Mastering One-One and Onto Functions: NCERT Ex 1.2 Explained

Understanding One-One and Onto Functions

Functions are a cornerstone of mathematics, mapping elements from one set (domain) to another (codomain). In Class 12, we refine our understanding by classifying functions based on how these mappings occur. Exercise 1.2 of NCERT focuses specifically on two crucial types: one-one functions and onto functions. A function f: A -> B is essentially a rule that assigns each element x from set A (the domain) to exactly one element y from set B (the codomain). The set of all actual output values f(x) is called the range, which is always a subset of the codomain.

One-one (Injective) Function: Imagine a scenario where each student in a class (domain) is assigned a unique roll number (codomain). No two different students can have the same roll number. This perfectly illustrates a one-one function. Mathematically, it means if f(x1) = f(x2), then it must imply x1 = x2. Conversely, if x1 ≠ x2, then f(x1) ≠ f(x2). Each distinct element in the domain maps to a distinct element in the codomain.

Onto (Surjective) Function: Now, consider a concert hall where every seat (codomain) is occupied by at least one person (domain). It means there are no empty seats. In function terms, for every element y in the codomain B, there exists at least one element x in the domain A such that f(x) = y. This implies that the range of the function is equal to its codomain. If there's even one element in the codomain that doesn't have a pre-image in the domain, the function is not onto.

Formal Definitions of Functions

One-one Function (Injective)

A function f: A → B is said to be one-one (or injective) if distinct elements of A have distinct images in B. That is, for all x1, x2 ∈ A, if f(x1) = f(x2), then x1 = x2. Equivalently, if x1 ≠ x2, then f(x1) ≠ f(x2).

Onto Function (Surjective)

A function f: A → B is said to be onto (or surjective) if every element of B is the image of some element of A under f. That is, for every y ∈ B, there exists an element x ∈ A such that f(x) = y. This implies that the range of f is equal to the codomain B.

Bijective Function

A function f: A → B is said to be bijective if it is both one-one (injective) and onto (surjective). Bijective functions are also known as one-to-one correspondence.

Step-by-Step Guide to Proving One-One and Onto

1. Proving a Function is One-One (Injective) — To prove f: A → B is one-one: 1. Assume f(x1) = f(x2): Start by taking two arbitrary elements x1, x2 from the domain A and assume their images under f are equal. 2. Algebraic Manipulation: Perform algebraic steps to simplify the equation f(x1) = f(x2). 3. Conclude x1 = x2: If you can successfully show that f(x1) = f(x2) necessarily implies x1 = x2, then the function is one-one. If you find a counterexample (i.e., f(x1) = f(x2) but x1 ≠ x2), then the function is not one-one.
2. Proving a Function is Onto (Surjective) — To prove f: A → B is onto: 1. Take an arbitrary y from the codomain B: Let y be any element in the codomain. 2. Set f(x) = y: Equate the function definition f(x) to y. 3. Solve for x in terms of y: Rearrange the equation to express x as a function of y. Let's say x = g(y). 4. Check if x is in the domain A: Verify that for every y in the codomain B, the corresponding x = g(y) exists and is an element of the domain A. If such an x always exists within the domain for every y in the codomain, then the function is onto. If there is any y in the codomain for which no such x exists in the domain, then the function is not onto.

Worked Examples: Applying One-One and Onto Proofs

• Example 1: Show that the function f: N → N given by f(x) = 2x is one-one but not onto, where N is the set of natural numbers. Solution: Step 1: Check for One-one (Injectivity) Let x1, x2 ∈ N such that f(x1) = f(x2). Then, 2x1 = 2x2. Dividing both sides by 2, we get x1 = x2. Since f(x1) = f(x2) implies x1 = x2, the function f(x) = 2x is one-one. Step 2: Check for Onto (Surjectivity) Let y be an arbitrary element in the codomain N. We need to find an x ∈ N such that f(x) = y. So, 2x = y. This implies x = y/2. Now, consider y = 1 (which is in the codomain N). Then x = 1/2. However, 1/2 is not a natural number, i.e., 1/2 ∉ N (the domain). Therefore, for y = 1 (and many other odd natural numbers), there is no x in the domain N such that f(x) = y. Hence, the function f(x) = 2x is not onto. Final Answer: The function f(x) = 2x is one-one but not onto.
• Example 2: Show that the function f: R → R defined by f(x) = x^3 + 1 is a bijection (both one-one and onto). Solution: Step 1: Check for One-one (Injectivity) Let x1, x2 ∈ R such that f(x1) = f(x2). Then, x1^3 + 1 = x2^3 + 1. Subtracting 1 from both sides: x1^3 = x2^3. Taking the cube root of both sides: x1 = x2. Since f(x1) = f(x2) implies x1 = x2, the function f(x) = x^3 + 1 is one-one. Step 2: Check for Onto (Surjectivity) Let y be an arbitrary element in the codomain R. We need to find an x ∈ R such that f(x) = y. So, x^3 + 1 = y. x^3 = y - 1. Taking the cube root of both sides: x = (y - 1)^(1/3). For any real number y, (y - 1)^(1/3) is always a real number. This means for every y in the codomain R, there exists a real number x = (y - 1)^(1/3) in the domain R such that f(x) = y. Hence, the function f(x) = x^3 + 1 is onto. Final Answer: Since f(x) = x^3 + 1 is both one-one and onto, it is a bijective function.

Exam Tips and Common Pitfalls

When solving problems from Relations and Function Ex 1.2, paying attention to the details can make a huge difference. Here are some critical exam tips and common pitfalls to avoid:

1. Always Check Domain and Codomain: This is the most crucial aspect! A function f(x) = x^2 is not one-one if its domain is R (since f(2) = 4 and f(-2) = 4). However, if its domain is N (natural numbers), it is one-one. Similarly, f(x) = x^2 with codomain R is not onto (negative numbers in R have no pre-image), but with codomain [0, ∞), it can be onto. The nature of the domain and codomain fundamentally determines if a function is one-one or onto.
2. Clear Steps for Proof: For board exams, present your one-one and onto proofs systematically. Clearly state your assumption (f(x1) = f(x2) for one-one, and f(x) = y for onto), show all algebraic manipulations, and state your conclusion explicitly.
3. Counterexamples for 'Not One-one' or 'Not Onto': If a function is not one-one or not onto, providing a single, clear counterexample is often sufficient and easier than trying to algebraically disprove it in general terms. For example, for f: R → R, f(x) = x^2, state that f(2) = 4 and f(-2) = 4 but 2 ≠ -2, so it's not one-one. For not onto, show an element in the codomain with no pre-image (e.g., for f(x) = x^2, -1 ∈ R has no real x such that f(x) = -1).
4. Graphical Interpretation: While not always a formal proof, visualizing functions graphically can give you a strong intuition. For one-one, a horizontal line should intersect the graph at most once. For onto, the range of the function should cover the entire codomain (e.g., extend infinitely up and down if the codomain is R).
5. Don't Confuse Injective and Surjective: Ensure you understand the distinct requirements for each. A common mistake is to mix up the conditions or misapply the proof techniques.

Practice Questions with Solutions

• Q: Show that the function f: R → R defined by f(x) = |x| is neither one-one nor onto. A: Step 1: Check for One-one. Let x1 = 2 and x2 = -2. Then f(x1) = |2| = 2 and f(x2) = |-2| = 2. Here, f(x1) = f(x2) but x1 ≠ x2. Thus, f(x) = |x| is not one-one. Step 2: Check for Onto. Let y = -3 be an element in the codomain R. We need to find an x ∈ R such that f(x) = -3. So, |x| = -3. However, the absolute value of any real number is always non-negative. There is no real number x whose absolute value is -3. Thus, f(x) = |x| is not onto. Final answer: The function f(x) = |x| is neither one-one nor onto.
• Q: Let f: Z → Z be defined by f(x) = x^2. Determine if f is one-one and onto. (Z is the set of integers). A: Step 1: Check for One-one. Let x1 = 1 and x2 = -1. Then f(x1) = 1^2 = 1 and f(x2) = (-1)^2 = 1. Here, f(x1) = f(x2) but x1 ≠ x2. Thus, f(x) = x^2 is not one-one. Step 2: Check for Onto. Let y = 2 be an element in the codomain Z. We need to find an x ∈ Z such that f(x) = 2. So, x^2 = 2. This implies x = ±√2. Since ±√2 are not integers, there is no x ∈ Z such that f(x) = 2. Thus, f(x) = x^2 is not onto. Final answer: The function f(x) = x^2 is neither one-one nor onto.
• Q: Show that the function f: R → R given by f(x) = 5 - 4x is a bijection. A: Step 1: Check for One-one. Let x1, x2 ∈ R such that f(x1) = f(x2). Then 5 - 4x1 = 5 - 4x2. Subtracting 5 from both sides: -4x1 = -4x2. Dividing by -4: x1 = x2. Since f(x1) = f(x2) implies x1 = x2, f is one-one. Step 2: Check for Onto. Let y be an arbitrary element in the codomain R. We need to find x ∈ R such that f(x) = y. So, 5 - 4x = y. 4x = 5 - y. x = (5 - y)/4. For any y ∈ R, (5 - y)/4 is always a real number. Thus, for every y in the codomain, there exists an x in the domain such that f(x) = y. So, f is onto. Final answer: Since f(x) = 5 - 4x is both one-one and onto, it is a bijective function.
• Q: Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b, f(3) = a. Is f one-one? Is it onto? A: Step 1: Check for One-one. We have f(1) = a and f(3) = a. Here, f(1) = f(3) but 1 ≠ 3. Therefore, the function f is not one-one. Step 2: Check for Onto. The codomain is {a, b, c}. The range of f is {f(1), f(2), f(3)} = {a, b, a} = {a, b}. Since c is in the codomain but not in the range (i.e., c has no pre-image), the function f is not onto. Final answer: The function f is neither one-one nor onto.

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