Question

Assertion (A): Let \( A = \{ x \in \mathbb{R} : -1 \leq x \leq 1 \} \). If \( f : A \to A \) be defined as \( f(x) = x^2 \), then \( f \) is not an onto function.
Reason (R): If \( y = -1 \in A \), then \( x = \pm \sqrt{-1} \notin A \).

• A. Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A)
• B. Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A)
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

To determine if a function is onto, check if every element in the codomain has a corresponding element in the domain. If any element in the codomain is not attainable, the function is not onto.

Answer 1

The Correct Option is A

- The function \( f(x) = x^2 \) is defined from the set \( A = \{ x \in \mathbb{R} : -1 \leq x \leq 1 \} \) to \( A \).
- A function is said to be "onto" (surjective) if for every element \( y \) in the codomain, there exists at least one \( x \) in the domain such that \( f(x) = y \).
In this case, the range of \( f(x) = x^2 \) is \( [0, 1] \), because for \( x \in [-1, 1] \), \( f(x) = x^2 \) takes values between 0 and 1. 
However, \( f(x) \) never attains the value \( -1 \), which is part of the set \( A \). 
Thus, \( f \) is not onto, as it does not map to all values in the codomain.
- The reason (R) is also correct. If \( y = -1 \), we would need to solve \( x^2 = -1 \), but this does not have any real solutions. 
Therefore, \( x = \pm \sqrt{-1} \notin A \), confirming that \( f \) is not onto.
Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) correctly explains Assertion (A).