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प्रश्न
\[f : Z \to Z\] be given by
` f (x) = {(x/2, ", if x is even" ) ,(0 , ", if x is odd "):}`
Then, f is
विकल्प
onto but not one-one
one-one but not onto
one-one and onto
neither one-one nor onto
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उत्तर
Injectivity:
Let x and y be two elements in the domain (Z), such that
LaTeX
\[f\left( x \right) = f\left( y \right)\]
\[Case-1: \text{Let both x andybe even}.\]
\[\text{Then},\]
\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x}{2} = \frac{y}{2}\]
\[ \Rightarrow x = y\]
\[Case-2: \text{Let bothx andybe odd}.\]
\[\text{Then},\]
\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow 0 = 0\]
\[\text{Here, we cannot determine whether } x = y.\] So, f is not one-one.
Surjectivity:
Let y be an element in the co-domain (Z), such that
\[\text{Co-domain of f} = Z = \left\{ 0, \pm 1, \pm 2, \pm 3, \pm 4, . . . \right\} \]
\[\text{Range of f} = \left\{ 0, 0, \frac{\pm 2}{2}, 0, \frac{\pm 4}{2} , . . . \right\} = \left\{ 0, \pm 1, \pm 2, . . . \right\} \]
\[ \Rightarrow \text{Co-domain of f} = \text{Range of f}\]
\[\Rightarrow\] f is onto.
So, the answer is (a).
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