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


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



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

Case-1: Let x and y be two positive numbers, such that


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

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

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

$\Rightarrow x=y$

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


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

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


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

$\Rightarrow x=y$

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.

Let y be an element in the co-domain, such that y = f (x)

Case-1: Let $y>0 .$ Then, $0

$\Rightarrow y=f(x)=x|x|>0$

$\Rightarrow x>0$



$\Rightarrow x|x|=y$

$\Rightarrow x(x)=y$

$\Rightarrow x^{2}=y$

$\Rightarrow x=\sqrt{y} \in A($ We do not get $\pm$ because $x>0)$

Case-2: Let $y<0$. Then, $-1 \leq y<0$

$\Rightarrow y=f(x)=x|x|<0$

$\Rightarrow x<0$



$\Rightarrow x|x|=y$

$\Rightarrow x(-x)=y$


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

$\Rightarrow x=-\sqrt{-y} \in A$ (We do not get $\pm$ because $x>0$ )

$\Rightarrow f$ is onto.

$\Rightarrow f$ is a bijection.

So, the answer is (c).

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