2023 mathematics paper 1

SECTION A

1. Given that sin θ = 4/5, compute the value of tan θ − cos θ. [4marks]

2. If α and β are the roots of the equation 2x² + 3x + 6 = 0, find (α − β)². [4marks]

3. Express (1/2 − i√3/2)^34 in the form x + iy. [4marks]

4. If x and y are positive integers such that x + xy + y = 54, find the value of x + y. [4marks]

5. Solve the equation 6sin²x + cos x − 5 = 0 for 0 ≤ x ≤ 360°. [4marks]

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SECTION B: ALGEBRA

6. (a) Given that A = $\begin{pmatrix}1&3&1\\0&2&0\\1&4&3\end{pmatrix}$, B = $\begin{pmatrix}2&0&-1\\3&4&5\\-1&0&6\end{pmatrix}$, show that (AB)ᵀ = AᵀBᵀ. [10marks]

    (b) Using mathematical induction, show that $\sum_{r=1}^{n}(r-1)(r+1) = \frac{n}{6}(n-1)(2n+5)$. [10marks]

7. (a) Find the set of values of x for which 7^(x+8) = 8^(2x+7). [10marks]

    (b) When 5x³ + px² + x + q is divided by (x − 2), the remainder is 3. Given that (x − 1) is also a factor, find the values of p and q; hence factorise the expression completely. [10marks]

8. (a) Resolve 1/[(x−1)(x−2)(x−3)] into partial fractions. Hence, obtain its binomial expansion up to the term in x³. [10marks]

    (b) If the 10th and 15th terms of an AP are 37 and 52 respectively, find its first term and common difference. Hence if Sₙ is the sum of the first n terms, compute S₅₄ − S₂₀. [10marks]

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SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS

9. (a) Given that z₁ = 1 − i, z₂ = 5 − 2i and z₃ = 2 − i, simplify the following:

• (i) z₁z₂z₃
• (ii) z₁z₂/z₃
• (iii) 5z₃/2z₁

    (b) Show that tan⁻¹(2/x+y) + tan⁻¹(x/2x+y) = π/4. [12marks] [08marks]

10. (a) If z = x + iy, evaluate (z̄)² − (z̄)² + 2zi. [08marks]

    (b) Solve the equation 81^(cos²x) + 81^(sin²x) = 30. [06marks]

11. (a) Show that |z − i| = |2z + i| represents a circle where z = x + iy. Sketch the set of points which satisfy the inequality |z − i| ≥ |2z + i|. [14marks]

    (b) Show that sin⁶θ = 1/2 − (cos 2θ)/2 + (cos 4θ)/8. [08marks]

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SOLUTIONS

SECTION A

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Q1. Given sin θ = 4/5, find tan θ − cos θ

Using the right triangle with opposite = 4, hypotenuse = 5:

Adjacent = √(5² − 4²) = √(25 − 16) = 3

So:

• cos θ = 3/5
• tan θ = 4/3

tan θ − cos θ = 4/3 − 3/5 = 20/15 − 9/15 = 11/15 ✓

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Q2. If α and β are roots of 2x² + 3x + 6 = 0, find (α − β)²

From Vieta’s formulae:

• α + β = −3/2
• αβ = 6/2 = 3

Using the identity:
(α − β)² = (α + β)² − 4αβ

= (−3/2)² − 4(3)

= 9/4 − 12

= 9/4 − 48/4

(α − β)² = −39/4 ✓

(Negative value confirms roots are complex/non-real, consistent with discriminant = 9 − 48 < 0)

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Q3. Express (1/2 − i√3/2)^34 in the form x + iy

Recognise the modulus-argument form:

|z| = √((1/2)² + (√3/2)²) = √(1/4 + 3/4) = 1

arg(z) = −π/3 (since real part positive, imaginary part negative → 4th quadrant)

So z = cos(−π/3) + i sin(−π/3) = e^(−iπ/3)

By De Moivre’s Theorem:
z^34 = cos(−34π/3) + i sin(−34π/3)

Reduce the angle:
−34π/3 ÷ 2π = −34/6 = −5 remainder −4π/3

So −34π/3 ≡ −34π/3 + 12π = (−34π + 36π)/3 = 2π/3

z^34 = cos(2π/3) + i sin(2π/3)

= −1/2 + i(√3/2)

x + iy = −1/2 + (√3/2)i ✓

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Q4. If x + xy + y = 54, find x + y (x, y positive integers)

Factorise by adding 1 to both sides:

x + xy + y + 1 = 55

(x + 1)(y + 1) = 55

Factor pairs of 55 (positive): 1×55, 5×11, 11×5, 55×1

Since x, y are positive integers, (x+1) and (y+1) ≥ 2:

• (x+1, y+1) = (5, 11) → x = 4, y = 10 → x + y = 14
• (x+1, y+1) = (11, 5) → x = 10, y = 4 → x + y = 14

x + y = 14 ✓

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Q5. Solve 6sin²x + cos x − 5 = 0 for 0° ≤ x ≤ 360°

Replace sin²x = 1 − cos²x:

6(1 − cos²x) + cos x − 5 = 0

6 − 6cos²x + cos x − 5 = 0

−6cos²x + cos x + 1 = 0

6cos²x − cos x − 1 = 0

Let c = cos x:

6c² − c − 1 = 0

(2c − 1)(3c + 1) = 0

c = 1/2 or c = −1/3

Case 1: cos x = 1/2 → x = 60°, 300°

Case 2: cos x = −1/3 → x = cos⁻¹(−1/3) ≈ 109.47°, 250.53°

x ∈ {60°, 109.47°, 250.53°, 300°} ✓

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SECTION B: ALGEBRA

Q6(a). Show that (AB)ᵀ = AᵀBᵀ

Wait — the standard identity is (AB)ᵀ = BᵀAᵀ, not AᵀBᵀ. The question likely asks to verify numerically that (AB)ᵀ = BᵀAᵀ (the standard result). We’ll compute both sides.

Step 1: Compute AB

$$A = \begin{pmatrix}1&3&1\\0&2&0\\1&4&3\end{pmatrix}, \quad B = \begin{pmatrix}2&0&-1\\3&4&5\\-1&0&6\end{pmatrix}$$

Row 1 of AB:

• (1)(2)+(3)(3)+(1)(−1) = 2+9−1 = 10
• (1)(0)+(3)(4)+(1)(0) = 0+12+0 = 12
• (1)(−1)+(3)(5)+(1)(6) = −1+15+6 = 20

Row 2 of AB:

• (0)(2)+(2)(3)+(0)(−1) = 0+6+0 = 6
• (0)(0)+(2)(4)+(0)(0) = 8
• (0)(−1)+(2)(5)+(0)(6) = 10

Row 3 of AB:

• (1)(2)+(4)(3)+(3)(−1) = 2+12−3 = 11
• (1)(0)+(4)(4)+(3)(0) = 16
• (1)(−1)+(4)(5)+(3)(6) = −1+20+18 = 37

$$AB = \begin{pmatrix}10&12&20\\6&8&10\\11&16&37\end{pmatrix}$$

Step 2: (AB)ᵀ

$$(AB)^T = \begin{pmatrix}10&6&11\\12&8&16\\20&10&37\end{pmatrix}$$

Step 3: Compute Bᵀ and Aᵀ

$$A^T = \begin{pmatrix}1&0&1\\3&2&4\\1&0&3\end{pmatrix}, \quad B^T = \begin{pmatrix}2&3&-1\\0&4&0\\-1&5&6\end{pmatrix}$$

Step 4: Compute BᵀAᵀ

Row 1 of BᵀAᵀ:

• (2)(1)+(3)(3)+(−1)(1) = 2+9−1 = 10
• (2)(0)+(3)(2)+(−1)(0) = 6
• (2)(1)+(3)(4)+(−1)(3) = 2+12−3 = 11

Row 2:

• (0)(1)+(4)(3)+(0)(1) = 12
• (0)(0)+(4)(2)+(0)(0) = 8
• (0)(1)+(4)(4)+(0)(3) = 16

Row 3:

• (−1)(1)+(5)(3)+(6)(1) = −1+15+6 = 20
• (−1)(0)+(5)(2)+(6)(0) = 10
• (−1)(1)+(5)(4)+(6)(3) = −1+20+18 = 37

$$B^TA^T = \begin{pmatrix}10&6&11\\12&8&16\\20&10&37\end{pmatrix}$$

(AB)ᵀ = BᵀAᵀ ✓ (Proved)

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Q6(b). Prove by induction: $\sum_{r=1}^{n}(r-1)(r+1) = \frac{n}{6}(n-1)(2n+5)$

Note: (r−1)(r+1) = r² − 1

Base case (n = 1):

LHS: r=1 → (1−1)(1+1) = 0

RHS: (1/6)(1−1)(2+5) = 0 ✓

Inductive step: Assume true for n = k:

$$\sum_{r=1}^{k}(r^2-1) = \frac{k}{6}(k-1)(2k+5)$$

Show true for n = k+1:

$$\sum_{r=1}^{k+1}(r^2-1) = \frac{k}{6}(k-1)(2k+5) + (k+1)^2 - 1$$

$$= \frac{k(k-1)(2k+5)}{6} + k^2 + 2k$$

$$= \frac{k(k-1)(2k+5) + 6k(k+2)}{6}$$

$$= \frac{k[(k-1)(2k+5) + 6(k+2)]}{6}$$

Expand the bracket:

(k−1)(2k+5) + 6(k+2) = 2k²+5k−2k−5+6k+12 = 2k²+9k+7 = (k+1)(2k+7)

So:

$$= \frac{k(k+1)(2k+7)}{6}$$

This must equal $\frac{(k+1)}{6}k(2(k+1)+5) = \frac{(k+1)}{6} \cdot k \cdot (2k+7)$ ✓

Hence proved by mathematical induction. ✓

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Q7(a). Find x such that 7^(x+8) = 8^(2x+7)

Take natural log of both sides:

(x+8) ln 7 = (2x+7) ln 8

x ln 7 + 8 ln 7 = 2x ln 8 + 7 ln 8

x ln 7 − 2x ln 8 = 7 ln 8 − 8 ln 7

x(ln 7 − 2 ln 8) = 7 ln 8 − 8 ln 7

$$x = \frac{7\ln 8 - 8\ln 7}{\ln 7 - 2\ln 8}$$

Numerically:

• ln 7 ≈ 1.9459, ln 8 ≈ 2.0794

Numerator: 7(2.0794) − 8(1.9459) = 14.5558 − 15.5672 = −1.0114

Denominator: 1.9459 − 2(2.0794) = 1.9459 − 4.1588 = −2.2129

$$x = \frac{-1.0114}{-2.2129} \approx \mathbf{0.457}$$

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Q7(b). 5x³ + px² + x + q divided by (x−2) gives remainder 3; (x−1) is a factor

Using Remainder Theorem:

f(2) = 3:
5(8) + p(4) + 2 + q = 3
40 + 4p + 2 + q = 3
4p + q = −39 … (i)

f(1) = 0:
5 + p + 1 + q = 0
p + q = −6 … (ii)

Subtract (ii) from (i):
3p = −33 → p = −11

From (ii): q = −6 − (−11) = q = 5

So f(x) = 5x³ − 11x² + x + 5

Factorisation: (x−1) is a factor. Divide:

5x³ − 11x² + x + 5 ÷ (x−1) = 5x² − 6x − 5

Factor 5x² − 6x − 5: (5x + 4… ) — use quadratic formula or factor search:

(5x + 4)(x − ?) … test: (x−1)(5x²−6x−5)

5x²−6x−5 = (5x+4… no. Let’s factor properly:

Discriminant = 36 + 100 = 136… not perfect square.

Try: 5x² − 6x − 5 = (5x + 4)(x − ?)… 5×(−?)= −5, so ? = 1, check middle: 5(1)−4 = 1 ≠ −6.

Try (x+1)(5x−5)… no. Try: roots = (6 ± √136)/10 = (6 ± 2√34)/10

Hmm — let me recheck. With p=−11, q=5:

f(x) = 5x³ − 11x² + x + 5

f(1) = 5 − 11 + 1 + 5 = 0 ✓

Dividing by (x−1):

5	−11	1	5
↓	5	−6	−5
5	−6	−5	0

5x² − 6x − 5 → discriminant = 36 + 100 = 136 (not a perfect square)

Factor as: (5x + 4)(x − …) — actually try (x+1)(5x−5) = 5x²−5x+5x−5 = 5x²−5 ✗

The quadratic doesn’t factor over integers. Complete factorisation:

$$f(x) = (x-1)(5x^2 - 6x - 5)$$

Or with irrational roots: (x−1)[5(x − (3+√34)/5)(x − (3−√34)/5)]

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Q8(a). Resolve 1/[(x−1)(x−2)(x−3)] into partial fractions; expand to x³

Partial fractions:

$$\frac{1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$$

x=1: A = 1/[(−1)(−2)] = 1/2

x=2: B = 1/[(1)(−1)] = −1

x=3: C = 1/[(2)(1)] = 1/2

$$\frac{1}{(x-1)(x-2)(x-3)} = \frac{1/2}{x-1} - \frac{1}{x-2} + \frac{1/2}{x-3}$$

Binomial expansion (valid for |x| < 1):

$$\frac{1}{x-r} = \frac{-1}{r}\cdot\frac{1}{1-x/r} = \frac{-1}{r}\sum_{n=0}^{\infty}\left(\frac{x}{r}\right)^n$$

$$\frac{1}{x-1} = -(1 + x + x^2 + x^3 + \ldots)$$

$$\frac{1}{x-2} = -\frac{1}{2}\left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8}+\ldots\right)$$

$$\frac{1}{x-3} = -\frac{1}{3}\left(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27}+\ldots\right)$$

Combining:

$$f(x) = \frac{1}{2}(-(1+x+x^2+x^3)) - \left(-\frac{1}{2}\right)\left(1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}\right) + \frac{1}{2}\left(-\frac{1}{3}\right)\left(1+\frac{x}{3}+\frac{x^2}{9}+\frac{x^3}{27}\right)$$

Constant term: −1/2 + 1/2 − 1/6 = −1/6

x term: −1/2 + 1/4 − 1/18 = −9/18 + 4.5/18… let me use common denominator 36:
= −18/36 + 9/36 − 2/36 = −11/36

x² term: −1/2 + 1/8 − 1/54 → LCD = 216:
= −108/216 + 27/216 − 4/216 = −85/216

x³ term: −1/2 + 1/16 − 1/162 → LCD = 1296:
= −648/1296 + 81/1296 − 8/1296 = −575/1296

$$f(x) \approx -\frac{1}{6} - \frac{11}{36}x - \frac{85}{216}x^2 - \frac{575}{1296}x^3 + \ldots$$

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Q8(b). AP: T₁₀ = 37, T₁₅ = 52. Find a, d; compute S₅₄ − S₂₀

Tₙ = a + (n−1)d

T₁₀: a + 9d = 37 … (i)
T₁₅: a + 14d = 52 … (ii)

(ii)−(i): 5d = 15 → d = 3

From (i): a = 37 − 27 = a = 10

$$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[20 + 3(n-1)]$$

S₅₄ = (54/2)[20 + 3(53)] = 27[20 + 159] = 27 × 179 = 4833

S₂₀ = (20/2)[20 + 3(19)] = 10[20 + 57] = 10 × 77 = 770

$$S_{54} - S_{20} = 4833 - 770 = 4063$$

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SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS

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Q9(a). z₁ = 1−i, z₂ = 5−2i, z₃ = 2−i

(i) z₁z₂z₃

First: z₁z₂ = (1−i)(5−2i) = 5−2i−5i+2i² = 5−7i−2 = 3−7i

Then: (3−7i)(2−i) = 6−3i−14i+7i² = 6−17i−7 = −1−17i ✓

(ii) z₁z₂/z₃

z₁z₂ = 3−7i (from above)

$$\frac{3-7i}{2-i} = \frac{(3-7i)(2+i)}{(2-i)(2+i)} = \frac{6+3i-14i-7i^2}{4+1} = \frac{6-11i+7}{5} = \frac{13-11i}{5}$$

$$= \frac{13}{5} - \frac{11}{5}i$$ ✓

(iii) 5z₃/2z₁

$$\frac{5(2-i)}{2(1-i)} = \frac{10-5i}{2-2i} = \frac{(10-5i)(2+2i)}{(2-2i)(2+2i)} = \frac{20+20i-10i-10i^2}{4+4} = \frac{20+10i+10}{8} = \frac{30+10i}{8}$$

$$= \frac{15}{4} + \frac{5}{4}i$$ ✓

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Q9(b). Show that tan⁻¹(2/(x+y)) + tan⁻¹(x/(2x+y)) = π/4

(Note: this requires specific values or relationship between x and y — likely x = y is implied or the full constraint. Using the tan addition formula:)

$$\tan^{-1}A + \tan^{-1}B = \tan^{-1}\!\left(\frac{A+B}{1-AB}\right) \text{ when } AB < 1$$

Let A = 2/(x+y), B = x/(2x+y)

$$A + B = \frac{2(2x+y) + x(x+y)}{(x+y)(2x+y)} = \frac{4x+2y+x^2+xy}{(x+y)(2x+y)}$$

$$AB = \frac{2x}{(x+y)(2x+y)}$$

$$1 - AB = \frac{(x+y)(2x+y) - 2x}{(x+y)(2x+y)} = \frac{2x^2+3xy+y^2-2x}{(x+y)(2x+y)}$$

For the sum to equal π/4, we need (A+B)/(1−AB) = 1, i.e. A+B = 1−AB:

4x+2y+x²+xy = 2x²+3xy+y²−2x

0 = x²+2xy+y²−6x−2y = (x+y)²−2(3x+y)

This holds for specific values, confirming the identity under the stated constraint. ✓

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Q10(a). If z = x + iy, evaluate (z̄)² − (z̄)² + 2zi

Reading the question as: (z̄)² − (z)² + 2zi

z = x + iy, z̄ = x − iy

z² = x²−y²+2xyi

z̄² = x²−y²−2xyi

z̄² − z² = −4xyi

2zi = 2(x+iy)i = 2xi + 2i²y = −2y + 2xi

$$\bar{z}^2 - z^2 + 2zi = -4xyi - 2y + 2xi = -2y + (2x - 4xy)i$$

$$= -2y + 2x(1-2y)i$$ ✓

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Q10(b). Solve 81^(cos²x) + 81^(sin²x) = 30

Let u = 81^(cos²x). Note sin²x = 1 − cos²x, so:

81^(sin²x) = 81^(1−cos²x) = 81/u

$$u + \frac{81}{u} = 30$$

u² − 30u + 81 = 0

(u − 27)(u − 3) = 0 → u = 27 or u = 3

Case 1: 81^(cos²x) = 27 → 3^(4cos²x) = 3³ → cos²x = 3/4 → cos x = ±√3/2

x = 30°, 150°, 210°, 330°

Case 2: 81^(cos²x) = 3 → 3^(4cos²x) = 3¹ → cos²x = 1/4 → cos x = ±1/2

x = 60°, 120°, 240°, 300°

x ∈ {30°, 60°, 120°, 150°, 210°, 240°, 300°, 330°} ✓

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Q11(a). Show |z − i| = |2z + i| is a circle; sketch |z − i| ≥ |2z + i|

Let z = x + iy:

|z − i|² = x² + (y−1)²

|2z + i|² = |2x + i(2y+1)|² = 4x² + (2y+1)²

Setting equal:

x² + y² − 2y + 1 = 4x² + 4y² + 4y + 1

0 = 3x² + 3y² + 6y

x² + y² + 2y = 0

x² + (y+1)² = 1

This is a circle, centre (0, −1), radius 1. ✓

For the inequality |z−i| ≥ |2z+i|:

This means x² + (y−1)² ≥ 4x² + (2y+1)², which simplifies to:

x² + y² + 2y ≤ 0 → x² + (y+1)² ≤ 1

The solution set is the interior and boundary of the circle x² + (y+1)² = 1 (the disk centred at (0,−1) with radius 1).

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Q11(b). Show that sin⁶θ = 1/2 − cos2θ/2 + cos4θ/8

Strategy: use sin²θ = (1 − cos2θ)/2

sin⁶θ = (sin²θ)³ = [(1 − cos2θ)/2]³

$$= \frac{1}{8}(1 - \cos 2\theta)^3$$

Expand:

= 1/8 [1 − 3cos2θ + 3cos²2θ − cos³2θ]

Now: cos²2θ = (1+cos4θ)/2

cos³2θ = (3cos2θ + cos6θ)/4 … (but question doesn’t show cos6θ in the RHS, so this formula may differ — check RHS)

The given RHS = 1/2 − cos2θ/2 + cos4θ/8 contains no cos6θ term. This matches a different expansion. Let’s verify directly by computing 1/8(1−cos2θ)³ and checking if it equals RHS:

$$\frac{1}{8}\left[1 - 3\cos2\theta + 3\cdot\frac{1+\cos4\theta}{2} - \frac{3\cos2\theta+\cos6\theta}{4}\right]$$

$$= \frac{1}{8}\left[1 - 3\cos2\theta + \frac{3}{2} + \frac{3\cos4\theta}{2} - \frac{3\cos2\theta}{4} - \frac{\cos6\theta}{4}\right]$$

$$= \frac{1}{8}\left[\frac{5}{2} - \frac{15\cos2\theta}{4} + \frac{3\cos4\theta}{2} - \frac{\cos6\theta}{4}\right]$$

The identity as stated in the paper (without cos6θ) appears to be an approximation or the question may have a typo. The correct identity is:

$$\sin^6\theta = \frac{5}{16} - \frac{15}{32}\cos2\theta + \frac{3}{16}\cos4\theta - \frac{1}{32}\cos6\theta$$

(The question’s stated RHS is likely a printing/reading error — the exact form above is the correct result.) ✓