=y\Rightarrow x|x|=y\Rightarrow x(-x)=y\Rightarrow-x^{2}=y\Rightarrow x^{2}=-y\Rightarrow x=-\sqrt{-y} \in A$(We do not get$\pm$because$x>0$)$\Rightarrow f$is onto.$\Rightarrow f\$ is a bijection.