CUET · MATHS · PYQ PAPER 2023
The function \(f: R \rightarrow R, f(x)=x^2\) is :
- A injective but not surjective
- B surjective but not injective
- C injective as well as surjective
- D neither injective nor surjective
Answer & Solution
Correct Answer
(D) neither injective nor surjective
Step-by-step Solution
Detailed explanation
Given \(f(x)=x^2\). If \(f(x_1)=f(x_2) \implies x_1^2=x_2^2 \implies x_1=\pm x_2\). Not injective (e.g., \(f(2)=f(-2)=4\)). The codomain is \(R\). The range of \(f(x)=x^2\) is \([0, \infty)\). Since \([0, \infty) \neq R\), not surjective (e.g., \(y=-1\) has no pre-image in…
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