Practice Exercise 1.2

Let x₁, x₂ ∈ R* such that f(x₁) = f(x₂).
2/x₁ = 2/x₂
⇒ x₁ = x₂
Conclusion: f is one-one.

Let y ∈ R*. We need to find x ∈ R* such that f(x) = y.
2/x = y ⇒ x = 2/y.
Since y ≠ 0, 2/y is a defined non-zero real number.
So, x ∈ R* exists.
Conclusion: f is onto.

New function g: N → R* where g(x) = 2/x.
One-One: Still true (same logic).
Onto: Take y = 3 (which is in R*).
We need x ∈ N such that 2/x = 3 ⇒ x = 2/3.
But 2/3 is NOT a natural number (∉ N).
Conclusion: Not onto.

Let f(x₁) = f(x₂). x₁² + 1 = x₂² + 1 ⇒ x₁² = x₂².
Since domain is N, x₁, x₂ > 0, so x₁ = x₂.
One-One: Yes.

Range of f = {2, 5, 10, 17...}.
Numbers like 3, 4 ∈ N are not images of any x ∈ N (e.g., x² + 1 = 3 ⇒ x² = 2 ⇒ x = √2 ∉ N).
Onto: No.

Take x₁ = 1, x₂ = -1. Both are in Z.
f(1) = 1² + 1 = 2.
f(-1) = (-1)² + 1 = 2.
Different elements have same image.
One-One: No.

Range only includes positive integers ≥ 1. Negative integers in Z are not covered.
Onto: No.

Similar to Z case, f(1) = f(-1) = 2. Not One-One.
Range is [1, ∞). Negative numbers in Co-domain R are not covered. Not Onto.

x₁³ + 1 = x₂³ + 1 ⇒ x₁³ = x₂³ ⇒ x₁ = x₂.
One-One: Yes.

Range = {2, 9, 28...}. Many natural numbers (e.g., 3, 4) are skipped.
Onto: No.

Cube function preserves sign. Distinct integers have distinct cubes.
One-One: Yes.

We need x ∈ Z such that x³ + 1 = y. So x = ³√(y-1).
If y = 2, x = ³√(1) = 1 (Integer).
If y = 3, x = ³√(2) (Not an Integer).
Onto: No.

f(1.2) = [1.2] = 1.
f(1.9) = [1.9] = 1.
Different inputs (1.2 ≠ 1.9) give same output.
Not One-One.

Range of [x] is always Z (Integers).
Non-integers in Co-domain R (e.g., 2.5) have no pre-image.
Not Onto.

f(-2) = |-2| + 1 = 3.
f(2) = |2| + 1 = 3.
Two inputs map to 3.
Not One-One.

Since |x| ≥ 0, f(x) = |x| + 1 ≥ 1.
Range is [1, ∞). Numbers < 1 in Co-domain R are not covered.
Not Onto.

f(5) = 1.
f(100) = 1.
Many inputs map to single output '1'.
Not One-One (Many-One).

Range = {-1, 0, 1}.
Any real number other than these (e.g., 2, 5.5) has no pre-image.
Not Onto.

f(1) = 3.
f(2) = 4.
Distinct elements of A have distinct images in B.
Conclusion: f is one-one.

One-One: 2x₁ + 5 = 2x₂ + 5 ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂. (Yes)
Onto: y = 2x + 5 ⇒ x = (y-5)/2. For any real y, x is real. (Yes)
Result: Bijective.

One-One: f(1) = 2, f(-1) = 2. Not one-one.
Onto: Range is (-∞, 3]. Numbers > 3 not covered. Not onto.
Result: Not Bijective.

Let f(a₁, b₁) = f(a₂, b₂).
(b₁, a₁) = (b₂, a₂) ⇒ b₁ = b₂ and a₁ = a₂.
So (a₁, b₁) = (a₂, b₂). Yes.

For any (b, a) in Co-domain B × A, there exists (a, b) in Domain A × B such that f(a, b) = (b, a).
Yes.

f(1) = (1+1)/2 = 1.
f(2) = 2/2 = 1.
f(1) = f(2), but 1 ≠ 2.
Not One-One.

For any natural number y, consider 2y (even). f(2y) = 2y/2 = y.
So every y has a pre-image.
Is Onto.

(x₁-1)/(x₁-2) = (x₂-1)/(x₂-2)
Cross multiply: x₁x₂ - 2x₁ - x₂ + 2 = x₁x₂ - x₁ - 2x₂ + 2
-x₁ = -x₂ ⇒ x₁ = x₂.
Yes.

Let y = (x-1)/(x-2). Solve for x.
y(x-2) = x-1 ⇒ xy - 2y = x - 1
x(y-1) = 2y - 1 ⇒ x = (2y-1)/(y-1).
Since y ∈ B (y ≠ 1), x is defined. Also check if x ≠ 2 (if x=2, then 2y-1 = 2y-2 ⇒ -1=-2 impossible).
So x exists in A.
Yes.

One-One: f(1) = 1, f(-1) = 1. No.
Onto: x⁴ cannot be negative. Range [0, ∞) ≠ R. No.

One-One: 5x₁ = 5x₂ ⇒ x₁ = x₂. Yes.
Onto: y = 5x ⇒ x = y/5. For any real y, y/5 is real. Yes.