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प्रश्न
Let
\[A = \left\{ x : - 1 \leq x \leq 1 \right\} \text{and} f : A \to \text{A such that f}\left( x \right) = x|x|\]
विकल्प
a bijection
injective but not surjective
surjective but not injective
neither injective nor surjective
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उत्तर
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\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow x\left| x \right| = y\left| y \right|\]
\[ \Rightarrow x\left( x \right) = y\left( y \right)\]
\[ \Rightarrow x^2 = y^2 \]
\[ \Rightarrow x = y\]
\[ \Rightarrow x\left| x \right| = y\left| y \right|\]
\[ \Rightarrow x\left( - x \right) = y\left( - y \right)\]
\[ \Rightarrow - x^2 = - y^2 \]
\[ \Rightarrow x^2 = y^2 \]
\[ \Rightarrow x = y\]
\[ \Rightarrow f\left( x \right) = x\left| x \right| \text{is positive and }f\left( y \right) = y\left| y \right| \text{is negative}\]
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\]
\[So, x \neq y\]
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\]
Let y be an element in the co-domain, such that y = f (x)
\[y = f\left( x \right) = x\left| x \right| > 0\]
\[ \Rightarrow x > 0\]
\[ \Rightarrow \left| x \right| = x\]
\[ \Rightarrow f\left( x \right) = y\]
\[ \Rightarrow x\left| x \right| = y\]
\[ \Rightarrow x\left( x \right) = y\]
\[ \Rightarrow x^2 = y\]
\[ \Rightarrow x = \sqrt{y} \in A \left( \text{We do not get}\pm, \text{as } x > 0 \right)\] \[\] \[Case-2: Lety<0.\text{Then},-1\leq y<0\]
\[y = f\left( x \right) = x\left| x \right| < 0\]
\[ \Rightarrow x < 0\]
\[ \Rightarrow \left| x \right| = - x\]
\[ \Rightarrow f\left( x \right) = y\]
\[ \Rightarrow x\left| x \right| = y\]
\[ \Rightarrow x\left( - x \right) = y\]
\[ \Rightarrow - x^2 = y\]
\[ \Rightarrow x^2 = - y\]
\[ \Rightarrow x = - \sqrt{- y} \in A \left( \text{We do not get}\pm, \text{ as } x>0 \right)\]
So, the answer is (a).
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