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Are the following set of ordered pairs function?If so,examine whether the mapping is injective or subjective $\;\{(x,y)\;:\;x \;is \;a \;person,y\; is\; the\; mother\; of\; x\}$

Note: This is part 1 of a 2  part question, split as 2 separate questions here.

1 Answer

  • A function $f: A \rightarrow B$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
  • A function$ f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f:\{(x,y):$ x is a person, y is mother of $x\}$
Let $x_1\; and \; x_2$ be two persons
Step1: Injective or One-One function:
$f(x_1)=f(x_2) =>$ both $x_1 \;and\; x_2 $ have same mother
but this does not imply $x_1 \;and\; x_2$ are same $x_1 \;need\;not\;be\; x_2$ they can be brother or sisters
Hence f is not injective
Step 2: Surjective or On-to function:
For every mother y defined by (x,y) there exists a person x for whom y is mother .
Solution:Therefore f is a surjective function
Option 'c' is correct
answered Mar 1, 2013 by meena.p
edited Mar 25, 2013 by meena.p

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