CUET · MATHS · PYQ PAPER 2025

The function \(f: R \rightarrow R, f(x)=|x|\) (\(R\) is the set of real numbers) is

A injective but not surjective
B surjective but not injective
C both injective and surjective
D neither injective nor surjective

Verified Solution

Answer & Solution

Correct Answer

(D) neither injective nor surjective

Step-by-step Solution

Detailed explanation

The function is not injective because \(f(2) = 2\) and \(f(-2) = 2\) but \(2 \ne -2\). The function is not surjective because the codomain is \(R\) but the range is \([0, \infty)\). Neither injective nor surjective

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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List-l	List-ll
(A) \(A-A^T\)	(I) \(\|A\| A^{-1}\)
(B) \(A A^T\)	(II) Skew-symmetric
(C) \(\operatorname{det}\left(A^{-1}\right)\)	(III) Symmetric
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List-I	List-II
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(C) Angle between \(\hat{i}+2 \hat{j}\) and \(5 \hat{i}+10 \hat{j}\)	(III) \(\frac{\pi}{6}\)
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