(A) f(x)=1/x on R−{0} is bijective, IV.
(B) f(x)=x2 on N is injective but not surjective (e.g., 2 has no preimage), III.
(C) f(x)=x2 on R is neither (since f(1)=f(−1) and negatives missed), I.
(D) {(1,1),(2,2),(3,1)}: surjective, not injective, II.
Match List - I with List - II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | f(x)=x1,f:R−{0}→R−{0} | (I) | neither injective nor surjective |
| (B) | f(x)=x2,f:N→N | (II) | surjective but not injective |
| (C) | f(x)=x2,f:R→R | (III) | injective but not surjective |
| (D) | f:{1,2,3}→{1,2} defined as f:{(1,1),(2,2),(3,1)} | (IV) | injective and surjective |
Choose the correct answer from the options given below :
Held on 22 May 2023 · Verified 13 Jul 2026.
(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
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