Let $A=\{x:-1 \leq x \leq 1\}$ and $f: A \rightarrow A$ such that $f(x)=x|x|$, then $f$ is
(a) a bijection
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective
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 x 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, $x \neq y$
$\Rightarrow f(x) \neq f(y)$
So, $f$ 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
$y=f(x)=x|x|>0$
$\Rightarrow x>0$
$\Rightarrow|x|=x$
$\Rightarrow f(x)=y$
$\Rightarrow x|x|=y$
$\Rightarrow x(x)=y$
$\Rightarrow x^{2}=y$
$\Rightarrow x=\sqrt{y} \in A($ We do not get $\pm$, as $x>0)$
Case-2: Let $y<0 .$ Then, $-1 \leq y<0$
$y=f(x)=x|x|<0$
$\Rightarrow x<0$
$\Rightarrow|x|=-x$
$\Rightarrow f(x)=y$
$\Rightarrow x|x|=y$
$\Rightarrow x(-x)=y$
$\Rightarrow-x^{2}=y$
$\Rightarrow x^{2}=-y$
$\Rightarrow x=-\sqrt{-y} \in A($ We do not get $\pm$, as $x>0)$
$\Rightarrow f$ is onto
$\Rightarrow f$ is a bijection.
So, the answer is (a).
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