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Tamil Nadu SSLC Mathematics study materials
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Tamil Nadu SSLC Mathematics practice questions

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Question 1Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If A = {1, 2, 3} and B = {x, y}, then the number of elements in A × B is

  1. 6 Correct answer.
  2. 5
  3. 9
  4. 3

Explanation

The Cartesian product A × B is the set of all ordered pairs (a, b) with a ∈ A and b ∈ B, so its cardinality is the product of the cardinalities: n(A × B) = n(A) × n(B) = 3 × 2 = 6. Listing the pairs (1, x), (1, y), (2, x), (2, y), (3, x), (3, y) confirms the count.

Question 2Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

A relation R from set A to set B is, by definition, a subset of

  1. B × A
  2. A × B Correct answer.
  3. A − B
  4. A ∪ B

Explanation

Every relation from A to B is a collection of ordered pairs whose first components lie in A and whose second components lie in B, that is, a subset of the Cartesian product A × B. The other set operations produce sets of single elements or reverse the direction of the pairing.

Question 3Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

Which of the following is a function from A = {1, 2, 3} to B = {p, q}?

  1. {(1, q), (2, p), (3, q), (2, q)}
  2. {(1, p), (2, q)}
  3. {(1, p), (2, q), (3, p)} Correct answer.
  4. {(1, p), (1, q), (2, p)}

Explanation

A relation from A to B is a function only when every element of A appears exactly once as a first component. In {(1, p), (2, q), (3, p)} each of 1, 2, 3 occurs once, so it is a function; the repeated image value p is allowed.

Question 4Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If A = {1, 2, 3} and B = {3, 4}, then (A × B) ∩ (B × A) is

  1. {(3, 3), (4, 4)}
  2. {(1, 3), (3, 1)}
  3. {(3, 3)} Correct answer.
  4. {(3, 4), (4, 3)}

Explanation

A pair belongs to both products only when it can be written as (a, b) with a ∈ A, b ∈ B and also as (b, a) with b ∈ B, a ∈ A, which forces both coordinates to lie in A ∩ B = {3}. Hence the only common ordered pair is (3, 3).

Question 5Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

On A = {1, 2, 3}, the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3)} is

  1. neither reflexive nor symmetric
  2. symmetric and transitive
  3. reflexive and transitive Correct answer.
  4. reflexive and symmetric

Explanation

All three diagonal pairs are present, so R is reflexive. The pair (1, 2) is in R but (2, 1) is not, so R is not symmetric. Checking compositions: the only second components that appear as first components of other pairs lead back to existing pairs, e.g. (1, 2) with (2, 2) gives (1, 2), so R is transitive.

Question 6Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

The function f : ℝ → ℝ defined by f(x) = 2x + 1 is

  1. one-one but not onto
  2. onto but not one-one
  3. bijective Correct answer.
  4. constant

Explanation

If 2x₁ + 1 = 2x₂ + 1 then x₁ = x₂, so f is one-one. For any y ∈ ℝ the equation y = 2x + 1 is solved by x = (y − 1)/2, which is a real number, so every codomain element has a pre-image. Being both one-one and onto, f is bijective.

Question 7Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If f(x) = 2x + 3 and g(x) = x − 1, then (f ∘ g)(x) is

  1. 2x + 1 Correct answer.
  2. 2x − 1
  3. 2x + 2
  4. 2x + 3

Explanation

By definition (f ∘ g)(x) = f(g(x)), and g acts first, so substitute x − 1 into f: f(x − 1) = 2(x − 1) + 3 = 2x − 2 + 3 = 2x + 1.

Question 8Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If n(A) = 2 and n(B) = 3, then the number of relations from A to B is

  1. 8
  2. 64 Correct answer.
  3. 6
  4. 32

Explanation

A relation from A to B is any subset of A × B, and n(A × B) = 2 × 3 = 6. A set with 6 elements has 2⁶ subsets, so there are 64 relations from A to B.

Question 9Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If f : A → B and g : B → C are bijections, then (g ∘ f)⁻¹ equals

  1. g⁻¹ ∘ f⁻¹ Correct answer.
  2. f⁻¹ ∘ g⁻¹
  3. f ∘ g
  4. g ∘ f

Explanation

The inverse of a composition reverses the order of the operations: to undo g ∘ f we must first undo g and then undo f. Hence (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹, and both inverses exist because f and g are bijections.

Question 10Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If f(x) = 2x + 3 and (f ∘ g)(x) = 4x + 1, then g(x) is

  1. 2x − 1 Correct answer.
  2. 2x + 1
  3. 4x − 2
  4. x − 1

Explanation

The condition means f(g(x)) = 4x + 1, so 2g(x) + 3 = 4x + 1. Subtracting 3 and dividing by 2 gives g(x) = (4x − 2)/2 = 2x − 1, and checking: f(2x − 1) = 2(2x − 1) + 3 = 4x + 1.

Question 11Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If A × B = {(1, p), (1, q), (2, p), (2, q)}, then which of the following gives the sets A and B?

  1. A = {1, p}, B = {2, q}
  2. A = {p, q}, B = {1, 2}
  3. A = {1, 2, p, q}, B = ∅
  4. A = {1, 2}, B = {p, q} Correct answer.

Explanation

In a Cartesian product A × B, each first coordinate comes from A and each second coordinate comes from B. The first coordinates appearing are 1 and 2, and the second coordinates are p and q, so A = {1, 2} and B = {p, q}.

Question 12Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

For a relation R from set A to set B, the set of all first components of the ordered pairs in R is called the

  1. range
  2. Cartesian product
  3. co-domain
  4. domain Correct answer.

Explanation

For a relation R ⊆ A × B, the domain is the set of all first elements of the ordered pairs in R. The set of second elements is the range, while B is the co-domain.

Question 13Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

On A = {1, 2, 3}, the identity relation I_A is

  1. {1, 2, 3} × {1, 2, 3}
  2. {(1, 1), (2, 2), (3, 3)} Correct answer.
  3. {(1, 1), (1, 2), (1, 3)}
  4. {(1, 2), (2, 1), (3, 3)}

Explanation

The identity relation on A consists of all ordered pairs (a, a) with a in A. For A = {1, 2, 3}, this gives {(1,1), (2,2), (3,3)}.

Question 14Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

The number of reflexive relations on the set A = {1, 2, 3} is

  1. 512
  2. 9
  3. 64 Correct answer.
  4. 8

Explanation

A reflexive relation on A = {1,2,3} must contain (1,1), (2,2), and (3,3). The remaining 9 − 3 = 6 ordered pairs may be included or omitted independently, giving 2^6 = 64 reflexive relations.

Question 15Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

The function f : ℝ → ℝ defined by f(x) = |x| is

  1. neither one-one nor onto Correct answer.
  2. a bijection
  3. onto but not one-one
  4. one-one but not onto

Explanation

For f(x) = |x|, we have f(−1) = f(1), so the function is not one-one. Also, negative real numbers are never outputs, so f is not onto ℝ; hence it is neither one-one nor onto.

Question 16Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If f(x) = x² + 1 and g(x) = 2x − 3, then (g ∘ f)(2) equals

  1. 25
  2. 3
  3. 11
  4. 7 Correct answer.

Explanation

First find f(2) = 2² + 1 = 5. Then (g ∘ f)(2) = g(f(2)) = g(5) = 2(5) − 3 = 7.

Question 17Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If f : ℝ → ℝ is defined by f(x) = 3x − 5, then f⁻¹(x) is

  1. 3x + 5
  2. (x − 5)/3
  3. 1/(3x − 5)
  4. (x + 5)/3 Correct answer.

Explanation

Let y = 3x − 5. Solving for x gives x = (y + 5)/3, so f⁻¹(y) = (y + 5)/3; replacing y by x gives f⁻¹(x) = (x + 5)/3.

Question 18Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

If f(x) = x + 1 and g(x) = x², then (f ∘ g)(x) − (g ∘ f)(x) equals

  1. 0
  2. −2x Correct answer.
  3. −2x − 2
  4. 2x

Explanation

(f ∘ g)(x) = f(x²) = x² + 1, while (g ∘ f)(x) = g(x+1) = (x+1)² = x² + 2x + 1. Subtracting gives (x² + 1) − (x² + 2x + 1) = −2x.

Question 19Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

On A = {1, 2, 3}, let R = {(1, 2), (2, 3)}. Then R ∘ R is

  1. ∅
  2. {(1, 3)} Correct answer.
  3. {(1, 2), (2, 3)}
  4. {(2, 1), (3, 2)}

Explanation

By definition, (a,c) ∈ R ∘ R when there exists b with (a,b) ∈ R and (b,c) ∈ R. Here (1,2) and (2,3) give (1,3), and no other pair can be formed, so R ∘ R = {(1,3)}.

Question 20Relations and Functions (sets, ordered pairs, Cartesian product, types of relations, types of functions, composition)

The number of onto functions from a set with 3 elements to a set with 2 elements is

  1. 8
  2. 12
  3. 6 Correct answer.
  4. 9

Explanation

The total number of functions from a 3-element set to a 2-element set is 2³ = 8. Two of these are constant functions mapping every element to the same element, and they are not onto; hence there are 8 − 2 = 6 onto functions.

Question 21Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

According to Euclid’s division lemma, for positive integers a and b, there exist unique integers q and r such that a = bq + r. Which condition must r satisfy?

  1. 0 ≤ r ≤ b
  2. 0 < r ≤ b
  3. 0 ≤ r < b Correct answer.
  4. 1 ≤ r < b

Explanation

Euclid’s division lemma states that when a positive integer a is divided by a positive integer b, the remainder r is a non-negative integer strictly smaller than b. Hence the correct condition is 0 ≤ r < b.

Question 22Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

The prime factorisation 2^3 × 3^2 × 5 represents a positive integer. How many distinct prime factors does this integer have?

  1. 6
  2. 3 Correct answer.
  3. 5
  4. 2

Explanation

The distinct prime bases appearing in the factorisation are 2, 3 and 5, even though 2 and 3 have exponents greater than 1. Therefore the integer has three distinct prime factors.

Question 23Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

In the arithmetic progression 3, 7, 11, 15, …, what is the 10th term?

  1. 39 Correct answer.
  2. 35
  3. 37
  4. 43

Explanation

For an arithmetic progression, a_n = a + (n − 1)d. Here a = 3 and d = 4, so a_10 = 3 + 9 × 4 = 39.

Question 24Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

Use Euclid’s division algorithm to find the HCF of 867 and 255.

  1. 3
  2. 51 Correct answer.
  3. 255
  4. 17

Explanation

Applying Euclid’s algorithm: 867 = 255 × 3 + 102, then 255 = 102 × 2 + 51, and 102 = 51 × 2 + 0. The last non-zero remainder is 51, so the HCF is 51.

Question 25Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

What is the remainder when 7^100 is divided by 5?

  1. 4
  2. 2
  3. 3
  4. 1 Correct answer.

Explanation

Since 7 ≡ 2 (mod 5), we have 7^100 ≡ 2^100 (mod 5). The powers of 2 modulo 5 repeat every 4, and 100 is divisible by 4, so 2^100 ≡ 1 (mod 5).

Question 26Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

A sequence is defined by a_n = 3n − 2 for n = 1, 2, 3, … Which term of the sequence is equal to 31?

  1. 13th term
  2. 11th term Correct answer.
  3. 15th term
  4. 9th term

Explanation

Set 3n − 2 = 31 and solve: 3n = 33, so n = 11. Hence 31 is the 11th term of the sequence.

Question 27Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

Find the sum of the first 20 terms of the arithmetic progression 5, 8, 11, 14, …

  1. 670 Correct answer.
  2. 650
  3. 640
  4. 690

Explanation

For an AP, S_n = n/2[2a + (n − 1)d]. Here a = 5, d = 3 and n = 20, so S_20 = 10[10 + 19 × 3] = 10 × 67 = 670.

Question 28Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

In the geometric progression 2, 6, 18, 54, …, what is the 6th term?

  1. 243
  2. 324
  3. 486 Correct answer.
  4. 162

Explanation

In a geometric progression, a_n = ar^(n−1). Here a = 2 and r = 3, so the 6th term is 2 × 3^5 = 2 × 243 = 486.

Question 29Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

What is the smallest positive integer by which 1080 must be multiplied so that the product is a perfect square?

  1. 60
  2. 30 Correct answer.
  3. 15
  4. 10

Explanation

Factorising gives 1080 = 2^3 × 3^3 × 5. For a perfect square every prime exponent must be even, so we need to multiply by 2 × 3 × 5 = 30, making the exponents 4, 4 and 2.

Question 30Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

In an arithmetic progression, the 7th term is 34 and the 13th term is 64. What is the sum of the first 20 terms?

  1. 1050
  2. 1010
  3. 1030 Correct answer.
  4. 1070

Explanation

The difference between the 13th and 7th terms is 6d = 30, so d = 5. Then a + 6d = 34 gives a = 4, and S_20 = 20/2[2 × 4 + 19 × 5] = 10 × 103 = 1030.

Question 31Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

Using Euclid's division lemma, write 78 in the form 78 = 7q + r with 0 ≤ r < 7. What are q and r?

  1. q = 10, r = 8
  2. q = 11, r = 1 Correct answer.
  3. q = 11, r = 2
  4. q = 12, r = -6

Explanation

Since 7 × 11 = 77, we have 78 = 7 × 11 + 1, and 1 satisfies 0 ≤ 1 < 7. Thus the quotient is 11 and the remainder is 1.

Question 32Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

What is the common difference of the arithmetic progression 14, 9, 4, -1, ...?

  1. 4
  2. -5 Correct answer.
  3. -4
  4. 5

Explanation

The common difference is obtained by subtracting any term from the next term. Here 9 - 14 = -5, and 4 - 9 = -5, so the common difference is -5.

Question 33Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

What is the common ratio of the geometric progression 2, -6, 18, -54, ...?

  1. -3 Correct answer.
  2. -2
  3. 1/3
  4. 3

Explanation

Divide any term by the preceding term: (-6) ÷ 2 = -3, and 18 ÷ (-6) = -3. The common ratio is therefore -3.

Question 34Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

What is the remainder when 2^10 is divided by 5?

  1. 4 Correct answer.
  2. 3
  3. 1
  4. 2

Explanation

Powers of 2 modulo 5 repeat as 2, 4, 3, 1 with period 4. Since 10 leaves remainder 2 on division by 4, 2^10 has the same remainder as 2^2, namely 4.

Question 35Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

In an arithmetic progression, the 4th term is 11 and the 9th term is 26. What is the first term?

  1. 5
  2. 8
  3. 3
  4. 2 Correct answer.

Explanation

The 9th term equals the 4th term plus 5 common differences: 26 = 11 + 5d, so d = 3. Then a_4 = a + 3d gives 11 = a + 9, hence a = 2.

Question 36Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

Use Euclid's division algorithm to find the HCF of 405 and 252.

  1. 21
  2. 3
  3. 9 Correct answer.
  4. 27

Explanation

The Euclidean steps are 405 = 252 × 1 + 153, 252 = 153 × 1 + 99, 153 = 99 × 1 + 54, 99 = 54 × 1 + 45, 54 = 45 × 1 + 9, and 45 = 9 × 5 + 0. The last nonzero remainder is 9, so the HCF is 9.

Question 37Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

The prime factorisation of a positive integer is 2^4 × 3^2 × 5. How many positive divisors does this integer have?

  1. 20
  2. 24
  3. 30 Correct answer.
  4. 60

Explanation

For N = p^a q^b r^c, the number of positive divisors is (a + 1)(b + 1)(c + 1). Here the exponents are 4, 2, and 1, so the count is 5 × 3 × 2 = 30.

Question 38Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

Find the sum of the first 5 terms of the geometric progression 3, 6, 12, ...

  1. 90
  2. 45
  3. 48
  4. 93 Correct answer.

Explanation

The first term is 3 and the common ratio is 2. Using S_n = a(r^n - 1)/(r - 1), we get S_5 = 3(2^5 - 1)/(2 - 1) = 3 × 31 = 93.

Question 39Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

What is the remainder when 2^100 is divided by 7?

  1. 2 Correct answer.
  2. 1
  3. 6
  4. 4

Explanation

Since 2^3 = 8 ≡ 1 (mod 7), powers of 2 modulo 7 repeat every 3 exponents. Now 100 = 3 × 33 + 1, so 2^100 ≡ (2^3)^33 × 2 ≡ 2 (mod 7).

Question 40Numbers and Sequences (Euclid's division lemma, modular arithmetic, fundamental theorem of arithmetic, sequences, arithmetic and geometric progressions, series)

The sum of the first n terms of an arithmetic progression is S_n = 3n^2 + 5n. What is the 10th term?

  1. 68
  2. 62 Correct answer.
  3. 60
  4. 65

Explanation

For n ≥ 2, the nth term is S_n - S_{n-1}. Substituting gives a_n = (3n^2 + 5n) - [3(n - 1)^2 + 5(n - 1)] = 6n + 2. Therefore the 10th term is 6 × 10 + 2 = 62.

Question 41Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

p(x) = ax² + bx + c, a ≠ 0, is a quadratic polynomial. Its graph intersects the x-axis at two distinct points. Which statement about the discriminant Δ = b² − 4ac is necessarily true?

  1. Δ = 0
  2. Δ > 0 Correct answer.
  3. Δ is not defined for a quadratic
  4. Δ < 0

Explanation

A quadratic graph meets the x-axis at two distinct points exactly when the equation ax² + bx + c = 0 has two distinct real roots. That happens if and only if the discriminant b² − 4ac is positive. A zero discriminant gives one repeated root, and a negative discriminant gives no real root.

Question 42Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

The HCF and LCM of two polynomials are (x + 2) and (x² − 4)(x + 3) respectively. If one polynomial is (x + 2)(x + 3), what is the other polynomial?

  1. x² − 4 Correct answer.
  2. x² + 4
  3. (x + 2)²
  4. (x − 2)(x + 3)

Explanation

For two polynomials, HCF × LCM equals the product of the polynomials. Here HCF × LCM = (x + 2)(x² − 4)(x + 3), and dividing by the known polynomial (x + 2)(x + 3) leaves x² − 4. Thus the other polynomial is x² − 4.

Question 43Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

If A is a 2 × 2 matrix and det(A) = 5, then det(3A) equals

  1. 125
  2. 30
  3. 45 Correct answer.
  4. 15

Explanation

For an n × n matrix A, det(kA) = k^n det(A). Since A is 2 × 2, det(3A) = 3² det(A) = 9 × 5 = 45.

Question 44Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

The square root of the polynomial 4x⁴ − 12x³ + 25x² − 24x + 16 is

  1. 2x² − 3x − 4
  2. 2x² − 3x + 4 Correct answer.
  3. −2x² + 3x + 4
  4. 2x² + 3x + 4

Explanation

Check by squaring 2x² − 3x + 4: the x⁴, x³, x², x and constant terms become 4x⁴, −12x³, 25x², −24x and 16. Since the square matches the given polynomial and the leading coefficient is positive, the required square root is 2x² − 3x + 4.

Question 45Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

The solution of the system 2x + 3y = 13 and 5x − 4y = −2 is

  1. (2, −3)
  2. (2, 3) Correct answer.
  3. (3, 2)
  4. (−2, 3)

Explanation

Eliminate y by multiplying 2x + 3y = 13 by 4 and 5x − 4y = −2 by 3. Adding gives 23x = 46, so x = 2. Substituting back gives 4 + 3y = 13, hence y = 3.

Question 46Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

If α and β are the roots of x² − 6x + 8 = 0, then α² + β² equals

  1. 16
  2. 36
  3. 28
  4. 20 Correct answer.

Explanation

For x² − 6x + 8 = 0, the sum of roots is α + β = 6 and the product is αβ = 8. Therefore α² + β² = (α + β)² − 2αβ = 36 − 16 = 20.

Question 47Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

The vertex of the parabola y = (x − 1)(x + 3) is

  1. (−1, 0)
  2. (−3, 0)
  3. (−1, −4) Correct answer.
  4. (1, 0)

Explanation

The roots of (x − 1)(x + 3) = 0 are x = 1 and x = −3, so the axis of symmetry is halfway between them at x = −1. Substituting x = −1 gives y = (−2)(2) = −4, so the vertex is (−1, −4).

Question 48Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

The GCD of the polynomials x³ − 1 and x² − 1 is

  1. (x − 1)²
  2. x − 1 Correct answer.
  3. x + 1
  4. x² − 1

Explanation

Factor both polynomials: x³ − 1 = (x − 1)(x² + x + 1) and x² − 1 = (x − 1)(x + 1). The only common factor is x − 1, so the GCD is x − 1.

Question 49Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

For what value(s) of k does the system kx + 2y = 3 and 3x + ky = 5 have no solution?

  1. k = ±6
  2. k = ±√6 Correct answer.
  3. k = √6 only
  4. k = 6

Explanation

For no solution, the coefficient determinant must be zero: k² − 6 = 0, so k = ±√6. At either value the lines are parallel and not coincident because the constant ratio is not the same. Hence both k = √6 and k = −√6 give no solution.

Question 50Algebra (polynomials and their graphs, GCD and LCM of polynomials, square root of polynomials, quadratic equations and graphs, matrices and determinants, solving simultaneous linear equations)

If the roots of x² + kx − 6 = 0 differ by 5, then k equals

  1. k = 1 only
  2. k = ±5
  3. k = 5 only
  4. k = ±1 Correct answer.

Explanation

Let the roots be α and β. Then α + β = −k and αβ = −6, so (α − β)² = (α + β)² − 4αβ = k² + 24. Since the roots differ by 5, 25 = k² + 24, giving k² = 1 and hence k = ±1.