Assertion (A): Let A ={ x ∈ R:-1 ≤ x ≤ 1}. If f: A → A be defined as f ( x )= x^2, then f is not an onto function. Reason ( R ): If y =-1 ∈ A, then x…

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