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 for \( x \in A = \{ x \in \mathbb{R} : -1 \leq x \leq 1 \} \) and maps to \( A \).
- A function is "onto" (surjective) if every element \( y \) in the codomain has at least one corresponding element \( x \) in the domain such that \( f(x) = y \).
The range of \( f(x) = x^2 \) for \( x \in [-1, 1] \) is \( [0, 1] \).
Since \( f(x) \) does not produce the value \( -1 \), which is an element of the codomain \( A \), \( f \) is not onto. 
- The reason (R) is valid. Solving \( x^2 = -1 \) for \( y = -1 \) yields \( x = \pm \sqrt{-1} \), which are not real numbers and therefore not in \( A \). 
This confirms that \( f \) is not onto because it fails to map to all elements in the codomain. Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) provides a valid explanation for Assertion (A).