Assertion: A function \[f\,\,:\,\,Z\to Z\] defined as \[f\left( x \right)={{x}^{2}}\] is injective. |
Reason: A function \[f\,\,:\,\,A\to B\] is said to be injective 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: E
Solution :
Given \[f\left( x \right)={{x}^{2}}\] |
here \[f\left( -1 \right)={{\left( -1 \right)}^{2}}=1\,;\,\,f\left( 1 \right)={{\left( 1 \right)}^{2}}=1\] |
\[\Rightarrow \,\,f\left( -1 \right)=f\left( 1 \right)=1\] |
\[\Rightarrow \,\,f\left( x \right)\] is not one-one |
\[\therefore \]Assertion A is false |
We know that A function f (x) is said to be injective if corresponding to every element of A, there is one and only one image. |
\[\therefore \]Given Reason [R] is false |
Hence option [E] is the correct answer. |
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