Class 12 Mathematics | Chapter 1
Relations and Functions
Equivalence Relations • Injective & Surjective Functions • Composition
1. Introduction to Relations
A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. If A = B, then R is called a "relation on A".
R ⊆ A × A
If (a, b) ∈ R, we say that a is related to b under the relation R, and write it as a R b.
1.1 Types of Relations
- Empty Relation: A relation R on set A is empty if no element of A is related to any element of A, i.e., R = ∅ ⊆ A × A.
- Universal Relation: A relation R on set A is universal if each element of A is related to every element of A, i.e., R = A × A.
- Trivial Relations: Both the empty relation and the universal relation are sometimes called trivial relations.
2. Reflexive, Symmetric and Transitive Relations
A relation R on a set A is classified into three fundamental types:
1. Reflexive Relation:
R is reflexive if (a, a) ∈ R for every a ∈ A.
Example: On the set of real numbers, the relation "is equal to" (=) is reflexive because every number is equal to itself (a = a).
R is reflexive if (a, a) ∈ R for every a ∈ A.
Example: On the set of real numbers, the relation "is equal to" (=) is reflexive because every number is equal to itself (a = a).
2. Symmetric Relation:
R is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A.
Example: On the set of lines, the relation "is perpendicular to" is symmetric because if line L&sub1; ⊥ L&sub2;, then L&sub2; ⊥ L&sub1;.
R is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A.
Example: On the set of lines, the relation "is perpendicular to" is symmetric because if line L&sub1; ⊥ L&sub2;, then L&sub2; ⊥ L&sub1;.
3. Transitive Relation:
R is transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A.
Example: On the set of integers, the relation "is less than" (<) is transitive because if a < b and b < c, then a < c.
R is transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A.
Example: On the set of integers, the relation "is less than" (<) is transitive because if a < b and b < c, then a < c.
2.1 Equivalence Relation
A relation R on a set A is said to be an Equivalence Relation if it is Reflexive, Symmetric, AND Transitive.
Standard Example: Let T be the set of all triangles in a plane with R a relation in T given by R = {(T&sub1;, T&sub2;) : T&sub1; is congruent to T&sub2;}. Since congruence is reflexive, symmetric, and transitive, R is an equivalence relation.
2.2 Equivalence Classes
Given an equivalence relation R on a set A, the equivalence class of an element 'a' ∈ A, denoted by [a], is the set of all elements in A that are related to 'a'.
[a] = { x ∈ A : (x, a) ∈ R }
Properties of equivalence classes:
• All elements of [a] are related to each other.
• No element of [a] is related to any element of a different equivalence class [b].
• The union of all distinct equivalence classes gives the entire set A (partitioning of the set).
3. Types of Functions
A function f: X → Y relates each element of set X (Domain) to exactly one element of set Y (Codomain). The set of all images forms the Range.
Injective
One-One Function:
A function f: X → Y is defined to be one-one (or injective), if the images of distinct elements of X under f are distinct.
Mathematical Condition: For every x&sub1;, x&sub2; ∈ X,
If f(x&sub1;) = f(x&sub2;), then x&sub1; = x&sub2;.
Otherwise, f is called many-one.
One-One Function:
A function f: X → Y is defined to be one-one (or injective), if the images of distinct elements of X under f are distinct.
Mathematical Condition: For every x&sub1;, x&sub2; ∈ X,
If f(x&sub1;) = f(x&sub2;), then x&sub1; = x&sub2;.
Otherwise, f is called many-one.
Surjective
Onto Function:
A function f: X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f.
Mathematical Condition: For every y ∈ Y, there exists an element x ∈ X such that f(x) = y.
Equivalently, a function is onto if its Range = Codomain.
Onto Function:
A function f: X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f.
Mathematical Condition: For every y ∈ Y, there exists an element x ∈ X such that f(x) = y.
Equivalently, a function is onto if its Range = Codomain.
3.1 Bijective Function
A function f: X → Y is said to be bijective if it is BOTH one-one and onto. A function must be bijective to be invertible.
4. Composition of Functions
Let f: A → B and g: B → C be two functions. Then the composition of f and g, denoted by g o f, is defined as the function g o f: A → C.
gof(x) = g(f(x)), ∀ x ∈ A
Properties of Composition:
- Composition of functions is generally not commutative: f o g ≠ g o f.
- Composition of functions is associative: if f, g, h are three functions such that their composition is defined, then h o (g o f) = (h o g) o f.
5. Invertible Functions
A function f: X → Y is defined to be invertible, if there exists a function g: Y → X such that:
g o f = IX and f o g = IY
where IX and IY are identity functions on sets X and Y respectively.
The function g is called the inverse of f and is denoted by f−1.
Crucial Theorem: A function f: X → Y is invertible if and only if f is a bijective function (one-one and onto).
Steps to find the inverse of a function f(x):
- Verify that f(x) is one-one and onto (bijective).
- Put y = f(x).
- Solve the equation y = f(x) to express x in terms of y (i.e., x = g(y)).
- The expression g(y) gives the inverse function f−1(y). Replacing y with x gives f−1(x).
CLASS 12 MATHEMATICS | NCERT EXERCISE SOLUTIONS
Chapter 1 — Relations and Functions
22 Solved Questions — Equivalence Relations, Bijective Functions & Composition
📝 Part A: Equivalence Relations (Q1 - Q10)
2 MarksQ1. Determine whether the relation R in the
set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x - y = 0} is reflexive, symmetric and
transitive.
✓ Solution
R = {(x, y) : 3x = y} ⇒ R = {(1, 3), (2, 6), (3, 9), (4, 12)}.
Reflexivity: (1, 1) ∉ R because 3(1) ≠ 1. So, R is NOT reflexive.
Symmetry: Since (1, 3) ∈ R, but (3, 1) ∉ R (as 3(3) ≠ 1). So, R is NOT symmetric.
Transitivity: Since (1, 3) ∈ R and (3, 9) ∈ R, but (1, 9) ∉ R (as 3(1) ≠ 9). So, R is NOT transitive.
R = {(x, y) : 3x = y} ⇒ R = {(1, 3), (2, 6), (3, 9), (4, 12)}.
Reflexivity: (1, 1) ∉ R because 3(1) ≠ 1. So, R is NOT reflexive.
Symmetry: Since (1, 3) ∈ R, but (3, 1) ∉ R (as 3(3) ≠ 1). So, R is NOT symmetric.
Transitivity: Since (1, 3) ∈ R and (3, 9) ∈ R, but (1, 9) ∉ R (as 3(1) ≠ 9). So, R is NOT transitive.
3 MarksQ2. Show that the relation R in the set Z of
integers given by R = {(a, b) : 2 divides (a - b)} is an equivalence relation.
✓ Solution
Reflexivity: For any a ∈ Z, a - a = 0. Since 2 divides 0, (a, a) ∈ R. Thus, R is reflexive.
Symmetry: Let (a, b) ∈ R ⇒ 2 divides (a - b).
This means a - b = 2k for some integer k. Then, b - a = 2(-k). Since -k is an integer, 2 divides (b - a). So, (b, a) ∈ R. Thus, R is symmetric.
Transitivity: Let (a, b) ∈ R and (b, c) ∈ R.
a - b = 2x and b - c = 2y for integers x, y.
Adding both: (a - b) + (b - c) = 2x + 2y ⇒ a - c = 2(x + y).
Since x+y is an integer, 2 divides (a - c). So, (a, c) ∈ R. Thus, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Reflexivity: For any a ∈ Z, a - a = 0. Since 2 divides 0, (a, a) ∈ R. Thus, R is reflexive.
Symmetry: Let (a, b) ∈ R ⇒ 2 divides (a - b).
This means a - b = 2k for some integer k. Then, b - a = 2(-k). Since -k is an integer, 2 divides (b - a). So, (b, a) ∈ R. Thus, R is symmetric.
Transitivity: Let (a, b) ∈ R and (b, c) ∈ R.
a - b = 2x and b - c = 2y for integers x, y.
Adding both: (a - b) + (b - c) = 2x + 2y ⇒ a - c = 2(x + y).
Since x+y is an integer, 2 divides (a - c). So, (a, c) ∈ R. Thus, R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
3 MarksQ3. Check whether the relation R defined in
the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
✓ Solution
Set A = {1, 2, 3, 4, 5, 6}.
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.
Reflexivity: (1, 1) ∉ R. Not reflexive.
Symmetry: (1, 2) ∈ R but (2, 1) ∉ R. Not symmetric.
Transitivity: (1, 2) ∈ R and (2, 3) ∈ R, but (1, 3) ∉ R. Not transitive.
Set A = {1, 2, 3, 4, 5, 6}.
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.
Reflexivity: (1, 1) ∉ R. Not reflexive.
Symmetry: (1, 2) ∈ R but (2, 1) ∉ R. Not symmetric.
Transitivity: (1, 2) ∈ R and (2, 3) ∈ R, but (1, 3) ∉ R. Not transitive.
4 MarksQ4. Show that the relation R in the set A = {x
∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a - b| is a multiple of 4} is an equivalence
relation. Find the equivalence class [1].
✓ Solution
Reflexivity: |a - a| = 0, which is a multiple of 4. So, (a, a) ∈ R.
Symmetry: Let (a, b) ∈ R. Then |a - b| = 4k.
|b - a| = |-(a - b)| = |a - b| = 4k. Thus, (b, a) ∈ R.
Transitivity: Let (a, b) ∈ R and (b, c) ∈ R. Then a - b = ±4m and b - c = ±4n.
a - c = (a - b) + (b - c) = ±4m ± 4n = 4(±m ± n).
Thus |a - c| is a multiple of 4. So, (a, c) ∈ R.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Equivalence class [1]: Elements x ∈ A such that |x - 1| is a multiple of 4.
x - 1 = 0 ⇒ x=1; x - 1 = 4 ⇒ x=5; x - 1 = 8 ⇒ x=9; x - 1 = 12 ⇒ x=13 (not in A).
So, [1] = {1, 5, 9}.
Reflexivity: |a - a| = 0, which is a multiple of 4. So, (a, a) ∈ R.
Symmetry: Let (a, b) ∈ R. Then |a - b| = 4k.
|b - a| = |-(a - b)| = |a - b| = 4k. Thus, (b, a) ∈ R.
Transitivity: Let (a, b) ∈ R and (b, c) ∈ R. Then a - b = ±4m and b - c = ±4n.
a - c = (a - b) + (b - c) = ±4m ± 4n = 4(±m ± n).
Thus |a - c| is a multiple of 4. So, (a, c) ∈ R.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Equivalence class [1]: Elements x ∈ A such that |x - 1| is a multiple of 4.
x - 1 = 0 ⇒ x=1; x - 1 = 4 ⇒ x=5; x - 1 = 8 ⇒ x=9; x - 1 = 12 ⇒ x=13 (not in A).
So, [1] = {1, 5, 9}.
2 MarksQ5. Show that the relation R in R (Real
numbers) defined as R = {(a, b) : a ≤ b} is reflexive and transitive but not symmetric.
✓ Solution
Reflexivity: For any a ∈ R, a ≤ a is true. So (a, a) ∈ R.
Transitivity: Let (a, b) ∈ R ⇒ a ≤ b, and let (b, c) ∈ R ⇒ b ≤ c. From these, a ≤ c. Thus (a, c) ∈ R.
Symmetry: If (a, b) ∈ R, then a ≤ b. This does NOT mean b ≤ a.
Counterexample: (2, 5) ∈ R because 2 ≤ 5. But (5, 2) ∉ R because 5 is not ≤ 2. Therefore, R is not symmetric.
Reflexivity: For any a ∈ R, a ≤ a is true. So (a, a) ∈ R.
Transitivity: Let (a, b) ∈ R ⇒ a ≤ b, and let (b, c) ∈ R ⇒ b ≤ c. From these, a ≤ c. Thus (a, c) ∈ R.
Symmetry: If (a, b) ∈ R, then a ≤ b. This does NOT mean b ≤ a.
Counterexample: (2, 5) ∈ R because 2 ≤ 5. But (5, 2) ∉ R because 5 is not ≤ 2. Therefore, R is not symmetric.
3 MarksQ6. Show that the relation R in R (Real
numbers) defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.
✓ Solution
Reflexivity: Let a = 1/2. Is 1/2 ≤ (1/2)²? No, 0.5 is not less than 0.25.
Symmetry: Let a = 1, b = 4. (1, 4) ∈ R because 1 ≤ 16. But (4, 1) ∉ R because 4 ≤ 1² is false.
Transitivity: Let a = 3, b = 2, c = 1.5.
(3, 2) ∈ R because 3 ≤ 4. (2, 1.5) ∈ R because 2 ≤ 2.25.
But (3, 1.5) ∉ R because 3 ≤ (1.5)² = 2.25 is false.
Reflexivity: Let a = 1/2. Is 1/2 ≤ (1/2)²? No, 0.5 is not less than 0.25.
Symmetry: Let a = 1, b = 4. (1, 4) ∈ R because 1 ≤ 16. But (4, 1) ∉ R because 4 ≤ 1² is false.
Transitivity: Let a = 3, b = 2, c = 1.5.
(3, 2) ∈ R because 3 ≤ 4. (2, 1.5) ∈ R because 2 ≤ 2.25.
But (3, 1.5) ∉ R because 3 ≤ (1.5)² = 2.25 is false.
2 MarksQ7. Prove that the relation R defined on the
set A = {1, 2, 3} as R = {(1, 2), (2, 1)} is symmetric, but neither reflexive nor transitive.
✓ Solution
R is symmetric because (1, 2) ∈ R and (2, 1) ∈ R.
R is not reflexive because (1, 1), (2, 2), (3, 3) are not in R.
R is not transitive because (1, 2) ∈ R and (2, 1) ∈ R, but (1, 1) ∉ R.
R is symmetric because (1, 2) ∈ R and (2, 1) ∈ R.
R is not reflexive because (1, 1), (2, 2), (3, 3) are not in R.
R is not transitive because (1, 2) ∈ R and (2, 1) ∈ R, but (1, 1) ∉ R.
3 MarksQ8. Let L be the set of all lines in XY plane
and R be the relation in L defined as R = {(L&sub1;, L&sub2;) : L&sub1; is parallel to L&sub2;}. Show
that R is an equivalence relation.
✓ Solution
Reflexivity: Every line L&sub1; is parallel to itself. So (L&sub1;, L&sub1;) ∈ R.
Symmetry: If L&sub1; || L&sub2;, then L&sub2; || L&sub1;. So (L&sub1;, L&sub2;) ∈ R ⇒ (L&sub2;, L&sub1;) ∈ R.
Transitivity: If L&sub1; || L&sub2; and L&sub2; || L&sub3;, then L&sub1; || L&sub3; (lines parallel to the same line are parallel to each other). So (L&sub1;, L&sub3;) ∈ R.
Thus, R is an equivalence relation.
Reflexivity: Every line L&sub1; is parallel to itself. So (L&sub1;, L&sub1;) ∈ R.
Symmetry: If L&sub1; || L&sub2;, then L&sub2; || L&sub1;. So (L&sub1;, L&sub2;) ∈ R ⇒ (L&sub2;, L&sub1;) ∈ R.
Transitivity: If L&sub1; || L&sub2; and L&sub2; || L&sub3;, then L&sub1; || L&sub3; (lines parallel to the same line are parallel to each other). So (L&sub1;, L&sub3;) ∈ R.
Thus, R is an equivalence relation.
3 MarksQ9. Show that the relation R defined in the
set A of all polygons as R = {(P&sub1;, P&sub2;) : P&sub1; and P&sub2; have same number of sides}, is an
equivalence relation.
✓ Solution
Reflexive: Every polygon has the same number of sides as itself. So (P, P) ∈ R.
Symmetric: If P&sub1; has the same number of sides as P&sub2;, then P&sub2; has the same number of sides as P&sub1;. (P&sub1;, P&sub2;) ∈ R ⇒ (P&sub2;, P&sub1;) ∈ R.
Transitive: If P&sub1; and P&sub2; have the same number of sides, and P&sub2; and P&sub3; have the same number of sides, then P&sub1; and P&sub3; have the same number of sides. (P&sub1;, P&sub3;) ∈ R.
Hence, R is an equivalence relation.
Reflexive: Every polygon has the same number of sides as itself. So (P, P) ∈ R.
Symmetric: If P&sub1; has the same number of sides as P&sub2;, then P&sub2; has the same number of sides as P&sub1;. (P&sub1;, P&sub2;) ∈ R ⇒ (P&sub2;, P&sub1;) ∈ R.
Transitive: If P&sub1; and P&sub2; have the same number of sides, and P&sub2; and P&sub3; have the same number of sides, then P&sub1; and P&sub3; have the same number of sides. (P&sub1;, P&sub3;) ∈ R.
Hence, R is an equivalence relation.
2 MarksQ10. Let A = {1, 2, 3}. Then number of
relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is?
✓ Solution
To be reflexive, R must contain (1, 1), (2, 2), (3, 3).
It must contain (1, 2) and (1, 3).
To be symmetric, it must also contain (2, 1) and (3, 1).
So, minimum R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}.
Let us test for transitivity: (2, 1) ∈ R and (1, 3) ∈ R, but (2, 3) ∉ R. So this R is not transitive.
If we add (2, 3) and (3, 2) to R, it becomes the universal relation on A, which is transitive.
Therefore, there is exactly 1 such relation.
To be reflexive, R must contain (1, 1), (2, 2), (3, 3).
It must contain (1, 2) and (1, 3).
To be symmetric, it must also contain (2, 1) and (3, 1).
So, minimum R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)}.
Let us test for transitivity: (2, 1) ∈ R and (1, 3) ∈ R, but (2, 3) ∉ R. So this R is not transitive.
If we add (2, 3) and (3, 2) to R, it becomes the universal relation on A, which is transitive.
Therefore, there is exactly 1 such relation.
💡 Part B: Injective and Surjective Functions (Q11 - Q17)
3 MarksQ11. Check the injectivity and surjectivity of
the function f : N → N given by f(x) = x².
✓ Solution
Injectivity (One-One): Let f(x&sub1;) = f(x&sub2;).
x&sub1;² = x&sub2;² ⇒ x&sub1; = x&sub2; (since x&sub1;, x&sub2; ∈ N, they are positive). Thus, f is injective.
Surjectivity (Onto): Let y ∈ N (Codomain). Then f(x) = y ⇒ x² = y ⇒ x = √y.
If we take y = 2, then x = √2, which is not a natural number. So no element in domain maps to 2.
Thus, the function is NOT surjective.
Injectivity (One-One): Let f(x&sub1;) = f(x&sub2;).
x&sub1;² = x&sub2;² ⇒ x&sub1; = x&sub2; (since x&sub1;, x&sub2; ∈ N, they are positive). Thus, f is injective.
Surjectivity (Onto): Let y ∈ N (Codomain). Then f(x) = y ⇒ x² = y ⇒ x = √y.
If we take y = 2, then x = √2, which is not a natural number. So no element in domain maps to 2.
Thus, the function is NOT surjective.
3 MarksQ12. Check the injectivity and surjectivity of
the function f : Z → Z given by f(x) = x².
✓ Solution
Injectivity (One-One): f(2) = 4 and f(-2) = 4. Since two different elements have the same image, f is NOT injective (many-one).
Surjectivity (Onto): For y = -2 ∈ Z (Codomain), x² = -2 has no solution in Z (or even in Real numbers). Therefore, f is NOT surjective.
Injectivity (One-One): f(2) = 4 and f(-2) = 4. Since two different elements have the same image, f is NOT injective (many-one).
Surjectivity (Onto): For y = -2 ∈ Z (Codomain), x² = -2 has no solution in Z (or even in Real numbers). Therefore, f is NOT surjective.
3 MarksQ13. Show that the Modulus Function f : R
→ R, given by f(x) = |x|, is neither one-one nor onto.
✓ Solution
One-One: f(1) = |1| = 1 and f(-1) = |-1| = 1.
Since f(1) = f(-1) but 1 ≠ -1, the function is NOT one-one.
Onto: The codomain is R. However, the output/range of the modulus function is always non-negative ([0, ∞)). For any negative real number, say y = -3 in the codomain, there exists no x in R such that |x| = -3. Therefore, the function is NOT onto.
One-One: f(1) = |1| = 1 and f(-1) = |-1| = 1.
Since f(1) = f(-1) but 1 ≠ -1, the function is NOT one-one.
Onto: The codomain is R. However, the output/range of the modulus function is always non-negative ([0, ∞)). For any negative real number, say y = -3 in the codomain, there exists no x in R such that |x| = -3. Therefore, the function is NOT onto.
4 MarksQ14. Show that the Signum Function f : R
→ R, given by f(x) = 1 (if x > 0), 0 (if x = 0), and -1 (if x < 0), is neither one-one nor
onto.
✓ Solution
One-One: For x = 2 and x = 5, both are > 0, so f(2) = 1 and f(5) = 1.
Since distinct elements have the same image, it is NOT one-one.
Onto: The range of the signum function contains only three elements: {-1, 0, 1}. The codomain is R. Because the range ≠ codomain (e.g., there is no x such that f(x) = 2), the function is NOT onto.
One-One: For x = 2 and x = 5, both are > 0, so f(2) = 1 and f(5) = 1.
Since distinct elements have the same image, it is NOT one-one.
Onto: The range of the signum function contains only three elements: {-1, 0, 1}. The codomain is R. Because the range ≠ codomain (e.g., there is no x such that f(x) = 2), the function is NOT onto.
3 MarksQ15. Let A and B be sets. Show that f : A
× B → B × A such that f(a, b) = (b, a) is bijective.
✓ Solution
One-One: Let f(a&sub1;, b&sub1;) = f(a&sub2;, b&sub2;).
Then (b&sub1;, a&sub1;) = (b&sub2;, a&sub2;).
Equating coordinates: b&sub1; = b&sub2; and a&sub1; = a&sub2;.
Thus, (a&sub1;, b&sub1;) = (a&sub2;, b&sub2;). So f is one-one.
Onto: For any element (b, a) ∈ B × A, there exists an element (a, b) in A × B such that f(a, b) = (b, a).
Thus, f is onto.
Since it is both one-one and onto, it is bijective.
One-One: Let f(a&sub1;, b&sub1;) = f(a&sub2;, b&sub2;).
Then (b&sub1;, a&sub1;) = (b&sub2;, a&sub2;).
Equating coordinates: b&sub1; = b&sub2; and a&sub1; = a&sub2;.
Thus, (a&sub1;, b&sub1;) = (a&sub2;, b&sub2;). So f is one-one.
Onto: For any element (b, a) ∈ B × A, there exists an element (a, b) in A × B such that f(a, b) = (b, a).
Thus, f is onto.
Since it is both one-one and onto, it is bijective.
4 MarksQ16. Let A = R - {3} and B = R - {1}. Consider
the function f : A → B defined by f(x) = (x - 2) / (x - 3). Is f one-one and onto? Justify.
✓ Solution
One-One: Let f(x&sub1;) = f(x&sub2;).
x&sub1;x&sub2; - 3x&sub1; - 2x&sub2; + 6 = x&sub1;x&sub2; - 3x&sub2; - 2x&sub1; + 6
-3x&sub1; - 2x&sub2; = -3x&sub2; - 2x&sub1; ⇒ x&sub2; = x&sub1;. Thus, f is one-one.
Onto: Let y ∈ B. Then y = (x - 2) / (x - 3).
xy - 3y = x - 2
x(y - 1) = 3y - 2 ⇒ x = (3y - 2) / (y - 1).
Since y ≠ 1, x is a real number. Also, x cannot be 3 (if x=3, then 3y-3 = 3y-2 ⇒ -3=-2, absurd). So x ∈ A.
Thus, for every y ∈ B, there exists an x ∈ A. Hence f is onto (Bijective).
One-One: Let f(x&sub1;) = f(x&sub2;).
(x&sub1; - 2) / (x&sub1; - 3) = (x&sub2; - 2) / (x&sub2; - 3)
Cross-multiply: (x&sub1; - 2)(x&sub2; - 3) = (x&sub2; - 2)(x&sub1; - 3)x&sub1;x&sub2; - 3x&sub1; - 2x&sub2; + 6 = x&sub1;x&sub2; - 3x&sub2; - 2x&sub1; + 6
-3x&sub1; - 2x&sub2; = -3x&sub2; - 2x&sub1; ⇒ x&sub2; = x&sub1;. Thus, f is one-one.
Onto: Let y ∈ B. Then y = (x - 2) / (x - 3).
xy - 3y = x - 2
x(y - 1) = 3y - 2 ⇒ x = (3y - 2) / (y - 1).
Since y ≠ 1, x is a real number. Also, x cannot be 3 (if x=3, then 3y-3 = 3y-2 ⇒ -3=-2, absurd). So x ∈ A.
Thus, for every y ∈ B, there exists an x ∈ A. Hence f is onto (Bijective).
3 MarksQ17. Show that f : R → R defined as f(x)
= x³ is a bijection.
✓ Solution
One-One: Let f(x&sub1;) = f(x&sub2;) ⇒ x&sub1;³ = x&sub2;³ ⇒ x&sub1;³ - x&sub2;³ = 0
(x&sub1; - x&sub2;)(x&sub1;² + x&sub1;x&sub2; + x&sub2;²) = 0.
The quadratic part is never zero for real numbers unless both are zero. Thus x&sub1; = x&sub2;. So f is one-one.
Onto: Let y ∈ R. Then y = x³ ⇒ x = y^(1/3).
Since the cube root of a real number is always real, x ∈ R for all y ∈ R.
Thus, f is onto, hence bijective.
One-One: Let f(x&sub1;) = f(x&sub2;) ⇒ x&sub1;³ = x&sub2;³ ⇒ x&sub1;³ - x&sub2;³ = 0
(x&sub1; - x&sub2;)(x&sub1;² + x&sub1;x&sub2; + x&sub2;²) = 0.
The quadratic part is never zero for real numbers unless both are zero. Thus x&sub1; = x&sub2;. So f is one-one.
Onto: Let y ∈ R. Then y = x³ ⇒ x = y^(1/3).
Since the cube root of a real number is always real, x ∈ R for all y ∈ R.
Thus, f is onto, hence bijective.
📈 Part C: Composition of Functions & Inverses (Q18 - Q22)
3 MarksQ18. If f: R → R is defined by f(x) = (3
- x³)^(1/3), find fof(x).
✓ Solution
fof(x) = f(f(x)) = f( (3 - x³)^(1/3) )
Substitute the inner function into f in place of 'x':
fof(x) = ( 3 - [ (3 - x³)^(1/3) ]³ )^(1/3)
The cube cancels the 1/3 power:
= ( 3 - (3 - x³) )^(1/3)
= ( 3 - 3 + x³ )^(1/3)
= (x³)^(1/3) = x.
fof(x) = f(f(x)) = f( (3 - x³)^(1/3) )
Substitute the inner function into f in place of 'x':
fof(x) = ( 3 - [ (3 - x³)^(1/3) ]³ )^(1/3)
The cube cancels the 1/3 power:
= ( 3 - (3 - x³) )^(1/3)
= ( 3 - 3 + x³ )^(1/3)
= (x³)^(1/3) = x.
3 MarksQ19. Let f(x) = x / (1 + |x|) be defined from
R → (-1, 1). Is f one-one and onto?
✓ Solution
Case 1 (x ≥ 0): f(x) = x / (1 + x).
f(x&sub1;) = f(x&sub2;) ⇒ x&sub1;/(1+x&sub1;) = x&sub2;/(1+x&sub2;) ⇒ x&sub1; = x&sub2;.
For onto: y = x/(1+x) ⇒ x = y/(1-y). Since y ∈ [0, 1), x ≥ 0. Onto.
Case 2 (x < 0): f(x) = x / (1 - x).
f(x&sub1;) = f(x&sub2;) ⇒ x&sub1;/(1-x&sub1;) = x&sub2;/(1-x&sub2;) ⇒ x&sub1; = x&sub2;.
For onto: y = x/(1-x) ⇒ x = y/(1+y). Since y ∈ (-1, 0), x < 0. Onto.
It is both one-one and onto, thus bijective.
Case 1 (x ≥ 0): f(x) = x / (1 + x).
f(x&sub1;) = f(x&sub2;) ⇒ x&sub1;/(1+x&sub1;) = x&sub2;/(1+x&sub2;) ⇒ x&sub1; = x&sub2;.
For onto: y = x/(1+x) ⇒ x = y/(1-y). Since y ∈ [0, 1), x ≥ 0. Onto.
Case 2 (x < 0): f(x) = x / (1 - x).
f(x&sub1;) = f(x&sub2;) ⇒ x&sub1;/(1-x&sub1;) = x&sub2;/(1-x&sub2;) ⇒ x&sub1; = x&sub2;.
For onto: y = x/(1-x) ⇒ x = y/(1+y). Since y ∈ (-1, 0), x < 0. Onto.
It is both one-one and onto, thus bijective.
3 MarksQ20. Find gof and fog, given f(x) = |x| and
g(x) = |5x - 2|.
✓ Solution
gof(x): = g(f(x)) = g( |x| ).
Substitute |x| into g in place of x:
gof(x) = |5|x| - 2|.
fog(x): = f(g(x)) = f( |5x - 2| ).
Substitute |5x - 2| into f in place of x:
fog(x) = | |5x - 2| | = |5x - 2| (since the modulus of a modulus is just the modulus).
gof(x): = g(f(x)) = g( |x| ).
Substitute |x| into g in place of x:
gof(x) = |5|x| - 2|.
fog(x): = f(g(x)) = f( |5x - 2| ).
Substitute |5x - 2| into f in place of x:
fog(x) = | |5x - 2| | = |5x - 2| (since the modulus of a modulus is just the modulus).
2 MarksQ21. Are f and g inverses of each other if
fog(x) = x and gof(x) = x? If f(x) = x³ + 5, what is its inverse?
✓ Solution
Yes, if fog = I and gof = I, f and g are inverses of each other.
To find inverse of f(x) = x³ + 5:
Let y = x³ + 5
x³ = y - 5
x = (y - 5)^(1/3)
So, the inverse function f−1(x) = (x - 5)^(1/3).
Yes, if fog = I and gof = I, f and g are inverses of each other.
To find inverse of f(x) = x³ + 5:
Let y = x³ + 5
x³ = y - 5
x = (y - 5)^(1/3)
So, the inverse function f−1(x) = (x - 5)^(1/3).
3 MarksQ22. Consider f : R+ → [-5, ∞)
given by f(x) = 9x² + 6x - 5. Show that f is invertible with f−1(y) = (√(y +
6) - 1) / 3.
✓ Solution
Let y = 9x² + 6x - 5.
We rewrite it by completing the square:
y = (3x)² + 2(3x)(1) + 1 - 1 - 5
y = (3x + 1)² - 6.
Thus, (3x + 1)² = y + 6
3x + 1 = +√(y + 6) (since x ∈ R+, 3x+1 must be positive. We reject the negative root).
3x = √(y + 6) - 1
x = (√(y + 6) - 1) / 3.
Since each y ∈ [-5, ∞) gives a unique and valid x ∈ R+, f is one-one and onto. Thus, the derived expression is the inverse function f−1(y).
Let y = 9x² + 6x - 5.
We rewrite it by completing the square:
y = (3x)² + 2(3x)(1) + 1 - 1 - 5
y = (3x + 1)² - 6.
Thus, (3x + 1)² = y + 6
3x + 1 = +√(y + 6) (since x ∈ R+, 3x+1 must be positive. We reject the negative root).
3x = √(y + 6) - 1
x = (√(y + 6) - 1) / 3.
Since each y ∈ [-5, ∞) gives a unique and valid x ∈ R+, f is one-one and onto. Thus, the derived expression is the inverse function f−1(y).
✍ Math Problem Solving Guide
Questions carefully graded starting from simple equivalence relation verification up to composition of functions and finding inverse functions by completing the square, aligned directly with Class 12 board logic.
Questions carefully graded starting from simple equivalence relation verification up to composition of functions and finding inverse functions by completing the square, aligned directly with Class 12 board logic.
CLASS 12 MATHEMATICS | FORMULA CAPSULE
Relations and Functions
Chapter 1 — Complete Logic & Conditions Sheet for Boards & CUET
📅 Section A — Core Definitions
| Concept Type | Notation / Rule | Key Condition |
|---|---|---|
| Reflexive Relation | (a, a) ∈ R | Must hold ∀ a ∈ A. |
| Symmetric Relation | (a, b) ∈ R ⇒ (b, a) ∈ R | Must hold ∀ (a, b) ∈ R. |
| Transitive Relation | (a, b) ∈ R AND (b, c) ∈ R ⇒ (a, c) ∈ R | If (a, b) exists but NO (b, c) exists, it is trivially Transitive! |
| Equivalence Relation | Reflexive + Symmetric + Transitive | Partitions the set into disjoint equivalence classes. |
| One-One (Injective) | f(x&sub1;) = f(x&sub2;) ⇒ x&sub1; = x&sub2; | Distinct elements have distinct images. |
| Onto (Surjective) | y = f(x) for all y ∈ Codomain | Range of f = Codomain of f. |
| Invertible Function | f&supmin;¹ exists implies f is Bijective | Must be BOTH One-One and Onto. |
📅 Section B — Counting Formulas (Number of R & F)
Let set A have 'm' elements and set B have 'n' elements i.e. n(A) = m, n(B) = n.
- Total number of Relations from A to B = 2mn
- Number of Reflexive Relations on A (where B=A) = 2m(m-1)
- Total number of Functions from A to B = nm
- Number of One-One (Injective) Functions = nPm [if n ≥ m]. If n < m, = 0.
- Number of Bijective Functions (when m = n) = n!
🧠 Memory Tricks & Must-Knows
The "Dead End" Transitivity Trick
A relation is NOT transitive ONLY IF you find (a, b) and (b, c) but NO (a, c).
If you have (a, b) but no (b, c) starting point even exists, the relation is Trivially Transitive. Examples: R = {(1, 2), (3, 4)}. Focus closely on this!
If you have (a, b) but no (b, c) starting point even exists, the relation is Trivially Transitive. Examples: R = {(1, 2), (3, 4)}. Focus closely on this!
Injectivity Check Strategy
Algebraic: Start with f(a) = f(b) and solve to get a = b.
Graphical (Horizontal Line Test): Draw f(x). If any horizontal line cuts the graph at more than ONE point, it is NOT one-one (it is many-one) (e.g., y = x², modulus x).
Graphical (Horizontal Line Test): Draw f(x). If any horizontal line cuts the graph at more than ONE point, it is NOT one-one (it is many-one) (e.g., y = x², modulus x).
Surjectivity Check Strategy
Set y = f(x). Express x entirely in terms of y, i.e., x = g(y).
Check if for every possible y in the given Codomain, the calculated x exists back in the Domain.
If Yes = Onto. If No (e.g. x becomes complex or division by zero) = Not Onto.
Check if for every possible y in the given Codomain, the calculated x exists back in the Domain.
If Yes = Onto. If No (e.g. x becomes complex or division by zero) = Not Onto.
Identity Relation vs Reflexive Relation
Identity Relation on A only contains {(a,a) ∀ a ∈ A} and absolutely nothing
else.
Reflexive Relation MUST contain all {(a,a) ∀ a ∈ A}, but it CAN contain other extra pairs like (a, b).
Reflexive Relation MUST contain all {(a,a) ∀ a ∈ A}, but it CAN contain other extra pairs like (a, b).
❌ Common Mistakes to Avoid
- Symmetry vs Anti-Symmetry: Don't confuse "not symmetric" with anti-symmetric. A relation can be neither.
- f o g vs g o f: Matrix multiplication and function composition are generally NOT commutative. Calculate the inner function first: f(g(x)).
- Domain Checking during Inverses: When finding f&supmin;¹(x), always check if the plus/minus sign (like from a square root) aligns with the original domain of f(x) to ensure the inverse is a true function.
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App