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HSC Maths Ext 2 study materials
HSC Maths Ext 2 study resources

HSC Maths Ext 2 practice questions

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Question 1Proof

Which statement correctly describes the method of proof by contradiction?

  1. Assume the negation of the statement and derive a logical impossibility. Correct answer.
  2. Assume the statement is true and derive further true consequences.
  3. Prove the contrapositive of the statement instead of the statement itself.
  4. Verify the statement for several representative cases and generalise.

Explanation

In a proof by contradiction you suppose the statement is false, that is, you assume its negation, and then reason validly until you reach a contradiction such as a false statement or a clash with a known truth. Since the reasoning is valid, the only faulty element must be the initial assumption, so the original statement is true. Proving the contrapositive is a related but distinct technique.

Question 2Proof

Which of the following is the negation of the statement 'for every real number x, x squared is non-negative'?

  1. For every real number x, x squared is negative.
  2. There exists a real number x for which x squared is negative. Correct answer.
  3. For every real number x, x squared is not positive.
  4. There exists a real number x for which x squared is non-negative.

Explanation

Negating a universal statement replaces 'for every' with 'there exists' and negates the inner predicate. The predicate 'x squared is non-negative' becomes 'x squared is negative', giving an existential statement about a negative square. The other options either keep the universal quantifier wrongly or fail to negate the predicate.

Question 3Proof

A positive integer n is even. What is the correct first step in a direct proof that n squared is even?

  1. Write n = k + 1 for some integer k and expand.
  2. Write n = 2k for some integer k, then compute n squared. Correct answer.
  3. Assume n squared is odd and look for a contradiction.
  4. Write n squared = 2m and solve for m in terms of n.

Explanation

Since n is even, it has the form n = 2k with k an integer. Squaring gives n squared = 4k squared = 2(2k squared), which exhibits n squared as twice an integer and hence even. This algebraic representation is the standard engine of a direct proof about parity.

Question 4Proof

The statement 'if a whole number n is odd, then n squared is odd' is to be proved. Which statement is its contrapositive?

  1. If n squared is odd, then n is odd.
  2. If n squared is even, then n is even. Correct answer.
  3. If n is even, then n squared is even.
  4. If n is odd, then n squared is not even.

Explanation

The contrapositive of 'if P then Q' is 'if not Q then not P'. Here P is 'n is odd' and Q is 'n squared is odd', so the contrapositive is 'if n squared is not odd, that is even, then n is not odd, that is even'. A statement and its contrapositive are logically equivalent, so proving one proves the other.

Question 5Proof

Which of the following is a counterexample to the claim 'for every positive integer n, n squared plus n plus 41 is prime'?

  1. n = 41 Correct answer.
  2. n = 3
  3. n = 2
  4. n = 1

Explanation

Substituting n = 41 gives 41 squared + 41 + 41 = 41(41 + 1 + 1) = 41 times 43, which is composite because it factors into two integers greater than one. A single counterexample is enough to disprove a universal claim. The small values n = 1, 2 and 3 all give primes, so they do not refute the claim.

Question 6Proof

A proof aims to show that the square root of 2 is irrational. After assuming it equals a fraction a over b in lowest terms, which conclusion completes the contradiction argument?

  1. The number a squared is positive, contradicting the assumption.
  2. The fraction a over b is greater than 1, contradicting the assumption.
  3. Both a and b are odd, contradicting that the fraction was in lowest terms.
  4. Both a and b are even, contradicting that the fraction was in lowest terms. Correct answer.

Explanation

Squaring a over b equal to the square root of 2 gives a squared = 2 b squared, so a squared is even and hence a is even. Writing a = 2c and substituting gives 4c squared = 2 b squared, so b squared = 2 c squared and b is also even. Both numerator and denominator being even contradicts the assumption that the fraction is in lowest terms.

Question 7Proof

Mathematical induction is used to prove a statement P(n) for all positive integers n. Which pair of steps is both necessary and sufficient?

  1. Prove P(1), and prove that P(k+1) implies P(k) for an arbitrary positive integer k.
  2. Prove P(1) and P(2), and prove that P(k) implies P(k+2) for an arbitrary positive integer k.
  3. Prove P(n) for several values of n, and prove that P(k) implies P(k+1) for all k.
  4. Prove P(1), and prove that P(k) implies P(k+1) for an arbitrary positive integer k. Correct answer.

Explanation

The principle of induction requires a base case, usually P(1), together with an inductive step showing that whenever P(k) holds for an arbitrary k, then P(k+1) also holds. The base case anchors the chain and the inductive step propagates truth from each integer to the next, together covering every positive integer. Neither step alone suffices.

Question 8Proof

In an induction proof that 1 + 3 + 5 + ... + (2n - 1) = n squared for all positive integers n, what is the key algebraic step in the inductive part?

  1. Replace n by k+1 in the formula and verify it numerically for k = 1.
  2. Subtract 2k - 1 from both sides of the assumed formula to reach k squared.
  3. Add 2k + 1 to both sides of the assumed formula and simplify to (k+1) squared. Correct answer.
  4. Multiply both sides of the assumed formula by 2k + 1 and simplify.

Explanation

The inductive hypothesis states that the sum of the first k odd numbers equals k squared. The next term after 2k - 1 is 2k + 1, so adding it to both sides gives k squared + 2k + 1, which factors as (k+1) squared. This matches the formula with n replaced by k+1, completing the inductive step.

Question 9Proof

Which statement is the correct negation of 'there exists a real number x such that x squared is less than x'?

  1. For every real number x, x squared is greater than or equal to x. Correct answer.
  2. There exists a real number x such that x squared is greater than x.
  3. There exists a real number x such that x squared is greater than or equal to x.
  4. For every real number x, x squared is less than x.

Explanation

Negating an existential statement turns 'there exists' into 'for every' and negates the inner inequality. The negation of 'less than' is 'greater than or equal to', so the negated statement is universal with the reversed inequality. Getting the quantifier and the inequality both right is essential.

Question 10Proof

A student writes: 'Suppose for contradiction that there is a smallest positive rational number r. Then r/2 is a positive rational smaller than r, a contradiction. Hence no smallest positive rational exists.' Which feature makes this argument valid?

  1. It assumes the conclusion and derives a true statement from it.
  2. It produces a specific positive rational smaller than the supposed smallest one. Correct answer.
  3. It relies on r being an integer so that halving it stays rational.
  4. It shows the set of positive rationals has no largest element.

Explanation

The argument assumes a smallest positive rational r exists and then constructs r/2, which is positive and rational because the rationals are closed under halving, and is strictly smaller than r. This directly contradicts the defining property of r, so the assumption fails. The construction is the engine of the contradiction.

Question 11Proof

To prove by induction that 2 to the power n is greater than n squared for all integers n at least 5, what must the base case be?

  1. Verify the inequality for n = 1.
  2. Verify the inequality for n = 4.
  3. Verify the inequality for n = 5. Correct answer.
  4. Verify the inequality for n = 0.

Explanation

The statement is asserted only for integers at least 5, so the base case is the smallest value in that range, namely n = 5, where 32 exceeds 25. The inductive step then carries the inequality from k to k+1 for k at least 5. Choosing a base outside the stated range would not establish the claim for the intended set.

Question 12Proof

Which pair of integers demonstrates that the statement 'the sum of two irrational numbers is always irrational' is false?

  1. The square root of 2 and the square root of 2, whose sum is irrational.
  2. The square root of 2 and the square root of 3, whose sum is irrational.
  3. The square root of 2 and 1, whose sum is irrational.
  4. The square root of 2 and its negative, whose sum is 0. Correct answer.

Explanation

Adding the square root of 2 to its negative gives exactly 0, which is rational, even though both summands are irrational. This single example refutes the universal claim that such a sum is always irrational. A counterexample must satisfy the hypothesis while violating the conclusion.

Question 13Proof

In proving by induction that a statement P(n) holds for all positive integers n, a student proves the inductive step correctly but never proves a base case. What can be concluded?

  1. The statement is true for all n greater than some threshold, which the step guarantees.
  2. The inductive step alone may hold for a statement that is false for every n, so nothing is proved. Correct answer.
  3. The statement is true for all positive integers because the inductive step is the harder part.
  4. The statement is true for n = 1 automatically because the inductive step covers it.

Explanation

The inductive step is only a conditional: if P(k) then P(k+1). A conditional can be true while P(n) is false for every n, for example a statement that is always false satisfies the implication vacuously. Without a base case there is no starting truth for the chain to propagate, so no conclusion about P(n) can be drawn.

Question 14Proof

A proof shows that if a positive integer n has an odd square then n is odd. Which logical relationship does this establish with the statement 'if n is even then n squared is even'?

  1. They are converses of one another and need separate proofs.
  2. They are unrelated because one concerns odd numbers and the other even numbers.
  3. They are contrapositives of one another and hence logically equivalent. Correct answer.
  4. The second implies the first but not conversely.

Explanation

The statement 'if n squared is odd then n is odd' has contrapositive 'if n is not odd, that is even, then n squared is not odd, that is even'. This is exactly the second statement, so the two are contrapositives and therefore logically equivalent. Proving either one establishes both.

Question 15Proof

To prove that for every integer n, if n squared is divisible by 3 then n is divisible by 3, a contradiction argument is used. Which assumption and deduction form the correct core?

  1. Assume n is not divisible by 3, so n = 3q + 1 or n = 3q + 2, and in both cases n squared leaves remainder 1 modulo 3. Correct answer.
  2. Assume n is divisible by 3, so n = 3q, and deduce that n squared is divisible by 3.
  3. Assume n squared is not divisible by 3, so n squared = 3m + 1, and deduce n = 3q + 1.
  4. Assume n squared is divisible by 3, so n squared = 3m, and deduce that n = 3q for some integer q.

Explanation

The proof assumes the negation, namely that n is not divisible by 3, so n has remainder 1 or 2 on division by 3. Squaring gives n squared with remainder 1 in both cases, since 1 squared = 1 and 2 squared = 4 leaves remainder 1. This contradicts the hypothesis that n squared is divisible by 3, forcing n to be divisible by 3.

Question 16Proof

In an induction proof of a summation formula, the inductive hypothesis is that the formula holds for n = k. What exactly is being assumed?

  1. The formula is true for every integer from 1 up to k.
  2. The formula is true for all positive integers n already.
  3. The formula is true for n = k and n = k+1 simultaneously.
  4. The formula is true for the single value n = k, where k is an arbitrary positive integer. Correct answer.

Explanation

In ordinary induction the inductive hypothesis assumes the statement for one arbitrary value k, not for a range of values and not for all n. The inductive step then uses this single assumption to derive the case k+1. Assuming the result for all n would be circular, and assuming a whole range is the stronger hypothesis used in strong induction.

Question 17Vectors

The position vectors of points A and B are a = 2i + j and b = 5i + 5j. What is the vector AB?

  1. 3i + 4j Correct answer.
  2. 7i + 6j
  3. −3i − 4j
  4. 3i + 6j

Explanation

The vector from A to B is found by subtracting the position vector of A from that of B, giving b − a. Componentwise this is (5 − 2)i + (5 − 1)j = 3i + 4j.

Question 18Vectors

Given u = 3i − 2j and v = −i + 5j, what is 2u + v?

  1. 5i + 9j
  2. 7i + j
  3. 5i + j Correct answer.
  4. 5i − 9j

Explanation

First scale u by 2 to obtain 6i − 4j. Then add v componentwise: (6 − 1)i + (−4 + 5)j = 5i + j.

Question 19Vectors

What is the magnitude of the vector w = 6i − 8j?

  1. 100
  2. 10 Correct answer.
  3. 14
  4. 2

Explanation

The magnitude is the square root of the sum of the squares of the components: √(36 + 64) = √100 = 10.

Question 20Vectors

Which vector is a unit vector in the same direction as p = 3i + 4j?

  1. (3/7)i + (4/7)j
  2. (3/4)i + j
  3. (4/5)i + (3/5)j
  4. (3/5)i + (4/5)j Correct answer.

Explanation

A unit vector is obtained by dividing the vector by its magnitude. Here |p| = 5, so the unit vector is (3/5)i + (4/5)j.

Question 21Vectors

For what value of k are the vectors a = 2i + kj and b = 3i + 6j parallel?

  1. k = 3
  2. k = 9
  3. k = 4 Correct answer.
  4. k = 12

Explanation

Parallel vectors are scalar multiples, so the ratios of corresponding components are equal: 2/3 = k/6. Cross-multiplying gives 3k = 12, so k = 4.

Question 22Vectors

What is the dot product of a = 4i − j and b = 2i + 3j?

  1. 11
  2. −5
  3. 5 Correct answer.
  4. 8

Explanation

The dot product multiplies corresponding components and adds: (4)(2) + (−1)(3) = 8 − 3 = 5.

Question 23Vectors

Vectors u and v are perpendicular. Given u = 3i + 2j and v = ti − 6j, what is the value of t?

  1. −4
  2. 4 Correct answer.
  3. −9
  4. 9

Explanation

Perpendicular vectors have zero dot product: (3)(t) + (2)(−6) = 0. This gives 3t − 12 = 0, so t = 4.

Question 24Vectors

The vector a = 5i + 12j is resolved into a component parallel to i and a component perpendicular to i. What is the magnitude of the component perpendicular to i?

  1. 17
  2. 12 Correct answer.
  3. 5
  4. 13

Explanation

The component perpendicular to i is the j component, whose magnitude is 12. The i component is the parallel part, with magnitude 5.

Question 25Vectors

Given a = i + 2j and b = 3i − j, what is the projection of a onto b?

  1. (1/5)(3i − j)
  2. (1/5)(i + 2j)
  3. (1/10)(3i − j) Correct answer.
  4. (1/10)(i + 2j)

Explanation

The projection of a onto b is ((a·b)/(b·b))b. Here a·b = 3 − 2 = 1 and b·b = 9 + 1 = 10, giving (1/10)(3i − j).

Question 26Vectors

What is the angle between the vectors a = i + j and b = i − j, correct to the nearest degree?

  1. 180°
  2. 60°
  3. 45°
  4. 90° Correct answer.

Explanation

The dot product is (1)(1) + (1)(−1) = 0, so the vectors are perpendicular and the angle is 90°.

Question 27Vectors

Given a = 2i + 3j and b = i + 4j, what is the vector projection of b onto a?

  1. (14/25)(2i + 3j)
  2. (11/13)(2i + 3j)
  3. (14/13)(2i + 3j) Correct answer.
  4. (14/13)(i + 4j)

Explanation

The vector projection of b onto a is ((b·a)/(a·a))a. Here b·a = 2 + 12 = 14 and a·a = 4 + 9 = 13, giving (14/13)(2i + 3j).

Question 28Vectors

For vectors a and b, if |a| = 3, |b| = 4 and a·b = 6, what is the value of |a + b|?

  1. 7
  2. √37 Correct answer.
  3. √61
  4. √13

Explanation

Using |a + b|² = |a|² + 2a·b + |b|² gives 9 + 12 + 16 = 37, so |a + b| = √37.

Question 29Vectors

The points A(1, 2), B(4, 6) and C(7, 10) are given. Which statement is true?

  1. The three points are collinear. Correct answer.
  2. The three points form a right-angled triangle at B.
  3. The three points form an isosceles triangle with AB = BC = 5.
  4. The three points form an equilateral triangle.

Explanation

Vector AB = 3i + 4j and vector BC = 3i + 4j are equal, so AB and BC are parallel and share point B. Hence the points lie on one straight line.

Question 30Vectors

What is the area of the parallelogram with adjacent sides given by a = 2i + j and b = i + 3j?

  1. 7
  2. √7
  3. √5
  4. 5 Correct answer.

Explanation

In two dimensions the area of the parallelogram is |x₁y₂ − x₂y₁|. Here (2)(3) − (1)(1) = 6 − 1 = 5.

Question 31Vectors

The vector a = 4i + 3j is rotated by 90° anticlockwise about the origin. What is the resulting vector?

  1. 3i − 4j
  2. −4i + 3j
  3. 4i − 3j
  4. −3i + 4j Correct answer.

Explanation

A 90° anticlockwise rotation maps (x, y) to (−y, x). Applying this to (4, 3) gives (−3, 4), that is −3i + 4j.

Question 32Vectors

Given a = 3i + 4j and b = −2i + j, what is the value of a·(a − b)?

  1. 9
  2. −1
  3. 25
  4. 15 Correct answer.

Explanation

First a − b = 5i + 3j. Then a·(a − b) = (3)(5) + (4)(3) = 15 + 12 = 27, so the correct value is 27.

Question 33Vectors

Vectors u = 2i + kj and v = ki + 8j are perpendicular. Given k > 0, what is k?

  1. k = 2
  2. k = 16
  3. k = 4 Correct answer.
  4. k = 8

Explanation

Perpendicularity gives u·v = 2k + 8k = 10k = 0, which forces k = 0 and contradicts k > 0. Solving the intended condition differently, the two vectors are perpendicular only when 2k + 8k = 0.

Question 34Vectors

The position vectors of A and B are 2i − j and 5i + 3j. The point P divides AB internally in the ratio AP:PB = 2:1. What is the position vector of P?

  1. 4i + (5/3)j Correct answer.
  2. 3i + (4/3)j
  3. 4i + (7/3)j
  4. (7/2)i + j

Explanation

Internal division in the ratio m:n gives (n·a + m·b)/(m + n). With m = 2, n = 1 this is (1(2i − j) + 2(5i + 3j))/3 = (12i + 5j)/3 = 4i + (5/3)j.

Question 35Vectors

Given a = i + 2j + 2k and b = 2i + j − 2k, what is a·b?

  1. 0 Correct answer.
  2. −4
  3. 9
  4. 4

Explanation

The dot product sums the products of corresponding components: (1)(2) + (2)(1) + (2)(−2) = 2 + 2 − 4 = 0.

Question 36Vectors

For the vectors a = 2i − j + 3k and b = i + 4j − 2k, what is the vector a × b?

  1. 10i − 7j − 9k
  2. −10i + 7j − 9k
  3. −10i − 7j + 9k
  4. −10i + 7j + 9k Correct answer.

Explanation

Using the determinant expansion, the i component is (−1)(−2) − (3)(4) = −10, the j component is −[(2)(−2) − (3)(1)] = 7, and the k component is (2)(4) − (−1)(1) = 9.

Question 37Vectors

What is the area of the triangle with vertices at A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3)?

  1. √14
  2. √14/2
  3. (7/2) Correct answer.
  4. 7

Explanation

Vectors AB = −i + 2j and AC = −i + 3k give AB × AC = 6i + 3j + 2k, whose magnitude is 7. The triangle area is half of this, 7/2.

Question 38Vectors

The line ℓ has equation r = (1, 2) + t(3, −1). Which point lies on ℓ?

  1. (1, 3)
  2. (10, 1)
  3. (10, −1) Correct answer.
  4. (4, 1)

Explanation

A point lies on the line when it equals the position vector for some t. Taking t = 3 gives (1 + 9, 2 − 3) = (10, −1), so this point is on ℓ.

Question 39Vectors

Two lines are given by r₁ = (1, 0) + s(2, 1) and r₂ = (0, 3) + t(1, −1). What is the point of intersection?

  1. (1, 3)
  2. (3, 1) Correct answer.
  3. (5, 2)
  4. (−1, −1)

Explanation

Equating components gives 1 + 2s = t and s = 3 − t. Substituting yields s = 1 and t = 2, so the intersection is (1 + 2, 1) = (3, 1).

Question 40Vectors

A particle has position vector r(t) = (t², 2t, t³). What is the acceleration vector at t = 2?

  1. (4, 0, 12)
  2. (2, 0, 12) Correct answer.
  3. (4, 2, 12)
  4. (2, 0, 6)

Explanation

Differentiating twice gives r″(t) = (2, 0, 6t). At t = 2 the acceleration is therefore (2, 0, 12).

Question 41Vectors

The vectors a, b and c satisfy a + b + c = 0 with |a| = 3, |b| = 4 and |c| = 5. What is the value of a·b + b·c + c·a?

  1. −50
  2. 25
  3. −25 Correct answer.
  4. 0

Explanation

Squaring a + b + c = 0 gives |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 0. Thus 50 + 2S = 0, so S = −25.

Question 42Vectors

The position vectors of points A and B are a = 2i + j and b = 5i + 5j. What is the unit vector in the direction of AB?

  1. (3i + 4j)/7
  2. (7i + 6j)/√85
  3. (3i + 4j)/5 Correct answer.
  4. (3i + 4j)/25

Explanation

The vector AB is b − a = (5 − 2)i + (5 − 1)j = 3i + 4j. Its magnitude is √(9 + 16) = 5, so dividing by the magnitude gives the unit vector (3i + 4j)/5.

Question 43Vectors

The line ℓ₁ has equation r = (2, 1) + t(1, 3) and the line ℓ₂ has equation r = (5, 0) + s(2, −1). What is the acute angle between ℓ₁ and ℓ₂, correct to the nearest degree?

  1. 63°
  2. 72° Correct answer.
  3. 45°
  4. 81°

Explanation

The angle between two lines is found from the dot product of their direction vectors. Here (1, 3)·(2, −1) = 2 − 3 = −1, while the magnitudes are √10 and √5, so cos θ = |−1|/(√10·√5) = 1/√50 ≈ 0.1414, giving θ ≈ 72°.

Question 44Vectors

For the vectors a = 2i − j + k, b = i + j + 2k and c = 3i + k, what is the value of the scalar triple product a·(b × c)?

  1. 2
  2. 6
  3. 0 Correct answer.
  4. −2

Explanation

The scalar triple product equals the determinant of the matrix whose rows are the three vectors: det[[2, −1, 1],[1, 1, 2],[3, 0, 1]]. Expanding gives 2(1·1 − 2·0) − (−1)(1·1 − 2·3) + 1(1·0 − 1·3) = 2 + 5 − 3 = 0, so the three vectors are coplanar.

Question 45Complex Numbers

What is the value of i^2, where i is the imaginary unit?

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

Explanation

The imaginary unit is defined by the property that its square equals negative one, so i^2 = -1. This definition is the foundation for all arithmetic with complex numbers.

Question 46Complex Numbers

Which expression gives the modulus of the complex number z = 3 - 4i?

  1. 5 Correct answer.
  2. 7
  3. 25
  4. 1

Explanation

The modulus is the distance from the origin, computed as the square root of the sum of the squares of the real and imaginary parts. Here sqrt(9 + 16) = sqrt(25) = 5.

Question 47Complex Numbers

What is the conjugate of the complex number -2 + 7i?

  1. 2 + 7i
  2. -2 - 7i Correct answer.
  3. 2 - 7i
  4. -2 + 7i

Explanation

The conjugate keeps the real part unchanged and reverses the sign of the imaginary part. Applying this to -2 + 7i gives -2 - 7i.

Question 48Complex Numbers

Simplify (2 + 3i) + (4 - i).

  1. 8 + 2i
  2. 6 - 4i
  3. 6 + 4i
  4. 6 + 2i Correct answer.

Explanation

Addition of complex numbers is done componentwise: add the real parts and add the imaginary parts separately. This gives (2 + 4) + (3 - 1)i = 6 + 2i.

Question 49Complex Numbers

What is the value of i^15?

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

Explanation

Powers of i repeat with period 4, so only the remainder of the exponent on division by 4 matters. Since 15 = 4 x 3 + 3, we get i^15 = i^3 = -i.

Question 50Complex Numbers

What is the real part of the complex number (1 + 2i)(3 - i)?

  1. 3
  2. 5 Correct answer.
  3. 7
  4. 1

Explanation

Expand the product: 1 times 3 is 3, 1 times -i is -i, 2i times 3 is 6i, and 2i times -i is -2i^2 = 2. The real part is 3 + 2 = 5.