CBSE Class 12 Mathematics Relations and Functions Case Studies Set 02

Answer: (a) Neither Surjective nor Injective

Explanation:
1. f(−1) = 1 = f(1) but −1 ≠ 1, so f is not one-one (not injective).
2. x² ≥ 0 for every real x, so a negative number such as −2 in the co-domain R has no pre-image.
3. Hence f is not onto (not surjective) either: f is neither surjective nor injective.

Teacher's Note:
1. For f(x) = x² always test a pair x and −x first; it breaks one-one whenever both lie in the domain.
2. Onto fails whenever the co-domain contains negative numbers, because a square is never negative.

Answer: (c) Injective

Explanation:
1. If f(x₁) = f(x₂) then x₁² = x₂², and since x₁, x₂ are natural numbers (positive), x₁ = x₂. So f is one-one.
2. 2 is in the co-domain N, but no natural number x has x² = 2, so f is not onto.
3. Hence f is injective but not surjective.

Teacher's Note:
1. Restricting the domain to positive numbers removes the x, −x clash and makes x² one-one.
2. The range here is only the perfect squares {1, 4, 9, …}, a proper part of N, so onto fails.

Answer: (a) Bijective

Explanation:
1. For natural numbers, x₁² = x₂² gives x₁ = x₂, so f is one-one.
2. Every element of the co-domain is a perfect square n², and its pre-image is n, so f is onto.
3. One-one and onto together: f is bijective.

Teacher's Note:
1. Shrinking the co-domain to exactly the range always makes a function onto.
2. Compare with the previous part: the rule is the same, only the co-domain changed, and that changes the answer.

Answer: (a) {1, 4, 9, 16, …}

Explanation:
1. Range = set of images: f(1) = 1, f(2) = 4, f(3) = 9, f(4) = 16, …
2. Only perfect squares appear; 8, 10 and 15 are not squares of natural numbers.
3. Range = {1, 4, 9, 16, …}.

Teacher's Note:
1. The range depends only on the domain and the rule, not on the co-domain R.
2. Eliminate options quickly by spotting any non-square such as 8, 10 or 15.

Answer: (a) Neither Injective nor Surjective

Explanation:
1. f(−2) = 4 = f(2) with −2 ≠ 2, so f is not one-one.
2. −1 is in the co-domain Z, but x² = −1 has no integer solution, so f is not onto.
3. Hence f is neither injective nor surjective.

Teacher's Note:
1. Z contains both x and −x, so x² behaves exactly as it does on R here.
2. Even non-square positive integers like 2 have no pre-image, another reason onto fails.