2024 IJMB mathematics paper I

SECTION A

1. Given that sec θ = 25/7, compute the value of cot θ + cos θ. [4 marks]

2. If α and β are the roots of the equation x² − 5x + 6 = 0, find α − β. [4 marks]

3. Express ((√2 + i)/(√2 − 2i))² in the form a + ib. [4 marks]

4. If the 16th term of an AP is 3 times the 4th term, prove that the 23rd term is 5 times the 3rd term. [4 marks]

5. Solve the equation 1ˣ = 3. [4 marks]

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SECTION B: ALGEBRA

6. (a) Given that x − 1 is a factor of the polynomial px² + qx + r, and when the polynomial is divided by x + 1 and x − 2 leaves remainders of 2 and 8 respectively. Calculate the values of p, q and r. [10 marks]

(b) Obtain the first four terms of the binomial expansion of (x − y)¹⁰ and use it to estimate the value of (0.99)¹⁰, correct to 3 decimal places. [10 marks]

7. (a) Prove that 2⁴ⁿ − 1 is always divisible by 5. [10 marks]

(page continues — question 7b not visible)

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SECTION B (cont’d) / Page 2

7. (b) Solve the following simultaneous linear equations using Cramer’s rule:

x + 2y − 2z = 1

x − 3y + 8z = 6

2x − 3y − z = −2

[10 marks]

8. (a) Solve the equation log₅(x² + 9) − log₅(x + 1) = 1 [6 marks]

(b) Solve the equation 49ˣ − 42ˣ = 36ˣ. [14 marks]

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SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS

9. (a) Given that z₁ = 1 − 2i, z₂ = 3 + 2i and z₃ = 1 + i, simplify the following:

• (i) z₁z₂z₃

• (ii) z̄₁z₂ − z₂z̄₃

• (iii) z̄₂z₁ / z₁

[8 marks]

(b) Show that if ω is one complex cube root of unity then ω² is also another complex cube root of unity and 1 + ω + ω² = 0. [8 marks]

© Given that z = x + iy, show that |z + i| = |1 + i| is the circle x² + y² + 2y − 1 = 0. [4 marks]

10. (a) Given that tan A = 1/5, tan B = 1/2 and A + B + C = 45°, find the value of tan C. [8 marks]

(b) Find the square roots of −i and 3 + 4i. [8 marks]

© Show that (1 + cos x)/(1 − cos x) = cosec²x + 2cosec x cot x + cot²x [4 marks]

11. (a) Solve the equation sin²x − sin x − cos²x = 0 [8 marks]

(b) Given that sin 2x = 2 sin x cos x, cos 2x = cos²x − sin²x, obtain an expression for sin x, cos x in terms of tan(x/2). [8 marks]

© If cos(x + θ) = sin(x − θ), find the value of tan x in terms of θ and φ. [4 marks]

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SOLUTIONS

SECTION A

Q1. sec θ = 25/7, find cot θ + cos θ

sec θ = 25/7 → cos θ = 7/25

sin²θ = 1 − cos²θ = 1 − 49/625 = 576/625 → sin θ = 24/25

cot θ = cos θ / sin θ = (7/25)/(24/25) = 7/24

$$\cot\theta + \cos\theta = \frac{7}{24} + \frac{7}{25} = \frac{175 + 168}{600} = \boxed{\frac{343}{600}}$$

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Q2. α and β are roots of x² − 5x + 6 = 0, find α − β

By Vieta’s: α + β = 5, αβ = 6

(α − β)² = (α + β)² − 4αβ = 25 − 24 = 1

$$\alpha - \beta = \pm 1$$

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Q3. Express ((√2 + i)/(√2 − 2i))² in the form a + ib

First simplify (√2 + i)/(√2 − 2i). Multiply numerator and denominator by conjugate (√2 + 2i):

Numerator: (√2 + i)(√2 + 2i) = 2 + 2√2 i + √2 i + 2i² = 2 + 3√2 i − 2 = 3√2 i

Denominator: (√2)² + (2)² = 2 + 4 = 6

So the fraction = 3√2 i / 6 = (√2/2)i

Now square it: ((√2/2)i)² = (2/4)(i²) = (1/2)(−1) = −1/2

$$\boxed{a + ib = -\frac{1}{2} + 0i}$$

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Q4. 16th term = 3 × 4th term; prove 23rd term = 5 × 3rd term

Let first term = a, common difference = d.

T₁₆ = a + 15d, T₄ = a + 3d

Given: a + 15d = 3(a + 3d)

→ a + 15d = 3a + 9d

→ 6d = 2a → a = 3d

Now check: T₂₃ = a + 22d = 3d + 22d = 25d

5 × T₃ = 5(a + 2d) = 5(3d + 2d) = 5(5d) = 25d ✓

$$T_{23} = 5 \times T_3 \quad \blacksquare$$

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Q5. Solve 1ˣ = 3

1ˣ = 1 for all real x. Since 1 ≠ 3, there is no real solution. The equation has no solution in ℝ.

(Note: This is likely a misprint in the paper — possibly intended as 13ˣ = 3 or similar. As written, no solution exists.)

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SECTION B: ALGEBRA

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Q6(a). px² + qx + r, with x − 1 as factor; remainders 2 and 8 when divided by x + 1 and x − 2

Let f(x) = px² + qx + r.

Condition 1 (x − 1 is a factor → f(1) = 0):

p + q + r = 0 … (i)

Condition 2 (f(−1) = 2):

p − q + r = 2 … (ii)

Condition 3 (f(2) = 8):

4p + 2q + r = 8 … (iii)

From (i) and (ii):

(ii) − (i): −2q = 2 → q = −1

Sub into (i): p − 1 + r = 0 → p + r = 1 … (iv)

From (iii): 4p + 2(−1) + r = 8 → 4p + r = 10 … (v)

(v) − (iv): 3p = 9 → p = 3

From (iv): r = 1 − 3 = −2

$$\boxed{p = 3,\quad q = -1,\quad r = -2}$$

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Q6(b). First four terms of (x − y)¹⁰; estimate (0.99)¹⁰

Using binomial theorem:

$$(x-y)^{10} = \binom{10}{0}x^{10} - \binom{10}{1}x^9y + \binom{10}{2}x^8y^2 - \binom{10}{3}x^7y^3 + \cdots$$

First four terms:

$$= x^{10} - 10x^9y + 45x^8y^2 - 120x^7y^3 + \cdots$$

Estimating (0.99)¹⁰: Set x = 1, y = 0.01:

$$= 1 - 10(0.01) + 45(0.0001) - 120(0.000001)$$

$$= 1 - 0.1 + 0.0045 - 0.00012$$

$$= \boxed{0.904}$$

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Q7(a). Prove 2⁴ⁿ − 1 is divisible by 5

By induction:

Base case n = 1: 2⁴ − 1 = 16 − 1 = 15 = 5 × 3 ✓

Inductive step: Assume 2⁴ᵏ − 1 = 5m for some integer m.

Then: 2⁴⁽ᵏ⁺¹⁾ − 1 = 2⁴ᵏ⁺⁴ − 1 = 16 · 2⁴ᵏ − 1

= 16(5m + 1) − 1 = 80m + 16 − 1 = 80m + 15 = 5(16m + 3)

This is divisible by 5. ∎

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Q7(b). Simultaneous equations by Cramer’s Rule

$$x + 2y - 2z = 1$$

$$x - 3y + 8z = 6$$

$$2x - 3y - z = -2$$

Coefficient matrix:

$$D = \begin{vmatrix} 1 & 2 & -2 \\ 1 & -3 & 8 \\ 2 & -3 & -1 \end{vmatrix}$$

Expanding along row 1:

= 1[(−3)(−1) − (8)(−3)] − 2[(1)(−1) − (8)(2)] + (−2)[(1)(−3) − (−3)(2)]

= 1[3 + 24] − 2[−1 − 16] + (−2)[−3 + 6]

= 27 − 2(−17) + (−2)(3)

= 27 + 34 − 6 = 55

Dₓ (replace col 1 with constants):

$$D_x = \begin{vmatrix} 1 & 2 & -2 \\ 6 & -3 & 8 \\ -2 & -3 & -1 \end{vmatrix}$$

= 1[(−3)(−1) − (8)(−3)] − 2[(6)(−1) − (8)(−2)] + (−2)[(6)(−3) − (−3)(−2)]

= 1[3 + 24] − 2[−6 + 16] + (−2)[−18 − 6]

= 27 − 20 + 48 = 55

D_y (replace col 2 with constants):

$$D_y = \begin{vmatrix} 1 & 1 & -2 \\ 1 & 6 & 8 \\ 2 & -2 & -1 \end{vmatrix}$$

= 1[(6)(−1) − (8)(−2)] − 1[(1)(−1) − (8)(2)] + (−2)[(1)(−2) − (6)(2)]

= 1[−6 + 16] − 1[−1 − 16] + (−2)[−2 − 12]

= 10 + 17 + 28 = 55

D_z (replace col 3 with constants):

$$D_z = \begin{vmatrix} 1 & 2 & 1 \\ 1 & -3 & 6 \\ 2 & -3 & -2 \end{vmatrix}$$

= 1[(−3)(−2) − (6)(−3)] − 2[(1)(−2) − (6)(2)] + 1[(1)(−3) − (−3)(2)]

= 1[6 + 18] − 2[−2 − 12] + 1[−3 + 6]

= 24 + 28 + 3 = 55

$$x = \frac{D_x}{D} = \frac{55}{55} = 1, \quad y = \frac{D_y}{D} = \frac{55}{55} = 1, \quad z = \frac{D_z}{D} = \frac{55}{55} = 1$$

$$\boxed{x = 1,\quad y = 1,\quad z = 1}$$

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Q8(a). log₅(x² + 9) − log₅(x + 1) = 1

$$\log_5\frac{x^2+9}{x+1} = 1 \implies \frac{x^2+9}{x+1} = 5$$

x² + 9 = 5x + 5 → x² − 5x + 4 = 0 → (x − 1)(x − 4) = 0

$$\boxed{x = 1 \quad \text{or} \quad x = 4}$$

(Both give positive arguments — valid.)

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Q8(b). Solve 49ˣ − 42ˣ = 36ˣ

Divide through by 49ˣ:

$$1 - \left(\frac{42}{49}\right)^x = \left(\frac{36}{49}\right)^x$$

$$1 - \left(\frac{6}{7}\right)^x = \left(\frac{6}{7}\right)^{2x}$$

Let t = (6/7)ˣ:

$$t^2 + t - 1 = 0 \implies t = \frac{-1 + \sqrt{5}}{2}$$

(taking positive root since t > 0)

$$\left(\frac{6}{7}\right)^x = \frac{\sqrt{5}-1}{2}$$

$$x = \frac{\ln\left(\frac{\sqrt{5}-1}{2}\right)}{\ln\left(\frac{6}{7}\right)}$$

$$\boxed{x = \frac{\ln\!\left(\frac{\sqrt5 - 1}{2}\right)}{\ln 6 - \ln 7} \approx \frac{-0.9624}{-0.1542} \approx 6.24}$$

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SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS

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Q9(a). z₁ = 1−2i, z₂ = 3+2i, z₃ = 1+i

(i) z₁z₂z₃

z₁z₂ = (1−2i)(3+2i) = 3 + 2i − 6i − 4i² = 3 − 4i + 4 = 7 − 4i

(7−4i)(1+i) = 7 + 7i − 4i − 4i² = 7 + 3i + 4 = 11 + 3i

(ii) z̄₁z₂ − z₂z̄₃

z̄₁ = 1+2i, z̄₃ = 1−i

z̄₁z₂ = (1+2i)(3+2i) = 3 + 2i + 6i + 4i² = 3 + 8i − 4 = −1 + 8i

z₂z̄₃ = (3+2i)(1−i) = 3 − 3i + 2i − 2i² = 3 − i + 2 = 5 − i

Result: (−1+8i) − (5−i) = −6 + 9i

(iii) z̄₂z₁ / z₁

z̄₂ = 3−2i

z̄₂z₁ = (3−2i)(1−2i) = 3 − 6i − 2i + 4i² = 3 − 8i − 4 = −1 − 8i

Divide by z₁ = 1−2i, multiply by conjugate (1+2i):

$$\frac{(-1-8i)(1+2i)}{(1-2i)(1+2i)} = \frac{-1-2i-8i-16i^2}{1+4} = \frac{-1-10i+16}{5} = \frac{15-10i}{5} = \boxed{3-2i}$$

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Q9(b). Show ω is a cube root of unity → ω² also is, and 1 + ω + ω² = 0

ω is a complex cube root of unity → ω³ = 1 and ω ≠ 1.

From ω³ − 1 = 0: (ω − 1)(ω² + ω + 1) = 0. Since ω ≠ 1:

$$\omega^2 + \omega + 1 = 0 \implies 1 + \omega + \omega^2 = 0 \quad \blacksquare$$

Now show ω² is also a cube root of unity:

(ω²)³ = ω⁶ = (ω³)² = 1² = 1 ✓, and ω² ≠ 1 (since ω ≠ 1 and ω ≠ 0). ∎

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Q9©. Show |z + i| = |1 + i| gives x² + y² + 2y − 1 = 0

z = x + iy, so z + i = x + (y+1)i

|z + i| = √(x² + (y+1)²)

|1 + i| = √(1 + 1) = √2

Setting equal and squaring:

x² + (y+1)² = 2

x² + y² + 2y + 1 = 2

$$\boxed{x^2 + y^2 + 2y - 1 = 0} \quad \blacksquare$$

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Q10(a). tan A = 1/5, tan B = 1/2, A + B + C = 45°; find tan C

C = 45° − A − B, so:

$$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} = \frac{\frac{1}{5}+\frac{1}{2}}{1-\frac{1}{10}} = \frac{\frac{7}{10}}{\frac{9}{10}} = \frac{7}{9}$$

$$\tan C = \tan(45° - (A+B)) = \frac{1 - \frac{7}{9}}{1 + \frac{7}{9}} = \frac{\frac{2}{9}}{\frac{16}{9}} = \boxed{\frac{1}{8}}$$

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Q10(b). Square roots of −i and 3 + 4i

√(−i): Let a + bi = √(−i), so a² − b² = 0 and 2ab = −1.

From a² = b²: a = ±b. With 2ab = −1: 2a(a) = −1 is impossible; use a = −b:

2(−b)(b) = −1 → b² = 1/2 → b = ±1/√2

$$\sqrt{-i} = \pm\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i\right) = \pm\frac{1-i}{\sqrt{2}}$$

√(3 + 4i): Let a + bi = √(3+4i): a² − b² = 3, 2ab = 4 → b = 2/a

a² − 4/a² = 3 → a⁴ − 3a² − 4 = 0 → (a² − 4)(a² + 1) = 0 → a² = 4 → a = ±2

$$\sqrt{3+4i} = \pm(2 + i)$$

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Q10©. Show (1 + cos x)/(1 − cos x) = cosec²x + 2cosec x cot x + cot²x

RHS = (cosec x + cot x)²

$$= \left(\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right)^2 = \left(\frac{1+\cos x}{\sin x}\right)^2 = \frac{(1+\cos x)^2}{\sin^2 x}$$

$$= \frac{(1+\cos x)^2}{1 - \cos^2 x} = \frac{(1+\cos x)^2}{(1-\cos x)(1+\cos x)} = \frac{1+\cos x}{1-\cos x} = \text{LHS} \quad \blacksquare$$

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Q11(a). Solve sin²x − sin x − cos²x = 0

Replace cos²x = 1 − sin²x:

sin²x − sin x − (1 − sin²x) = 0

2sin²x − sin x − 1 = 0

(2sin x + 1)(sin x − 1) = 0

sin x = −1/2 → x = 210°, 330° (i.e. 7π/6, 11π/6)

sin x = 1 → x = 90° (i.e. π/2)

$$\boxed{x = 90°,\ 210°,\ 330°}$$

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Q11(b). Express sin x and cos x in terms of tan(x/2)

Let t = tan(x/2). Using half-angle identities:

$$\sin x = \frac{2t}{1+t^2}, \qquad \cos x = \frac{1-t^2}{1+t^2}$$

Derivation:

sin 2θ = 2 sin θ cos θ → with θ = x/2:

sin x = 2·(t/√(1+t²))·(1/√(1+t²)) = 2t/(1+t²) ✓

cos x = cos²(x/2) − sin²(x/2) = (1−t²)/(1+t²) ✓

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Q11©. cos(x + θ) = sin(x − θ), find tan x in terms of θ

Expand both sides:

cos x cos θ − sin x sin θ = sin x cos θ − cos x sin θ

cos x cos θ + cos x sin θ = sin x cos θ + sin x sin θ

cos x(cos θ + sin θ) = sin x(cos θ + sin θ)

Dividing both sides by cos x(cos θ + sin θ) [assuming cos θ + sin θ ≠ 0]:

$$\tan x = 1$$

$$\boxed{\tan x = 1}$$

(tan x = 1 regardless of θ, provided cos θ + sin θ ≠ 0)