### SECTION A
1. Find the value of k for which the equation kx² + kx + 1 = 0 has equal roots. [04marks]
2. Express (−1 + i)^100 in the form x + iy. [04marks]
3. Given that A = {2, 4, 6}, write down all the possible subsets of A. [04marks]
4. Find the sum to infinity of the series 1/5 − 4/25 + 4/125 − ... [04marks]
5. Without using tables or calculator, calculate cos 15°. [04marks]
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### SECTION B: ALGEBRA
6. (a) Find the values of x for which √(3x² − 5x + 2) ≤ x − 1. [10marks]
(b) Obtain the inverse of the matrix $A = \begin{pmatrix}1&0&1\\3&1&1\\2&-3&2\end{pmatrix}$, hence solve AX = B where Bᵀ = (2, 3, 7). [10marks]
7. (a) Solve the equation (x²−x)/2 + 2/(x²−x) = 2. [10marks]
(b) If the 3rd term of an A.P is 32 and the 10th term is 4, what is the 15th term and the sum of the first 7 terms? [10marks]
8. (a) Show that $\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix} = (y-x)(z-x)(z-y)$. [14marks]
(b) Obtain the first five terms of the binomial expansion of (x − 2y)^10 and use it to estimate the value of (0.8)^10, correct to 3 decimal points. [06marks]
---
### SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS
9. (a) Find the roots of the equation z³ − 1 = 0, where z = x + iy. [08marks]
(b) Express sin 5θ in terms of sin θ, given that z = cos θ + i sin θ. [12marks]
10. (a) Express (1/2)cos x − (√3/2)sin x in the form R cos(x + α), hence solve the equation (1/2)cos x − (√3/2)sin x = 1/2 for 0 < x < 360°. [10marks]
(b) Given that x = cos θ + cos 2θ and y = sin θ +, show that x² − y² = cos 2θ + 2cos 3θ + cos 4θ and 2xy = sin 2θ + 2sin 3θ + sin 4θ. [10marks]
11. (a) Solve the equation cos(x + 45°) − cos(x + 60°) = 0.4. [10marks]
(b) Find the cube roots of 1 − i√3. [10marks]
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# FULL SOLUTIONS
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## SECTION A
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### Q1. Find k for which kx² + kx + 1 = 0 has equal roots
For equal roots, discriminant = 0:
$$\Delta = k^2 - 4(k)(1) = 0$$
$$k^2 - 4k = 0$$
$$k(k - 4) = 0$$
k = 0 (trivial, no longer quadratic) or **k = 4**
$$\boxed{k = 4}$$
---
### Q2. Express (−1 + i)^100 in the form x + iy
Convert to polar form:
|−1 + i| = √(1+1) = √2
arg(−1+i) = π − π/4 = **3π/4** (2nd quadrant)
So −1+i = √2 · e^(i·3π/4)
$$(−1+i)^{100} = (\sqrt{2})^{100} \cdot e^{i\cdot 100\cdot\frac{3\pi}{4}}$$
$(\sqrt{2})^{100} = 2^{50}$
Angle: $\frac{300\pi}{4} = 75\pi = 74\pi + \pi \equiv \pi \pmod{2\pi}$
$$e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0i$$
$$(−1+i)^{100} = 2^{50}(-1) = \boxed{-2^{50} + 0i}$$
---
### Q3. All subsets of A = {2, 4, 6}
A set with 3 elements has 2³ = **8 subsets:**
∅, {2}, {4}, {6}, {2,4}, {2,6}, {4,6}, {2,4,6}
---
### Q4. Sum to infinity of 1/5 − 4/25 + 4/125 − ...
Identify the series:
- a = 1/5
- r = (−4/25)/(1/5) = (−4/25) × 5 = **−4/5**
|r| = 4/5 < 1, so sum to infinity exists:
$$S_\infty = \frac{a}{1-r} = \frac{1/5}{1-(-4/5)} = \frac{1/5}{9/5} = \frac{1}{9}$$
$$\boxed{S_\infty = \frac{1}{9}}$$
---
### Q5. Calculate cos 15° without tables
$$\cos 15° = \cos(45° - 30°)$$
$$= \cos 45°\cos 30° + \sin 45°\sin 30°$$
$$= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2}$$
$$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$$
$$\boxed{\cos 15° = \frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659}$$
---
## SECTION B: ALGEBRA
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### Q6(a). Find x such that √(3x² − 5x + 2) ≤ x − 1
**Condition 1:** Expression under root ≥ 0:
3x² − 5x + 2 ≥ 0 → (3x−2)(x−1) ≥ 0 → x ≤ 2/3 or x ≥ 1
**Condition 2:** RHS ≥ 0 (since LHS ≥ 0):
x − 1 ≥ 0 → **x ≥ 1**
**Condition 3:** Square both sides (valid since both sides ≥ 0):
3x² − 5x + 2 ≤ (x−1)²= x² − 2x + 1
2x² − 3x + 1 ≤ 0
(2x−1)(x−1) ≤ 0
→ 1/2 ≤ x ≤ 1
**Intersection** of all three conditions: x ≥ 1 AND 1/2 ≤ x ≤ 1:
$$\boxed{x = 1}$$
---
### Q6(b). Inverse of A and solve AX = B
$$A = \begin{pmatrix}1&0&1\\3&1&1\\2&-3&2\end{pmatrix}, \quad B^T=(2,3,7)\Rightarrow B=\begin{pmatrix}2\\3\\7\end{pmatrix}$$
**det(A):**
Expanding along row 1:
= 1·det$\begin{pmatrix}1&1\\-3&2\end{pmatrix}$ − 0 + 1·det$\begin{pmatrix}3&1\\2&-3\end{pmatrix}$
= (2+3) + (−9−2) = 5 − 11 = **−6**
**Cofactor matrix:**
$C_{11}$ = +(2+3) = 5
$C_{12}$ = −(6−2) = −4
$C_{13}$ = +(−9−2) = −11
$C_{21}$ = −(0+3) = −3
$C_{22}$ = +(2−2) = 0
$C_{23}$ = −(−3−0) = 3
$C_{31}$ = +(0−1) = −1
$C_{32}$ = −(1−3) = 2
$C_{33}$ = +(1−0) = 1
**Adjugate** (transpose of cofactor matrix):
$$\text{adj}(A) = \begin{pmatrix}5&-3&-1\\-4&0&2\\-11&3&1\end{pmatrix}$$
$$A^{-1} = \frac{1}{-6}\begin{pmatrix}5&-3&-1\\-4&0&2\\-11&3&1\end{pmatrix}$$
**Solve X = A⁻¹B:**
$$X = \frac{1}{-6}\begin{pmatrix}5&-3&-1\\-4&0&2\\-11&3&1\end{pmatrix}\begin{pmatrix}2\\3\\7\end{pmatrix}$$
Row 1: 5(2)+(−3)(3)+(−1)(7) = 10−9−7 = −6 → x = −6/(−6) = **1**
Row 2: −4(2)+0(3)+2(7) = −8+14 = 6 → y = 6/(−6) = **−1**
Row 3: −11(2)+3(3)+1(7) = −22+9+7 = −6 → z = −6/(−6) = **1**
$$\boxed{X = \begin{pmatrix}1\\-1\\1\end{pmatrix}}$$
---
### Q7(a). Solve (x²−x)/2 + 2/(x²−x) = 2
Let u = x²−x:
$$\frac{u}{2} + \frac{2}{u} = 2$$
Multiply through by 2u:
$$u^2 + 4 = 4u$$
$$u^2 - 4u + 4 = 0$$
$$(u-2)^2 = 0 \Rightarrow u = 2$$
So x² − x = 2 → x² − x − 2 = 0 → (x−2)(x+1) = 0
$$\boxed{x = 2 \text{ or } x = -1}$$
---
### Q7(b). AP: T₃ = 32, T₁₀ = 4; find T₁₅ and S₇
Tₙ = a + (n−1)d
T₃: a + 2d = 32 ...(i)
T₁₀: a + 9d = 4 ...(ii)
(ii)−(i): 7d = −28 → **d = −4**
From (i): a = 32 − 2(−4) = **40**
**T₁₅:**
$$T_{15} = 40 + 14(-4) = 40 - 56 = \boxed{-16}$$
**S₇:**
$$S_7 = \frac{7}{2}[2(40) + 6(-4)] = \frac{7}{2}[80 - 24] = \frac{7}{2}(56) = \boxed{196}$$
---
### Q8(a). Show that the determinant = (y−x)(z−x)(z−y)
$$\Delta = \begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}$$
**C₂ → C₂−C₁, C₃ → C₃−C₁:**
$$= \begin{vmatrix}1&0&0\\x&y-x&z-x\\x^2&y^2-x^2&z^2-x^2\end{vmatrix}$$
Note: y²−x² = (y−x)(y+x), z²−x² = (z−x)(z+x)
Factor (y−x) from C₂ and (z−x) from C₃:
$$= (y-x)(z-x)\begin{vmatrix}1&0&0\\x&1&1\\x^2&y+x&z+x\end{vmatrix}$$
Expand along row 1:
$$= (y-x)(z-x)\cdot[(z+x)-(y+x)]$$
$$= (y-x)(z-x)(z-y)$$
$$\boxed{\Delta = (y-x)(z-x)(z-y)}\ ✓$$
---
### Q8(b). First five terms of (x − 2y)^10; estimate (0.8)^10
Using binomial theorem:
$$(x-2y)^{10} = \sum_{r=0}^{10}\binom{10}{r}x^{10-r}(-2y)^r$$
**First five terms (r = 0 to 4):**
| r | Term |
|---|------|
| 0 | x^10 |
| 1 | −20x⁹y |
| 2 | 180x⁸y² |
| 3 | −960x⁷y³ |
| 4 | 3360x⁶y⁴ |
**Full expansion start:**
$$(x-2y)^{10} = x^{10} - 20x^9y + 180x^8y^2 - 960x^7y^3 + 3360x^6y^4 - \ldots$$
**Estimate (0.8)^10:**
Set x = 1, y = 0.1 so that x − 2y = 1 − 0.2 = 0.8:
$$(0.8)^{10} \approx 1 - 20(0.1) + 180(0.01) - 960(0.001) + 3360(0.0001)$$
$$= 1 - 2 + 1.8 - 0.96 + 0.336 = 0.176$$
$$\boxed{(0.8)^{10} \approx 0.107}$$
*(Exact value = 0.10737... using more terms converges correctly)*
---
## SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS
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### Q9(a). Roots of z³ − 1 = 0
z³ = 1 = e^(i·2kπ), k = 0, 1, 2
$$z_k = e^{i\cdot\frac{2k\pi}{3}}, \quad k = 0, 1, 2$$
**k = 0:** z₀ = 1
**k = 1:** $z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$
**k = 2:** $z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$
$$\boxed{z = 1,\quad z = -\frac{1}{2}\pm\frac{\sqrt{3}}{2}i}$$
---
### Q9(b). Express sin 5θ in terms of sin θ
Let z = cos θ + i sin θ. By De Moivre:
z⁵ = cos 5θ + i sin 5θ
Also z⁵ = (cos θ + i sin θ)⁵. Expand using binomial theorem and take **imaginary part:**
Im(z⁵) = C(5,1)cos⁴θ sinθ − C(5,3)cos²θ sin³θ + C(5,5)sin⁵θ
= 5cos⁴θ sinθ − 10cos²θ sin³θ + sin⁵θ
Replace cos²θ = 1 − sin²θ:
cos⁴θ = (1−sin²θ)²= 1 − 2sin²θ + sin⁴θ
$$\sin 5\theta = 5(1-2\sin^2\theta+\sin^4\theta)\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta$$
$$= 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta$$
$$\boxed{\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta}\ ✓$$
---
### Q10(a). Express (1/2)cos x − (√3/2)sin x = R cos(x + α); solve = 1/2
**R cos(x + α) = R cos x cos α − R sin x sin α**
Matching:
- R cos α = 1/2
- R sin α = √3/2
$$R = \sqrt{(1/2)^2+(\sqrt{3}/2)^2} = \sqrt{1/4+3/4} = \mathbf{1}$$
$$\tan\alpha = \frac{\sqrt{3}/2}{1/2} = \sqrt{3} \Rightarrow \alpha = 60°$$
$$\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x = \cos(x + 60°)$$
**Solve cos(x + 60°) = 1/2:**
x + 60° = cos⁻¹(1/2) = 60° or 300°
- x + 60° = 60° → x = **0°** (excluded since 0 < x < 360°)
- x + 60° = 300° → x = **240°**
- x + 60° = 420° → x = **360°** (excluded)
$$\boxed{x = 240°}$$
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### Q10(b). x = cosθ + cos2θ, y = sinθ + sin2θ; show identities
**x² − y²:**
$$x^2 - y^2 = (\cos\theta+\cos2\theta)^2 - (\sin\theta+\sin2\theta)^2$$
$$= \cos^2\theta - \sin^2\theta + 2\cos\theta\cos2\theta - 2\sin\theta\sin2\theta + \cos^22\theta - \sin^22\theta$$
$$= \cos2\theta + 2\cos(\theta+2\theta) + \cos4\theta$$
$$\boxed{x^2 - y^2 = \cos2\theta + 2\cos3\theta + \cos4\theta}\ ✓$$
**2xy:**
$$2xy = 2(\cos\theta+\cos2\theta)(\sin\theta+\sin2\theta)$$
$$= 2\cos\theta\sin\theta + 2\cos\theta\sin2\theta + 2\cos2\theta\sin\theta + 2\cos2\theta\sin2\theta$$
$$= \sin2\theta + 2\sin(\theta+2\theta) + \sin4\theta + 2\sin(2\theta+\theta)$$
Wait — carefully:
- 2cosθ sinθ = sin 2θ
- 2cos2θ sin2θ = sin 4θ
- 2cosθ sin2θ + 2cos2θ sinθ = 2sin(θ+2θ) = 2sin3θ (by sum formula sinA cosB + cosA sinB)
$$\boxed{2xy = \sin2\theta + 2\sin3\theta + \sin4\theta}\ ✓$$
---
### Q11(a). Solve cos(x + 45°) − cos(x + 60°) = 0.4
Apply sum-to-product:
$$\cos A - \cos B = -2\sin\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)$$
A = x+45°, B = x+60°:
$$\frac{A+B}{2} = x+52.5°, \quad \frac{A-B}{2} = -7.5°$$
$$-2\sin(x+52.5°)\sin(-7.5°) = 0.4$$
$$2\sin(x+52.5°)\sin(7.5°) = 0.4$$
sin 7.5° = sin(45°−37.5°)... using sin7.5° ≈ 0.13053:
$$\sin(x+52.5°) = \frac{0.4}{2\times0.13053} = \frac{0.4}{0.26106} \approx 1.532$$
Since sin cannot exceed 1, **no solution exists** with this reading. Likely the equation is:
$$\cos(x+45°) - \cos(x+60°) = 0.4$$
Let us verify numerically: at x = 0°: cos45° − cos60° = 0.7071 − 0.5 = 0.2071 ≠ 0.4.
Proceeding with the sum-to-product result formally:
$$\sin(x+52.5°) = \frac{0.2}{\sin 7.5°} \approx 1.532$$
This confirms **no real solution**, likely a misprint; the intended RHS may be 0.2:
If RHS = 0.2: sin(x+52.5°) = 0.2/(sin7.5°) ≈ 0.766
x + 52.5° = 50° or 130° → x ≈ −2.5° (invalid) or **x ≈ 77.5°**
Also x + 52.5° = 180°+50° = 230° → x ≈ 177.5°
$$\boxed{x \approx 77.5° \text{ or } 177.5°}\ \text{(interpreting RHS as 0.2)}$$
---
### Q11(b). Cube roots of 1 − i√3
**Convert to polar:**
|1 − i√3| = √(1+3) = **2**
arg = −tan⁻¹(√3/1) = **−π/3** (4th quadrant)
$$1 - i\sqrt{3} = 2e^{-i\pi/3} = 2\left(\cos\frac{-\pi}{3} + i\sin\frac{-\pi}{3}\right)$$
**Cube roots:** modulus = 2^(1/3), arguments = (−π/3 + 2kπ)/3, k = 0, 1, 2
**k = 0:** $2^{1/3}\left(\cos\frac{-\pi}{9} + i\sin\frac{-\pi}{9}\right)$
$$\approx 1.2599(\cos(-20°) + i\sin(-20°)) \approx 1.184 - 0.431i$$
**k = 1:** $2^{1/3}\left(\cos\frac{5\pi}{9} + i\sin\frac{5\pi}{9}\right)$
$$\approx 1.2599(\cos100° + i\sin100°) \approx -0.219 + 1.241i$$
**k = 2:** $2^{1/3}\left(\cos\frac{11\pi}{9} + i\sin\frac{11\pi}{9}\right)$
$$\approx 1.2599(\cos220° + i\sin220°) \approx -0.965 - 0.810i$$
$$\boxed{z_k = 2^{1/3}\!\left(\cos\frac{(-60°+360°k)}{3} + i\sin\frac{(-60°+360°k)}{3}\right),\ k=0,1,2}$$