| Assertion: A mapping shown in the following figure is not surjective |
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| Reason: A function \[f\,\,:\,\,A\to B\] is said to be surjective if every element of B has a pre-image in A. |
A) Both A and R are individually true and R is the correct explanation of A.
B) Both A and R are individually true and R is not the correct explanation of A.
C) 'A' is true but 'R' is false
D) 'A' is false but 'R' is true
E) Both A and R are false.
Correct Answer: A
Solution :
We know that A function \[y=f\left( x \right)\]is said to be surjective if for every 'y', there exist an x such that \[f\left( x \right)=y\]. Here set B has an extra element \[\therefore \]Given mapping is not surjective \[\Rightarrow \]Assertion is true Also Reason [R] is true and R is correct explanation of A. Hence option [A] is the correct answer.
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