### SECTION A
1. Calculate the gradient of the line joining the points (7, 50) and (−7, 8). [03 marks]
2. Calculate the mean of the distribution f(x) = e⁻³ˣ where 0 ≤ x ≤ ∞. [04 marks]
3. Find the equation of the line through the mid-point of (2,3), (−5,1) and perpendicular to the line 2x + y = 700. [05 marks]
4. Find the equation of the circle whose end points of diameter are (7, −3) and (−1, 2). [04 marks]
5. Find the coordinates of the vertex and focus of the parabola x² = 16 − 4y. [04 marks]
---
### SECTION B: COORDINATE GEOMETRY
6. (a) Obtain the equation of the conic whose parametric equations are given by x = 1 + 2tanθ and y = −1 + 3secθ, and describe it completely. [08 marks]
(b) Sketch the curve r = (1 + sinθ) in polar form for 0 ≤ θ ≤ 360°. [08 marks]
(c) Show that r = 1/(1 + cosθ) is the polar form of a parabola. [04 marks]
7. (a) Show that the equation 9x² + 25y² − 54x − 100y − 44 = 0 represents an Ellipse. Find the coordinates of its center. [08 marks]
(b) Find the equation of a straight line parallel to the line x − 3y = 5 which passes through the mid-point of the line joining the points (3, −2) and (7, 8). [08 marks]
(c) Find the value of 2x + y, if (2,3) is the mid-point of (x, 3) and (−3, y). [04 marks]
---
### SECTION B (continued): COORDINATE GEOMETRY / CALCULUS
8. (a) Determine the condition for the line y = mx to be tangent to the parabola y² = 4a(x − 5). Write down the general equation of the tangent to the parabola. [08 marks]
(b) Let A, B and C be the midpoints of the sides ST, RT and SR of triangle △RST respectively, whose coordinates are R(2, −7), S(4, −5), T(6, 3). Find the coordinates of the point of intersection P of the lines RA and SB. Show that TC passes through point P. [08 marks]
(c) Calculate the slope of the line parallel to the line joining the points (3,7) and (−2, −5). [04 marks]
---
### SECTION C: STATISTICS
9. (a) A biased coin has the probability for a head of 2/3. The coin is tossed six times. Calculate the probability that:
(i) all six trials result in a tail;
(ii) none of the six results in a tail;
(iii) at least three heads appear in the six trails;
(iv) not more than three tails appear;
(v) the first two results in a head, and the remaining four in a tail;
(vi) The heads and tails appear alternatively. (you may not simplify the answer) [14 marks]
(b) Find the Harmonic and Geometric means of 3, 5 and 7. [06 marks]
10. (a) Two dice are tossed once, what is the probability that the:
(i) sum is less than 7; (ii) same number appeared; (iii) difference is 2;
(iv) product is greater than 8; (v) first number is even. [12 marks]
(b) If the diameters of ball bearings are normally distributed with mean 60.5cm and standard deviation 2.5cm, determine the probability of the ball bearing with diameter between 58.5cm and 63.5cm. [08 marks]
11. (a) If f(x) = mx² for 0 ≤ x ≤ 2 is a probability density function for a continuous random variable x, calculate:
(i) the value of m; (ii) p(0 ≤ x ≤ 1); (iii) the mean; (iv) the variance of the distribution. [10 marks]
(b) On an average, two accidents occur every day on a busy highway and it is known that the number of accidents per day follows Poisson law. Find the probability that on a certain day:
(i) there are 2 accidents; (ii) there is no accident; (iii) Find the least value, n such that the probability for n accidents on a particular day is <0.1.
[Approximately take e⁻² = 1/7] [10 marks]
---
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## FULL SOLUTIONS
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### SECTION A
**Question 1**
Gradient = (y₂ − y₁)/(x₂ − x₁) = (8 − 50)/(−7 − 7) = −42/−14 = **3**
---
**Question 2**
Mean = ∫₀^∞ x·f(x)dx = ∫₀^∞ x·e⁻³ˣ dx
Using integration by parts or the gamma formula: ∫₀^∞ x·e⁻³ˣ dx = 1/3²= **1/9**
---
**Question 3**
Mid-point of (2,3) and (−5,1):
M = ((2−5)/2, (3+1)/2) = (−3/2, 2)
Slope of 2x + y = 700 → y = −2x + 700, so slope = −2.
Perpendicular slope = **1/2**
Equation: y − 2 = ½(x − (−3/2))
y − 2 = ½x + 3/4
**y = ½x + 11/4** or equivalently **4y = 2x + 11**
---
**Question 4**
Endpoints of diameter: (7, −3) and (−1, 2).
Centre = ((7−1)/2, (−3+2)/2) = (3, −1/2)
Radius² = (7−3)² + (−3+1/2)² = 16 + 6.25 = 22.25
Using the diameter form:
(x − 7)(x + 1) + (y + 3)(y − 2) = 0
x² − 6x − 7 + y² + y − 6 = 0
**x² + y² − 6x + y − 13 = 0**
---
**Question 5**
x² = 16 − 4y → x² = −4(y − 4)
This is x² = −4(y − 4), parabola opens downward.
- **Vertex: (0, 4)**
- 4p = 4 → p = 1, focus is 1 unit below vertex:
- **Focus: (0, 3)**
---
### SECTION B: COORDINATE GEOMETRY
**Question 6(a)**
Given: x = 1 + 2tanθ, y = −1 + 3secθ
(x − 1) = 2tanθ → tanθ = (x−1)/2
(y + 1) = 3secθ → secθ = (y+1)/3
Using sec²θ − tan²θ = 1:
((y+1)/3)² − ((x−1)/2)² = 1
**(y+1)²/9 − (x−1)²/4 = 1**
This is a **Hyperbola** with center (1, −1), transverse axis along y-axis, a² = 9, b² = 4.
- a = 3, b = 2
- c² = a² + b² = 13, c = √13
- Foci at (1, −1 ± √13)
- Eccentricity e = √13/3
---
**Question 6(b)**
r = 1 + sinθ is a **Cardioid**.
Key points:
| θ | r |
|---|---|
| 0° | 1 |
| 90° | 2 |
| 180° | 1 |
| 270° | 0 |
| 360° | 1 |
The curve is a heart-shaped cardioid symmetric about the 90°–270° axis, passing through the pole at θ = 270°, with maximum r = 2 at θ = 90°.
---
**Question 6(c)**
Given r = 1/(1 + cosθ)
Multiply both sides by (1 + cosθ):
r + r·cosθ = 1
Since x = r·cosθ and r = √(x² + y²):
r = 1 − r·cosθ = 1 − x
√(x² + y²) = 1 − x
x² + y² = (1 − x)² = 1 − 2x + x²
**y² = 1 − 2x = −2(x − 1/2)**
This is the equation of a **parabola** (y² = −2(x − ½)), hence proven. ∎
---
**Question 7(a)**
9x² + 25y² − 54x − 100y − 44 = 0
Complete the square:
9(x² − 6x) + 25(y² − 4y) = 44
9(x−3)² − 81 + 25(y−2)² − 100 = 44
9(x−3)² + 25(y−2)² = 225
Divide by 225:
**(x−3)²/25 + (y−2)²/9 = 1**
This is an **Ellipse** with **center (3, 2)**, a² = 25, b² = 9 (major axis along x-axis).
---
**Question 7(b)**
Line parallel to x − 3y = 5 has slope: x − 3y = 5 → y = x/3 − 5/3, slope = **1/3**
Mid-point of (3, −2) and (7, 8):
M = (5, 3)
Equation: y − 3 = 1/3(x − 5)
3y − 9 = x − 5
**x − 3y + 4 = 0**
---
**Question 7(c)**
Mid-point of (x, 3) and (−3, y) = (2, 3)
(x + (−3))/2 = 2 → x − 3 = 4 → **x = 7**
(3 + y)/2 = 3 → 3 + y = 6 → **y = 3**
**2x + y = 2(7) + 3 = 17**
---
**Question 8(a)**
Line: y = mx. Parabola: y² = 4a(x − 5)
Substitute x = y/m:
y² = 4a(y/m − 5)
my² = 4ay − 20am
my² − 4ay + 20am = 0
For tangency, discriminant = 0:
(4a)² − 4(m)(20am) = 0
16a² − 80am² = 0
16a = 80m²
**a = 5m²** ← condition for tangency
General tangent to y² = 4a(x − 5):
The standard tangent to y² = 4aX (where X = x−5) is Y = mX + a/m:
**y = m(x − 5) + a/m**
---
**Question 8(b)**
R(2,−7), S(4,−5), T(6,3)
A = midpoint of ST = ((4+6)/2, (−5+3)/2) = (5, −1)
B = midpoint of RT = ((2+6)/2, (−7+3)/2) = (4, −2)
C = midpoint of SR = ((4+2)/2, (−5−7)/2) = (3, −6)
**Line RA** (through R(2,−7) and A(5,−1)):
Slope = (−1+7)/(5−2) = 6/3 = 2
y + 7 = 2(x − 2) → **y = 2x − 11** ... (i)
**Line SB** (through S(4,−5) and B(4,−2)):
Both have x = 4 → vertical line **x = 4** ... (ii)
From (ii) into (i): y = 2(4) − 11 = −3
**P = (4, −3)**
**Line TC** (through T(6,3) and C(3,−6)):
Slope = (−6−3)/(3−6) = −9/−3 = 3
y − 3 = 3(x − 6) → y = 3x − 15
Check P(4,−3): y = 3(4) − 15 = −3 ✓
**TC passes through P(4,−3). ∎**
---
**Question 8(c)**
Points: (3,7) and (−2,−5)
Slope = (−5−7)/(−2−3) = −12/−5 = **12/5**
A parallel line has the same slope = **12/5**
---
### SECTION C: STATISTICS
**Question 9(a)**
P(Head) = 2/3, P(Tail) = 1/3, n = 6
**(i) All six = tails:**
P = (1/3)⁶ = **1/729**
**(ii) None is a tail (all heads):**
P = (2/3)⁶ = **64/729**
**(iii) At least 3 heads:**
P(X ≥ 3) = 1 − P(X ≤ 2)
P(X=0) = (1/3)⁶ = 1/729
P(X=1) = C(6,1)(2/3)¹(1/3)⁵ = 6 × 2/243 = 12/729
P(X=2) = C(6,2)(2/3)²(1/3)⁴ = 15 × 4/81 × 1/81...
Let me compute directly:
P(X=2) = 15 × (4/729) × ... = 15 × 4/729 = 60/729
P(X ≤ 2) = (1 + 12 + 60)/729 = 73/729
**P(X ≥ 3) = 1 − 73/729 = 656/729**
**(iv) Not more than 3 tails (T ≤ 3, i.e., H ≥ 3):**
Same as (iii): **P = 656/729**
**(v) First two = heads, remaining four = tails:**
P = (2/3)²×(1/3)⁴ = (4/9)×(1/81) = **4/729**
**(vi) Heads and tails alternately:**
Two patterns: HTHTHT or THTHTH
P(HTHTHT) = (2/3)³(1/3)³ = 8/729
P(THTHTH) = (1/3)³(2/3)³ = 8/729
**P = 16/729**
---
**Question 9(b)**
Numbers: 3, 5, 7
**Harmonic Mean:**
HM = n / (1/3 + 1/5 + 1/7) = 3 / (35+21+15)/105 = 3 × 105/71 = **315/71 ≈ 4.44**
**Geometric Mean:**
GM = (3 × 5 × 7)^(1/3) = 105^(1/3) ≈ **4.72**
---
**Question 10(a)**
Total outcomes = 36
**(i) Sum < 7:**
Pairs: (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1) = **15**
P = **15/36 = 5/12**
**(ii) Same number:**
(1,1),(2,2),(3,3),(4,4),(5,5),(6,6) = 6
P = **6/36 = 1/6**
**(iii) Difference = 2:**
(1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4) = 8
P = **8/36 = 2/9**
**(iv) Product > 8:**
Pairs: (2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,2),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6) = **20**
P = **20/36 = 5/9**
**(v) First number is even:**
First die = 2, 4, or 6 → 18 outcomes
P = **18/36 = 1/2**
---
**Question 10(b)**
μ = 60.5cm, σ = 2.5cm
P(58.5 < X < 63.5)
Z₁ = (58.5 − 60.5)/2.5 = −2/2.5 = **−0.8**
Z₂ = (63.5 − 60.5)/2.5 = 3/2.5 = **1.2**
P(−0.8 < Z < 1.2) = Φ(1.2) − Φ(−0.8)
= 0.8849 − 0.2119
= **0.6730 (67.30%)**
---
**Question 11(a)**
f(x) = mx², 0 ≤ x ≤ 2
**(i) Find m:**
∫₀² mx² dx = 1
m[x³/3]₀² = 1
m(8/3) = 1
**m = 3/8**
**(ii) P(0 ≤ x ≤ 1):**
∫₀¹ (3/8)x² dx = (3/8)[x³/3]₀¹ = (3/8)(1/3) = **1/8**
**(iii) Mean:**
E(X) = ∫₀² x·(3/8)x² dx = (3/8)∫₀² x³ dx = (3/8)[x⁴/4]₀² = (3/8)(4) = **3/2**
**(iv) Variance:**
E(X²) = ∫₀² x²·(3/8)x² dx = (3/8)∫₀² x⁴ dx = (3/8)[x⁵/5]₀² = (3/8)(32/5) = 12/5
Var(X) = E(X²) − [E(X)]² = 12/5 − (3/2)² = 12/5 − 9/4 = 48/20 − 45/20 = **3/20**
---
**Question 11(b)**
λ = 2, e⁻² = 1/7, P(X=k) = e⁻²·2^k/k!
**(i) P(X = 2):**
= e⁻²·2²/2! = (1/7)(4/2) = (1/7)(2) = **2/7**
**(ii) P(X = 0):**
= e⁻²·2⁰/0! = **1/7**
**(iii) Least n such that P(X = n) < 0.1:**
P(X=n) = (1/7)·2ⁿ/n! < 0.1 = 1/10
So 2ⁿ/n! < 7/10
- n=0: 1/1 = 1 > 0.7 ✗
- n=1: 2/1 = 2 > 0.7 ✗
- n=2: 4/2 = 2 > 0.7 ✗
- n=3: 8/6 = 1.33 > 0.7 ✗
- n=4: 16/24 = 0.667 < 0.7 ✓
**Least n = 4**