Match the LIST-I with LIST-II
[ Where R and Z denotes set of real numbers and set of integers respectively]
| LIST-I | LIST-II |
| Function | Nature |
| A. \( f:\mathrm{R}\to\mathrm{R},f(x)=\sin^2x+\cos^2x \) | I. Bijective |
| B. \( f:\mathrm{Z}\to\mathrm{Z},f(x)=x^3 \) | II. Surjective but not injective |
| C. \( f:\mathrm{R}\to\mathrm{R},f(x)=x^3-x \) | III. Injective but not surjective |
| D. \( f:\mathrm{R}\to\mathrm{R},f(x)=3-4x \) | IV. Neither injective nor surjective |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-III, C-II, D-I
(A) \( f:\mathrm{R}\to\mathrm{R},f(x)=\sin^2x+\cos^2x \)
This matches IV. Neither injective nor surjective. f(x) = 1 for every x: many inputs give the same value, and no other value is reached.
(B) \( f:\mathrm{Z}\to\mathrm{Z},f(x)=x^3 \)
This matches III. Injective but not surjective. Different integers have different cubes, but 2 is not the cube of any integer.
(C) \( f:\mathrm{R}\to\mathrm{R},f(x)=x^3-x \)
This matches II. Surjective but not injective. A cubic takes every real value, but f(0) = f(1) = 0.
(D) \( f:\mathrm{R}\to\mathrm{R},f(x)=3-4x \)
This matches I. Bijective. A non-constant linear function is one-one and onto R.
So the matching is A-IV, B-III, C-II, D-I.
Note: It seems that there is a mistake in the original NTA paper, which prints Option (1) and Option (3) exactly the same (A-I, B-II, C-III, D-IV). Neither of them is the correct matching, so the question can still be solved and the answer, Option (4), is unaffected.