question_answer

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.