Match the following in List-I and List-II, for a function \(π\) :β¦
2018
Match the following inΒ List-IΒ andΒ List-II, for a functionΒ \(π\)Β :
\(\begin{array}{clcl} \text{} & \textbf{List-I} & & \textbf{List-II} \\ \text{(a)} & \text{$\forall \: x \: \forall \: y \: (f(x)=f(y) \rightarrow x=y)$} & \text{(i)} & \text{Constant} \\ \text{(b)} & \text{$\forall \: y \: \exists \: x \: (f(x) =y)$} & \text{(ii)} & \text{Injective} \\ \text{(c)} & \text{$\forall \: x \: f(x)=k$} & \text{(iii)} &\text{Surjective} \\ \end{array}\)
\(Code :\)
- A.
(a)-(i), (b)-(ii), (c)-(iii)
- B.
(a)-(iii), (b)-(ii), (c)-(i)
- C.
(a)-(ii), (b)-(i), (c)-(iii)
- D.
(a)-(ii), (b)-(iii), (c)-(i)
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Correct answer: D
Correct matching: (a) is Injective, (b) is Surjective, (c) is Constant.
For (a): The statement is βx βy (f(x)=f(y) β x=y). This means that whenever two outputs are equal the corresponding inputs must be equal, which is the definition of injective (one-to-one).
For (b): The statement is βy βx (f(x)=y). This means every element y of the codomain has at least one preimage x in the domain, which is the definition of surjective (onto).
For (c): The statement is βx f(x)=k for some fixed k. This says the function assigns the same value k to every input, which defines a constant function.
Therefore the correct correspondence is: (a) β Injective, (b) β Surjective, (c) β Constant. Other matchings fail because they swap these definitions.
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