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Relations and Functions

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Chapter Notes

Overview
This chapter introduces the foundational concepts of ordered pairs, Cartesian products, relations, and functions. It exists in the syllabus to formalize the idea of a mathematically precise correspondence between two quantities, which is essential for all higher mathematics. The one big idea a student must walk away with is that a function is a special type of relation where every element of the domain has exactly one and only one image in the codomain.
Key Concepts (20)
Ordered pair
A pair of elements written in small brackets and grouped together in a particular order, i.e., (p, q), where p ∈ P and q ∈ Q.
e.g. (red, b) is an ordered pair where 'red' is from set A and 'b' is from set B.
Cartesian product
Given two non-empty sets P and Q, the Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, i.e., P × Q = { (p, q) : p ∈ P, q ∈ Q }.
e.g. If A = {red, blue} and B = {b, c, s}, then A × B = {(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)}.
Equality of ordered pairs
Two ordered pairs are equal if and only if the corresponding first elements are equal and the second elements are also equal.
e.g. If (x + 1, y – 2) = (3, 1), then x + 1 = 3 and y – 2 = 1, giving x = 2 and y = 3.
Relation
A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B, derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
e.g. R = { (x, y) : x is the first letter of the name y, x ∈ P, y ∈ Q } = {(a, Ali), (b, Bhanu), (b, Binoy), (c, Chandra)}.
Domain of a relation
The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R.
e.g. For R = {(1,2), (2,3), (3,4), (4,5), (5,6)}, domain = {1, 2, 3, 4, 5}.
Range of a relation
The set of all second elements in a relation R from a set A to a set B is called the range of the relation R.
e.g. For R = {(1,2), (2,3), (3,4), (4,5), (5,6)}, range = {2, 3, 4, 5, 6}.
Codomain of a relation
The whole set B is called the codomain of the relation R. Note that range ⊂ codomain.
e.g. For R = {(1,2), (2,3), (3,4), (4,5), (5,6)} from A = {1,2,3,4,5,6} to A, codomain = {1,2,3,4,5,6}.
Function
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, a function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element.
e.g. f = {(1,1), (2,3), (0,–1), (–1,–3)} is a function because each first element has a unique image.
Real function
A function which has either R or one of its subsets as its range is called a real valued function. Further, if its domain is also either R or a subset of R, it is called a real function.
e.g. f : N → N defined by f(x) = 2x + 1 is a real function.
Identity function
The real valued function f : R → R defined by y = f(x) = x for each x ∈ R is called the identity function. Its domain and range are R.
e.g. f(5) = 5, f(–3) = –3.
Constant function
The function f: R → R defined by y = f(x) = c, x ∈ R, where c is a constant, is called a constant function. Its domain is R and its range is {c}.
e.g. f(x) = 3 for all x ∈ R.
Polynomial function
A function f : R → R is said to be a polynomial function if for each x in R, y = f(x) = a₀ + a₁x + a₂x² + ... + aₙxⁿ, where n is a non-negative integer and a₀, a₁, a₂, ..., aₙ ∈ R.
e.g. f(x) = x³ – x² + 2 and g(x) = x⁴ + √2 x are polynomial functions.
Rational function
Functions of the type f(x)/g(x), where f(x) and g(x) are polynomial functions of x defined in a domain where g(x) ≠ 0.
e.g. f(x) = 1/x, x ∈ R – {0}.
Modulus function
The function f: R → R defined by f(x) = |x| for each x ∈ R is called the modulus function. For non-negative x, f(x) = x; for negative x, f(x) = –x.
e.g. |3| = 3, |–5| = 5.
Signum function
The function f: R → R defined by f(x) = 1 if x > 0, 0 if x = 0, –1 if x < 0 is called the signum function. Its domain is R and range is {–1, 0, 1}.
e.g. f(10) = 1, f(0) = 0, f(–4) = –1.
Greatest integer function
The function f: R → R defined by f(x) = [x], x ∈ R, assumes the value of the greatest integer less than or equal to x.
e.g. [1.5] = 1, [–0.5] = –1, [2] = 2.
Arrow diagram
A visual representation of a relation where arrows connect elements of the domain to their corresponding images in the codomain.
e.g. In Fig 2.4, arrows connect 'a' to 'Ali', 'b' to 'Bhanu' and 'Binoy', and 'c' to 'Chandra'.
Image and preimage
If f is a function from A to B and (a, b) ∈ f, then f(a) = b, where b is called the image of a under f and a is called the preimage of b under f.
e.g. In f = {(1,5), (2,9)}, the image of 1 is 5, and the preimage of 9 is 2.
Ordered triplet
A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered triplet.
e.g. If P = {1, 2}, then P × P × P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)}.
Linear function
The function f defined by f(x) = mx + c, x ∈ R, is called a linear function, where m and c are constants.
e.g. f(x) = x + 10 is a linear function.
Formulas & Equations (11)
n(A × B) = pq
Reads as: The number of elements in the Cartesian product of A and B equals p times q.
Variables: n(A) = p, n(B) = q, where p and q are the number of elements in sets A and B respectively.
Use when: Use when asked to find the number of elements in a Cartesian product or the number of possible ordered pairs.
⚠ Students often forget that this formula applies only when both sets are finite and non-empty.
Total number of relations from A to B = 2^(pq)
Reads as: The total number of relations from set A to set B is 2 raised to the power of p times q.
Variables: n(A) = p, n(B) = q, so n(A × B) = pq. The number of subsets of A × B is 2^(pq).
Use when: Use when asked to find the number of possible relations between two finite sets.
⚠ Students confuse this with the number of functions. The number of functions from A to B is q^p, not 2^(pq).
f(x) = |x| = { x, if x ≥ 0; –x, if x < 0 }
Reads as: f of x equals the modulus of x, which is x if x is greater than or equal to zero, and negative x if x is less than zero.
Variables: x ∈ R
Use when: Use when defining or evaluating the modulus function for any real number.
⚠ Students write |–5| = –5 instead of 5. The modulus always gives a non-negative output.
Signum function: f(x) = 1 if x > 0; 0 if x = 0; –1 if x < 0
Reads as: The signum function of x is 1 if x is greater than zero, 0 if x equals zero, and –1 if x is less than zero.
Variables: x ∈ R
Use when: Use when asked to define or evaluate the signum function.
⚠ Students forget the case x = 0, which gives f(0) = 0, not 1 or –1.
Greatest integer function: f(x) = [x]
Reads as: f of x equals the greatest integer less than or equal to x.
Variables: x ∈ R
Use when: Use when asked to evaluate the greatest integer function for a given real number.
⚠ For negative numbers, students often give the wrong integer. For example, [–0.5] = –1, not 0.
(f + g)(x) = f(x) + g(x)
Reads as: f plus g of x equals f of x plus g of x.
Variables: f : X → R, g : X → R, x ∈ X ⊂ R
Use when: Use when asked to find the sum of two real functions.
⚠ Students forget to specify the domain, which is the intersection of the domains of f and g.
(f – g)(x) = f(x) – g(x)
Reads as: f minus g of x equals f of x minus g of x.
Variables: f : X → R, g : X → R, x ∈ X ⊂ R
Use when: Use when asked to find the difference of two real functions.
⚠ Students incorrectly apply subtraction only to the constant terms, not the entire expressions.
(αf)(x) = α f(x)
Reads as: alpha times f of x equals alpha times f of x.
Variables: α is a scalar (real number), f : X → R, x ∈ X
Use when: Use when asked to multiply a real function by a scalar.
⚠ Students forget to multiply every term of the function by the scalar.
(fg)(x) = f(x) g(x)
Reads as: f g of x equals f of x times g of x.
Variables: f : X → R, g : X → R, x ∈ X
Use when: Use when asked to find the product of two real functions.
⚠ Students confuse this with composition of functions, which is different.
(f/g)(x) = f(x)/g(x), provided g(x) ≠ 0
Reads as: f over g of x equals f of x divided by g of x, provided g of x is not equal to zero.
Variables: f : X → R, g : X → R, x ∈ X, g(x) ≠ 0
Use when: Use when asked to find the quotient of two real functions.
⚠ Students forget to state the condition g(x) ≠ 0, which is essential for the definition.
t(C) = (9C/5) + 32
Reads as: t of C equals nine C over five plus thirty-two.
Variables: C is temperature in degree Celsius, t(C) is temperature in degree Fahrenheit.
Use when: Use when converting temperature from Celsius to Fahrenheit.
⚠ Students incorrectly write the formula as t(C) = (5C/9) + 32, which is the inverse conversion.
How It Works
Forming a Cartesian product
  1. Step 1 — Identify the two non-empty sets, say P and Q.
  2. Step 2 — Take each element p from set P and pair it with each element q from set Q to form an ordered pair (p, q).
  3. Step 3 — Collect all such ordered pairs into a set. This set is P × Q.
💡 The order matters: (p, q) is not the same as (q, p). The number of elements in P × Q is n(P) × n(Q).
Defining a relation from a set A to a set B
  1. Step 1 — Write the Cartesian product A × B.
  2. Step 2 — Describe a specific relationship between the first and second elements of the ordered pairs.
  3. Step 3 — Select only those ordered pairs from A × B that satisfy the relationship. This subset is the relation R.
💡 A relation is simply a subset of the Cartesian product. The total number of possible relations from A to B is 2^(n(A)×n(B)).
Checking if a relation is a function
  1. Step 1 — Check that every element of set A (the domain) appears as the first element in at least one ordered pair.
  2. Step 2 — Check that no element of set A appears as the first element in more than one ordered pair with different second elements.
  3. Step 3 — If both conditions are satisfied, the relation is a function.
💡 A function must have exactly one image for each element of the domain. If any element has zero or more than one image, it is not a function.
Algebra of real functions — addition
  1. Step 1 — Identify the two real functions f : X → R and g : X → R, where X ⊂ R.
  2. Step 2 — For each x ∈ X, compute f(x) and g(x).
  3. Step 3 — Add the two values: (f + g)(x) = f(x) + g(x).
💡 The domain of f + g is the intersection of the domains of f and g.
Algebra of real functions — subtraction
  1. Step 1 — Identify the two real functions f : X → R and g : X → R.
  2. Step 2 — For each x ∈ X, compute f(x) and g(x).
  3. Step 3 — Subtract: (f – g)(x) = f(x) – g(x).
💡 The domain of f – g is the intersection of the domains of f and g.
Algebra of real functions — multiplication by a scalar
  1. Step 1 — Identify the real function f : X → R and the scalar α (a real number).
  2. Step 2 — For each x ∈ X, compute f(x).
  3. Step 3 — Multiply: (αf)(x) = α f(x).
💡 The scalar multiplies every output value of the function.
Algebra of real functions — multiplication of two functions
  1. Step 1 — Identify the two real functions f : X → R and g : X → R.
  2. Step 2 — For each x ∈ X, compute f(x) and g(x).
  3. Step 3 — Multiply: (fg)(x) = f(x) g(x).
💡 This is pointwise multiplication, not composition.
Algebra of real functions — quotient of two functions
  1. Step 1 — Identify the two real functions f : X → R and g : X → R.
  2. Step 2 — For each x ∈ X, compute f(x) and g(x).
  3. Step 3 — Divide: (f/g)(x) = f(x)/g(x), provided g(x) ≠ 0.
💡 The domain excludes all x where g(x) = 0.
Exam Traps — Don't Make These Mistakes
⚠ Confusing range and codomain
✗ The range of the relation is the set B.
→ The range is the set of all second elements actually appearing in the relation, which is a subset of the codomain B, not necessarily the whole set B.
✓ The range of the relation is the set of all second elements of the ordered pairs in the relation, and it is a subset of the codomain.
⚠ Assuming all relations are functions
✗ Every relation is a function because it pairs elements from two sets.
→ A relation is a function only if every element of the domain has exactly one image. Many relations have elements with zero or multiple images.
✓ A function is a special type of relation where every element of the domain has one and only one image in the codomain.
⚠ Mistaking Cartesian product order
✗ P × Q = Q × P.
→ Cartesian product is not commutative. The ordered pair (p, q) is different from (q, p) unless p = q.
✓ In general, P × Q ≠ Q × P, though the number of elements in both sets is the same.
⚠ Incorrect evaluation of greatest integer function for negative numbers
✗ [–0.5] = 0.
→ The greatest integer less than or equal to –0.5 is –1, not 0, because –1 is the largest integer that is ≤ –0.5.
✓ [–0.5] = –1.
⚠ Forgetting the condition for quotient of functions
✗ (f/g)(x) = f(x)/g(x) for all x in the domain.
→ The quotient is defined only when g(x) ≠ 0. The domain must exclude points where the denominator is zero.
✓ (f/g)(x) = f(x)/g(x), provided g(x) ≠ 0.
Compare & Contrast
Definition and condition for mapping
Relation
Function
A relation is any subset of A × B, while a function is a relation where every element of set A has exactly one image in set B. In a function, no two ordered pairs can have the same first element.
Set of elements in a relation or function
Domain
Codomain
The domain is the set of all first elements of the ordered pairs, while the codomain is the entire set B to which the relation is defined. The range is a subset of the codomain.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks), 1 LA (5 marks)
Most asked: Identifying whether a given relation is a function · Finding domain and range of relations and functions · Evaluating modulus, signum, and greatest integer functions · Algebra of real functions (addition, subtraction, multiplication, division) · Cartesian products and number of relations
Don't skip: Long answer questions on algebra of real functions (5 marks); Questions involving domain and range of functions (3-5 marks); Problems on greatest integer and modulus functions (3 marks)
One-Line Revision (read aloud before the exam)
  1. An ordered pair (p, q) is a pair of elements where order matters; (p, q) ≠ (q, p) unless p = q.
  2. The Cartesian product P × Q is the set of all ordered pairs (p, q) with p ∈ P and q ∈ Q.
  3. If n(A) = p and n(B) = q, then n(A × B) = pq.
  4. A relation from A to B is a subset of A × B.
  5. The domain of a relation is the set of all first elements of its ordered pairs.
  6. The range of a relation is the set of all second elements of its ordered pairs, and it is a subset of the codomain.
  7. A function is a relation where every element of the domain has exactly one image.
  8. The total number of relations from A to B is 2^(n(A) × n(B)).
  9. The identity function is f(x) = x, with domain and range both R.
  10. The constant function is f(x) = c, with range {c}.
  11. The modulus function is f(x) = |x|, which is x for x ≥ 0 and –x for x < 0.
  12. The signum function gives 1 for x > 0, 0 for x = 0, and –1 for x < 0.
  13. The greatest integer function [x] gives the largest integer less than or equal to x.
  14. For real functions, (f + g)(x) = f(x) + g(x), (f – g)(x) = f(x) – g(x), (fg)(x) = f(x)g(x), and (f/g)(x) = f(x)/g(x) provided g(x) ≠ 0.
  15. A linear function is of the form f(x) = mx + c.

MCQs — 10 Questions with Answers

Q1. If (2x + 1, y – 3) = (5, 2), find the value of x + y.

  • A. 6
  • B. 7 ✓
  • C. 5
  • D. 8

Answer: B — From equality of ordered pairs: 2x + 1 = 5 gives x = 2, and y – 3 = 2 gives y = 5; therefore x + y = 7.

Q2. If A = {1, 2} and B = {a, b, c}, how many elements are in A × B?

  • A. 5
  • B. 6 ✓
  • C. 8
  • D. 2

Answer: B — n(A × B) = n(A) × n(B) = 2 × 3 = 6 elements.

Q3. Which of the following statements is true?

  • A. A × B = B × A for all non-empty sets A and B
  • B. If A and B are non-empty, then A × B is always non-empty ✓
  • C. n(A × B) = n(B × A) only when n(A) = n(B)
  • D. P × φ = P for any set P

Answer: B — If A and B are non-empty sets, then A × B contains at least one ordered pair (a, b), so it is non-empty; options A and D are false because order matters and empty set rule applies.

Q4. If A = {2, 3}, B = {3, 4} and C = {4, 5}, find A × (B ∩ C).

  • A. {(2, 4), (3, 4)} ✓
  • B. {(2, 3), (3, 3), (2, 4), (3, 4)}
  • C. {(4, 2), (4, 3)}
  • D. {(2, 4), (3, 4), (4, 2), (4, 3)}

Answer: A — B ∩ C = {4}, so A × (B ∩ C) = {(2, 4), (3, 4)} with 2 elements.

Q5. The set P × P × P has 8 elements. How many elements does P have?

  • A. 2 ✓
  • B. 3
  • C. 4
  • D. 8

Answer: A — If n(P) = k, then n(P × P × P) = k³ = 8, so k = 2.

Q6. Let A = {p, q} and B = {1, 2, 3}. Which ordered pair is NOT in A × B?

  • A. (p, 1)
  • B. (1, p) ✓
  • C. (q, 3)
  • D. (p, 2)

Answer: B — In A × B, the first element must come from A = {p, q} and second from B = {1, 2, 3}; (1, p) has 1 as first element, which is not in A.

Q7. If A × B = {(a, x), (a, y), (b, x), (b, y)}, which of the following is correct?

  • A. A = {x, y} and B = {a, b}
  • B. A = {a, b} and B = {x, y} ✓
  • C. A = {a, x, b, y} and B = {}
  • D. A = {} and B = {a, b, x, y}

Answer: B — A contains all first elements {a, b} and B contains all second elements {x, y}, so A = {a, b} and B = {x, y}.

Q8. State whether true or false: P × Q = Q × P if and only if P = Q. Explanation:

  • A. True; they are equal only when sets are identical
  • B. False; they are never equal regardless of P and Q
  • C. False; P × Q may equal Q × P even if P ≠ Q when both are empty ✓
  • D. True; order is irrelevant in Cartesian products

Answer: C — If P = φ and Q = φ, then P × Q = Q × P = φ even though P = Q is not the only case; the statement is too restrictive.

Q9. If M = {2, 4, 6} and N = {1, 3}, verify: n((M × N)) = n(M) × n(N). What is n(M × N)?

  • A. 5
  • B. 6 ✓
  • C. 8
  • D. 9

Answer: B — n(M) = 3 and n(N) = 2, so n(M × N) = 3 × 2 = 6.

Q10. For sets A, B, C: A × (B ∩ C) = (A × B) ∩ (A × C). If A = {1, 2}, B = {2, 3, 4}, C = {3, 4, 5}, find (A × B) ∩ (A × C).

  • A. {(1, 3), (1, 4), (2, 3), (2, 4)} ✓
  • B. {(1, 2), (2, 2)}
  • C. {(3, 1), (3, 2), (4, 1), (4, 2)}
  • D. {(1, 4), (2, 4)}

Answer: A — B ∩ C = {3, 4}, so A × (B ∩ C) = {(1, 3), (1, 4), (2, 3), (2, 4)}, which equals (A × B) ∩ (A × C).

Flashcards

What is the Cartesian product P × Q?

The set of all ordered pairs (p, q) where p ∈ P and q ∈ Q, written as P × Q = {(p, q) : p ∈ P, q ∈ Q}.

When are two ordered pairs equal?

Two ordered pairs (a, b) and (c, d) are equal if and only if a = c AND b = d.

If n(A) = 3 and n(B) = 5, what is n(A × B)?

n(A × B) = n(A) × n(B) = 3 × 5 = 15 elements.

Is P × Q always equal to Q × P?

No; P × Q = Q × P only if P = Q, because ordered pairs have direction and (p, q) ≠ (q, p) in general.

What happens to the Cartesian product if one set is empty?

If P = φ or Q = φ, then P × Q = φ (the product is the empty set).

What does R × R represent geometrically?

R × R represents all points in the two-dimensional plane with coordinates (x, y) where x, y ∈ R.

What is an ordered triplet and how does A × A × A relate to it?

An ordered triplet is a set of three elements written as (a, b, c); A × A × A = {(a, b, c) : a, b, c ∈ A}.

State the distributive property: A × (B ∪ C) = ?

A × (B ∪ C) = (A × B) ∪ (A × C).

State the distributive property: A × (B ∩ C) = ?

A × (B ∩ C) = (A × B) ∩ (A × C).

If A × B = {(p, q), (p, r), (m, q), (m, r)}, what are A and B?

A is the set of all first elements {p, m} and B is the set of all second elements {q, r}.

Important Board Questions

If (x – 2, y + 1) = (1, 3), find x and y. Also state the condition for equality of two ordered pairs. [2 marks]

Use equality of ordered pairs: first elements equal AND second elements equal; solve x – 2 = 1 and y + 1 = 3 separately.

Let A = {1, 2, 3}, B = {4, 5} and C = {5, 6}. Show that A × (B ∪ C) = (A × B) ∪ (A × C) by computing both sides. [5 marks]

Find B ∪ C first; compute A × (B ∪ C) directly; separately compute (A × B) and (A × C), then find their union; verify both sides give the same set.

If A × B has 12 elements, n(A) = 4, find n(B). If A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2), (d, 1), (d, 2), (e, 1), (e, 2), (f, 1), (f, 2)}, verify that the given product is NOT A × B and explain why. Also, extract the sets A and B from the given product. [6 marks]

Use cardinality formula n(A × B) = n(A) × n(B) to find n(B); count the actual elements in the given set — there are 12, but A has 6 elements {a, b, c, d, e, f} not 4; extract A as the set of first elements and B as the set of second elements.

Assertion-Reason Questions for Relations and Functions

Q1. Assertion (A):

If A = {1, 2} and B = {3}, then A × B ≠ B × A.

Reason (R):

The Cartesian product is commutative for all non-empty sets.

Answer: C — A is true but R is false — The assertion is true: (1,3), (2,3) ≠ (3,1), (3,2), so A × B ≠ B × A. However, the reason is false because Cartesian product is NOT commutative. The correct explanation is that ordered pairs require order, making (a,b) ≠ (b,a) in general.

Q2. Assertion (A):

If n(A) = 5 and n(B) = 3, then n(A × B) = 15.

Reason (R):

The number of elements in A × B equals the product of the number of elements in A and B.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the assertion holds. This is the fundamental property of Cartesian product cardinality: n(A × B) = n(A) × n(B) = 5 × 3 = 15.

Q3. Assertion (A):

An ordered pair (a, b) is equal to an ordered pair (c, d) if and only if a = c and b = d.

Reason (R):

Ordered pairs represent positions in the Cartesian plane where the first component is the x-coordinate.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true. The assertion is the definition of ordered pair equality. The reason is also true—ordered pairs do represent coordinates in the plane. However, the reason doesn't explain why equality requires both components to match; it merely describes an application. Thus, the reason doesn't provide the logical foundation for the assertion.

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Case-Based Questions (CBQ) for Relations and Functions

Case Study:

A restaurant menu offers 3 types of cuisines (Indian, Chinese, Continental) and 4 types of dishes for each cuisine (Appetizers, Main Course, Dessert, Beverage). The restaurant wants to create unique meal combinations by pairing one cuisine type with one dish type. A data analyst represents this as a Cartesian product where the first component is the cuisine and the second is the dish category. The restaurant has space to display all possible combinations on their digital menu board.

Q1. How many unique combinations can the restaurant display?

  • A. 7
  • B. 12 ✓
  • C. 16
  • D. 21

Answer: B

Q2. Which ordered pair represents the combination of Indian cuisine with Appetizers?

  • A. (Appetizers, Indian)
  • B. (Indian, Appetizers) ✓
  • C. (1, 1)
  • D. (A, I)

Answer: B

Q3. If the restaurant later decides to exclude Beverage from the second component, how many combinations remain?

  • A. 9 ✓
  • B. 8
  • C. 12
  • D. 6

Answer: A

Q4. Are the pairs (Chinese, Main Course) and (Main Course, Chinese) the same combination?

  • A. Yes, both represent the same pairing
  • B. They are equivalent only if the restaurant treats them as unordered
  • C. No, order matters in ordered pairs and they represent different combinations ✓
  • D. Cannot determine without more information

Answer: C

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