Year 11 Specialist Maths — Full Year Review

This review covers all topics from Year 11 Specialist Mathematics: Combinatorics, Proof by Induction, Vectors in the Plane, Complex Numbers, Trigonometry & Functions, and Matrices. Click each answer box to reveal the worked solution.

Review Questions

A team of 4 is to be chosen from 6 men and 5 women. How many teams contain at least 2 women?

Count teams with 2W+2M, 3W+1M, and 4W+0M separately.

2W, 2M: C(5,2) × C(6,2) = 10 × 15 = 150

3W, 1M: C(5,3) × C(6,1) = 10 × 6 = 60

4W, 0M: C(5,4) × C(6,0) = 5 × 1 = 5

Total = 150 + 60 + 5 = 215 teams
How many 5-letter arrangements can be formed from the letters of the word MATHS if no letter is repeated? How many of these begin with a vowel?

Total arrangements: P(5,5) = 5! = 120

Beginning with a vowel: The only vowel in MATHS is A. Fix A in position 1; arrange remaining 4 letters in 4! = 24 ways.

Arrangements beginning with A = 24
Use the binomial theorem to expand (2x − 3)&sup4;. State the coefficient of x².

Using (a + b)⁴ = ∑ C(4,k) a^4−k b^k with a = 2x, b = −3:

k=0: C(4,0)(2x)⁴(−3)⁰ = 16x⁴

k=1: C(4,1)(2x)³(−3)¹ = 4 × 8x³ × (−3) = −96x³

k=2: C(4,2)(2x)²(−3)² = 6 × 4x² × 9 = 216x²

k=3: C(4,3)(2x)¹(−3)³ = 4 × 2x × (−27) = −216x

k=4: C(4,4)(2x)⁰(−3)⁴ = 81

(2x − 3)⁴ = 16x⁴ − 96x³ + 216x² − 216x + 81

Coefficient of x² = 216
Find the term independent of x in the expansion of (x + 2/x)⁶.

General term: T_k+1 = C(6,k) · x^6−k · (2/x)^k = C(6,k) · 2^k · x^6−2k

For the term independent of x: 6 − 2k = 0 ⇒ k = 3

T₄ = C(6,3) · 2³ = 20 × 8 = 160
Prove by mathematical induction that for all integers n ≥ 1: 1 + 3 + 5 + … + (2n − 1) = n².

Base case (n = 1): LHS = 1; RHS = 1² = 1. True.

Inductive hypothesis: Assume true for n = k: 1 + 3 + … + (2k−1) = k².

Inductive step: Need to show the result for n = k + 1, i.e., LHS = (k+1)².

1 + 3 + … + (2k−1) + (2(k+1)−1)

= k² + (2k + 1) [by inductive hypothesis]

= (k + 1)²

This is exactly the statement for n = k + 1. By the principle of mathematical induction, the result holds for all n ≥ 1. ■
Prove by mathematical induction that 3ⁿ − 1 is divisible by 2 for all integers n ≥ 1.

Base case (n = 1): 3¹ − 1 = 2, which is divisible by 2. True.

Inductive hypothesis: Assume 3^k − 1 = 2m for some integer m (i.e., 3^k = 2m + 1).

Inductive step: Consider 3^k+1 − 1:

3^k+1 − 1 = 3 · 3^k − 1 = 3(2m + 1) − 1 = 6m + 3 − 1 = 6m + 2 = 2(3m + 1)

Since 3m + 1 is an integer, 3^k+1 − 1 is divisible by 2. By mathematical induction, the result holds for all n ≥ 1. ■
Prove by mathematical induction: ∑_r=1ⁿ r(r+1) = n(n+1)(n+2)/3 for all n ≥ 1.

Base case (n = 1): LHS = 1 × 2 = 2; RHS = 1 × 2 × 3 / 3 = 2. True.

Inductive hypothesis: Assume ∑_r=1^k r(r+1) = k(k+1)(k+2)/3.

Inductive step:

∑_r=1^k+1 r(r+1) = k(k+1)(k+2)/3 + (k+1)(k+2)

= (k+1)(k+2) [k/3 + 1]

= (k+1)(k+2) · (k+3)/3

= (k+1)(k+2)(k+3)/3

This is the formula for n = k + 1. By mathematical induction, the result holds for all n ≥ 1. ■
Given vectors a = (3, −4) and b = (1, 2): (a) Find a · b and hence the angle between them. (b) Find the scalar projection of a onto b.

(a) a · b = (3)(1) + (−4)(2) = 3 − 8 = −5

|a| = √(9+16) = 5; |b| = √(1+4) = √5

cosθ = −5 / (5√5) = −1/√5

θ = cos⁻¹(−1/√5) ≈ 116.6°

(b) Scalar projection of a onto b: comp_ba = (a · b) / |b| = −5/√5 = −√5
A particle moves along a straight line. At time t = 0 it is at position (1, 2) and its velocity vector is (3, −1). Write the parametric equations of the path. Find where the path crosses the line y = 0.

Parametric equations:

x = 1 + 3t

y = 2 − t

Where y = 0:

2 − t = 0 ⇒ t = 2

x = 1 + 3(2) = 7

The path crosses y = 0 at the point (7, 0).
Find the vector projection of u = (4, 1) onto v = (2, −3). Hence find the component of u perpendicular to v.

u · v = (4)(2) + (1)(−3) = 8 − 3 = 5

|v|² = 4 + 9 = 13

proj_vu = (5/13)(2, −3) = (10/13, −15/13)

Component perpendicular to v: u − proj_vu = (4 − 10/13, 1 + 15/13) = (42/13, 28/13)

Check: (42/13, 28/13) · (2,−3) = 84/13 − 84/13 = 0 ✓
Express z = −1 + √3 i in polar (modulus-argument) form. Hence evaluate z⁶ using De Moivre’s theorem.

|z| = √(1 + 3) = 2

arg(z): Re = −1 < 0, Im = √3 > 0 ⇒ second quadrant.

Reference angle = arctan(√3/1) = π/3. So arg(z) = π − π/3 = 2π/3

Polar form: z = 2(cos(2π/3) + i sin(2π/3))

By De Moivre: z⁶ = 2⁶(cos(6 × 2π/3) + i sin(6 × 2π/3))

= 64(cos(4π) + i sin(4π))

= 64(1 + 0i) = 64
Find all cube roots of 8i. Express your answers in Cartesian form a + bi.

Write 8i in polar form: |8i| = 8, arg(8i) = π/2

8i = 8 cis(π/2)

Cube roots: ³√8 cis((π/2 + 2πk)/3) for k = 0, 1, 2

= 2 cis((π/6 + 2πk/3))

k = 0: 2 cis(π/6) = 2(cos30° + i sin30°) = 2(√3/2 + i/2) = √3 + i

k = 1: 2 cis(π/6 + 2π/3) = 2 cis(5π/6) = 2(−√3/2 + i/2) = −√3 + i

k = 2: 2 cis(π/6 + 4π/3) = 2 cis(3π/2) = 2(0 − i) = −2i
Let z = 1 + i. (a) Find z⁴ using De Moivre’s theorem. (b) Show that z⁴ is real and negative.

(a) |z| = √2, arg(z) = π/4

z = √2 cis(π/4)

z⁴ = (√2)⁴ cis(4 × π/4) = 4 cis(π) = 4(cosπ + i sinπ) = 4(−1 + 0i) = −4

(b) z⁴ = −4, which has imaginary part 0 (real) and real part −4 < 0 (negative). ✓
The complex number z satisfies z² + 2z + 5 = 0. Find both roots and plot their positions on the Argand diagram, identifying symmetry.

Using the quadratic formula: z = (−2 ± √(4 − 20)) / 2 = (−2 ± √(−16)) / 2

= (−2 ± 4i) / 2

z = −1 + 2i and z = −1 − 2i

On the Argand diagram, both roots lie on the vertical line Re(z) = −1, at heights +2 and −2 respectively. They are reflections of each other in the real axis (complex conjugate pair), as expected for a quadratic with real coefficients.
Prove the identity: cos(A + B) + cos(A − B) = 2 cos A cos B.

Expanding the compound angle formulas:

cos(A + B) = cos A cos B − sin A sin B

cos(A − B) = cos A cos B + sin A sin B

Adding:

cos(A + B) + cos(A − B) = 2 cos A cos B − sin A sin B + sin A sin B = 2 cos A cos B ✓
Solve for x ∈ [0, 2π]: 2 sin x cos x = sin x.

Rearrange: 2 sin x cos x − sin x = 0

sin x (2 cos x − 1) = 0

Case 1: sin x = 0 ⇒ x = 0, π, 2π

Case 2: 2 cos x − 1 = 0 ⇒ cos x = 1/2 ⇒ x = π/3, 5π/3

Solutions: x = 0, π/3, π, 5π/3, 2π
Find the exact value of sin(75°) using the compound angle formula.

Write 75° = 45° + 30°:

sin(75°) = sin(45° + 30°) = sin45°cos30° + cos45°sin30°

= (√2/2)(√3/2) + (√2/2)(1/2)

= √6/4 + √2/4

= (√6 + √2)/4
Find the inverse function of f(x) = 2 sin(x) − 1, for x ∈ [−π/2, π/2]. State the domain and range of f⁻¹.

Set y = 2 sin(x) − 1 and solve for x:

y + 1 = 2 sin(x)

sin(x) = (y + 1)/2

x = arcsin((y + 1)/2)

So f⁻¹(x) = arcsin((x + 1)/2)

Domain of f⁻¹: Range of f. When x ∈ [−π/2, π/2], sin(x) ∈ [−1, 1], so f(x) = 2sin(x)−1 ∈ [−3, 1]. Domain of f⁻¹ = [−3, 1].

Range of f⁻¹: Domain of f = [−π/2, π/2].
Let A = [3 1 ; −2 4]. Find det(A), A⁻¹, and verify AA⁻¹ = I.

A = [[3, 1], [−2, 4]]

det(A) = (3)(4) − (1)(−2) = 12 + 2 = 14

A⁻¹ = (1/14)[[4, −1], [2, 3]] = [[4/14, −1/14], [2/14, 3/14]] = [[2/7, −1/14], [1/7, 3/14]]

Verification: AA⁻¹:

Row 1 of A × Col 1 of A⁻¹: 3(4/14) + 1(2/14) = 12/14 + 2/14 = 14/14 = 1 ✓

Row 1 of A × Col 2 of A⁻¹: 3(−1/14) + 1(3/14) = −3/14 + 3/14 = 0 ✓

Row 2 of A × Col 1 of A⁻¹: −2(4/14) + 4(2/14) = −8/14 + 8/14 = 0 ✓

Row 2 of A × Col 2 of A⁻¹: −2(−1/14) + 4(3/14) = 2/14 + 12/14 = 1 ✓

AA⁻¹ = I ✓
Solve the system using matrices (or Gaussian elimination): 2x + y − z = 3, x − y + 2z = −1, 3x + 2y + z = 7.

Write as augmented matrix and row reduce:

[2 1 −1 | 3]

[1 −1 2 | −1]

[3 2 1 | 7]

Swap R1 and R2: [1 −1 2 | −1], [2 1 −1 | 3], [3 2 1 | 7]

R2 ← R2 − 2R1: [0 3 −5 | 5]

R3 ← R3 − 3R1: [0 5 −5 | 10]

R3 ← R3 − (5/3)R2: [0 0 (25/3−5) | 10−25/3] = [0 0 10/3 | 5/3]

z = (5/3)/(10/3) = 1/2

From R2: 3y − 5(1/2) = 5 ⇒ 3y = 15/2 ⇒ y = 5/2

From R1: x − 5/2 + 2(1/2) = −1 ⇒ x = −1 + 5/2 − 1 = 1/2

x = 1/2, y = 5/2, z = 1/2

Verify in original equations: 2(1/2) + 5/2 − 1/2 = 1 + 2 = 3 ✓
The transformation matrix T = [[0, −1], [1, 0]] represents a rotation. (a) What angle does T rotate by? (b) Find T⁴. (c) Find the image of point (3, −2) under T.

(a) T = [[cosθ, −sinθ], [sinθ, cosθ]] ⇒ cosθ = 0, sinθ = 1 ⇒ θ = 90° anticlockwise

(b) T⁴ represents a rotation of 4 × 90° = 360°, so T⁴ = I = [[1,0],[0,1]]

(c) Image of (3, −2):

T [[3], [−2]] = [[0×3 + (−1)(−2)], [1×3 + 0×(−2)]] = [[2], [3]]

Image = (2, 3)