Exercise 1.2 answers: Relations and Functions

Open NCERT p. 10Matches NCERT’s answer

1. One-one on R*: suppose f(x1) = f(x2), i.e. 1/x1 = 1/x2. Cross-multiplying gives x1 = x2, so f is one-one.
2. Onto R*: given any non-zero real number y, the number 1/y is also non-zero real, and f(1/y) = 1/(1/y) = y.
3. So every y in R* is reached, and f is onto — the original result holds.
4. Now replace the domain by N (only 1, 2, 3, ...), keeping the codomain R*.
5. One-one still holds, by the same cross-multiplication argument, which never used the domain being all of R*.
6. Onto fails now: take the target y = 2. We would need an input x in N with 1/x = 2, i.e. x = 1/2.
7. But 1/2 is not a natural number, so 2 is never reached — the function is no longer onto.

AnswerNo — with domain N, f is still one-one, but it is no longer onto (for example, 2 is never reached).

Watch this explained “Shrink the domain instead”, 10:49 into Onto as the demand that nothing in the codomain be left unhit

Open NCERT p. 10Matches NCERT’s answer

(i) f : N → N given by f(x) = x²

1. On N, all inputs are positive, so x1² = x2² forces x1 = x2 (squaring never turns two different positive numbers into the same value): injective.
2. 2 is not a perfect square of any natural number, so 2 is never reached: not surjective.

Answerf is injective but not surjective.

(ii) f : Z → Z given by f(x) = x²

1. On Z, negatives are allowed: f(1) = 1 and f(−1) = 1, two different inputs with the same output: not injective.
2. Negative targets like −1 are never reached, since a square is never negative: not surjective.

Answerf is neither injective nor surjective.

(iii) f : R → R given by f(x) = x²

1. The same collision as on Z: f(1) = f(−1) = 1: not injective.
2. No real number squares to a negative number such as −4: not surjective.

Answerf is neither injective nor surjective.

(iv) f : N → N given by f(x) = x³

1. On N, x1³ = x2³ forces x1 = x2, since cubing keeps positive numbers in the same order: injective.
2. 2 is not a perfect cube of any natural number, so 2 is never reached: not surjective.

Answerf is injective but not surjective.

(v) f : Z → Z given by f(x) = x³

1. Cubing keeps its sign and never repeats a value: if x1³ = x2³ then x1 = x2, for any integers, so f is injective.
2. 2 is not a perfect cube of any integer, so 2 is never reached: not surjective.

Answerf is injective but not surjective.

Watch this explained “A column that never varies”, 7:25 into Onto as the demand that nothing in the codomain be left unhit

Open NCERT p. 10One way to think about it

1. The greatest integer function maps each real number to the greatest integer ≤ that number.
2. For example: [1.5] = 1, [1.3] = 1, [2.7] = 2.
3. To show not one-one: Find two different inputs with the same output.
4. f(1.5) = [1.5] = 1 and f(1.3) = [1.3] = 1. So 1.5 ≠ 1.3 but f(1.5) = f(1.3).
5. Therefore, f is not one-one.
6. To show not onto: Find an element in ℝ that is not in the range.
7. The range of f is ℤ (the set of integers), since f(x) is always an integer.
8. But 1.5 ∈ ℝ and 1.5 ∉ ℤ, so 1.5 is not in the range of f.
9. Therefore, f is not onto ℝ.
10. Conclusion: f is neither one-one nor onto.

In shortThe greatest integer function is not one-one because f(1.5) = f(1.3) = 1, and it is not onto because the range is ℤ while the codomain is ℝ.

Watch this explained “Three that fail, three different ways”, 13:43 into What one-one asks of distinct inputs, and what many-one allows

Open NCERT p. 11One way to think about it

1. |x| gives the distance of x from 0, so it is never negative.
2. Not one-one: f(2) = |2| = 2 and f(−2) = |−2| = 2. Two different inputs, 2 and −2, give the same output.
3. Not onto: |x| can never be negative, so an output like −1, which is in the codomain R, is never produced.
4. So f is neither one-one nor onto.

In shortf is neither one-one nor onto.

Watch this explained “Three that fail, three different ways”, 13:43 into What one-one asks of distinct inputs, and what many-one allows

Open NCERT p. 11One way to think about it

1. The signum function only ever gives one of three outputs: 1, 0 or −1.
2. Not one-one: f(2) = 1 and f(5) = 1. Two different inputs, 2 and 5, give the same output.
3. Not onto: the output is always 1, 0 or −1, so a value like 2, which is in the codomain R, is never produced.
4. So f is neither one-one nor onto.

In shortf is neither one-one nor onto.

Watch this explained “Three that fail, three different ways”, 13:43 into What one-one asks of distinct inputs, and what many-one allows

Open NCERT p. 11One way to think about it

1. f sends 1 → 4, 2 → 5, and 3 → 6.
2. There are only three inputs, so list their outputs: 4, 5 and 6.
3. All three outputs are different from each other, so no two different inputs share an output.
4. So f is one-one.

In shortf is one-one.

Watch this explained “What checking every pair costs”, 1:20 into What one-one asks of distinct inputs, and what many-one allows

Open NCERT p. 11Matches NCERT’s answer

(i) f : R → R defined by f(x) = 3 − 4x

1. One-one: suppose f(x₁) = f(x₂), so 3 − 4x₁ = 3 − 4x₂.
2. This gives −4x₁ = −4x₂, so x₁ = x₂. So f is one-one.
3. Onto: for any real y, solve y = 3 − 4x to get x = (3 − y)/4, which is always a real number.
4. So every y has an input that reaches it, and f is onto.
5. f is one-one and onto, so f is bijective.

Answerf is one-one, onto, and bijective.

(ii) f : R → R defined by f(x) = 1 + x²

1. Not one-one: f(1) = 1 + 1² = 2 and f(−1) = 1 + (−1)² = 2. Different inputs, same output.
2. Not onto: 1 + x² is always at least 1, so a value like 0, which is in the codomain R, is never produced.
3. So f is neither one-one nor onto, hence not bijective.

Answerf is neither one-one nor onto (not bijective).

Watch this explained “A column that never varies”, 7:25 into Onto as the demand that nothing in the codomain be left unhit

Open NCERT p. 11One way to think about it

1. One-one: suppose f(a₁, b₁) = f(a₂, b₂), that is (b₁, a₁) = (b₂, a₂).
2. Two ordered pairs are equal only when their matching entries are equal, so b₁ = b₂ and a₁ = a₂.
3. That means (a₁, b₁) = (a₂, b₂), so f is one-one.
4. Onto: take any element (b, a) in B × A. Then (a, b) is in A × B, and f(a, b) = (b, a).
5. So every element of B × A is reached, and f is onto.
6. f is one-one and onto, so f is bijective.

In shortf is bijective.

Watch this explained “A bijection between two different sets”, 15:45 into Bijections, and why on a finite set either half implies the other

Open NCERT p. 11Matches NCERT’s answer

1. Check f at n = 1 and n = 2: f(1) = (1+1)/2 = 1, and f(2) = 2/2 = 1.
2. Two different inputs, 1 and 2, give the same output 1, so f is not one-one.
3. Since f is not one-one, it cannot be bijective, no matter whether it is onto.

Answerf is not bijective (it is not one-one).

Watch this explained “A third one, and a warning”, 12:36 into Bijections, and why on a finite set either half implies the other

Open NCERT p. 11Matches NCERT’s answer

1. First, f really lands in B: f(x) = 1 would need x − 2 = x − 3, that is −2 = −3, which is impossible. So f(x) is never 1.
2. One-one: suppose f(x₁) = f(x₂) for x₁, x₂ in A (neither is 3): (x₁ − 2)/(x₁ − 3) = (x₂ − 2)/(x₂ − 3).
3. Cross-multiply: (x₁ − 2)(x₂ − 3) = (x₂ − 2)(x₁ − 3).
4. Expand: x₁x₂ − 3x₁ − 2x₂ + 6 = x₁x₂ − 3x₂ − 2x₁ + 6.
5. Cancel x₁x₂ and 6 from both sides: −3x₁ − 2x₂ = −3x₂ − 2x₁, so −x₁ = −x₂, that is x₁ = x₂. So f is one-one.
6. Onto: take any y in B, so y ≠ 1. Solve y = (x − 2)/(x − 3): y(x − 3) = x − 2, so xy − x = 3y − 2, so x(y − 1) = 3y − 2.
7. Since y ≠ 1, we can divide: x = (3y − 2)/(y − 1), which is a real number.
8. This x is never 3: (3y − 2)/(y − 1) = 3 would mean 3y − 2 = 3y − 3, that is −2 = −3, which is false. So x is in A.
9. Check: x − 2 = y/(y − 1) and x − 3 = 1/(y − 1), so f(x) = y. Every y in B is reached, so f is onto.
10. So f is both one-one and onto.

AnswerYes, f is one-one and onto.

Watch this explained “Two exclusions, each doing a job”, 15:33 into Onto as the demand that nothing in the codomain be left unhit

Open NCERT p. 11Matches NCERT’s answer

1. f(1) = 1⁴ = 1 and f(−1) = (−1)⁴ = 1, so two different inputs give the same output — f is not one-one.
2. x⁴ can never be negative, so a value like −1, which is in the codomain R, is never produced — f is not onto.
3. So f is neither one-one nor onto.

Answer(D) f is neither one-one nor onto.

Watch this explained “A bijection between two different sets”, 15:45 into Bijections, and why on a finite set either half implies the other

Open NCERT p. 11Matches NCERT’s answer

1. One-one: if 3x₁ = 3x₂, dividing both sides by 3 gives x₁ = x₂.
2. Onto: for any real y, x = y/3 is a real number, and f(y/3) = 3(y/3) = y.
3. So f is one-one and onto.

Answer(A) f is one-one onto.

Watch this explained “A bijection between two different sets”, 15:45 into Bijections, and why on a finite set either half implies the other