### 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]
**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 the 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 the 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 = 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 18; (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]
## SOLUTIONS
### SECTION A
---
**Q1. Gradient of line joining (7, 50) and (−7, 8)**
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 50}{-7 - 7} = \frac{-42}{-14} = \boxed{3}$$
---
**Q2. Mean of f(x) = e^(−3x), 0 ≤ x ≤ ∞**
First verify it's a valid PDF:
$$\int_0^\infty e^{-3x}dx = \left[\frac{-1}{3}e^{-3x}\right]_0^\infty = \frac{1}{3}$$
So the proper PDF is f(x) = 3e^(−3x).
$$E(x) = \int_0^\infty x \cdot 3e^{-3x}dx$$
Using integration by parts (or Gamma function: ∫₀^∞ x·e^(−3x)dx = 1/3²):
$$E(x) = 3 \cdot \frac{1}{9} = \boxed{\frac{1}{3}}$$
---
**Q3. Equation of line through mid-point of (2,3) and (−5,1), perpendicular to 2x + y = 700**
Mid-point: $M = \left(\frac{2+(-5)}{2}, \frac{3+1}{2}\right) = \left(-\frac{3}{2}, 2\right)$
Slope of 2x + y = 700 is **−2**.
Perpendicular slope = **1/2**
Equation: $y - 2 = \frac{1}{2}\left(x + \frac{3}{2}\right)$
$$2y - 4 = x + \frac{3}{2}$$
$$4y - 8 = 2x + 3$$
$$\boxed{2x - 4y + 11 = 0}$$
---
**Q4. Equation of circle with diameter endpoints (7, −3) and (−1, 2)**
Center = $\left(\frac{7-1}{2}, \frac{-3+2}{2}\right) = \left(3, -\frac{1}{2}\right)$
Radius² = $\left(\frac{7-(-1)}{2}\right)^2 + \left(\frac{-3-2}{2}\right)^2 = 16 + \frac{25}{4} = \frac{89}{4}$
Equation: $(x-3)^2 + \left(y+\frac{1}{2}\right)^2 = \frac{89}{4}$
Expanding:
$$\boxed{x^2 + y^2 - 6x + y - 15 = 0}$$
---
**Q5. Vertex and Focus of x² = 16 − 4y**
Rewrite: $x^2 = -4(y - 4)$
This is a downward parabola: x² = −4(y − 4)
Comparing with x² = −4p(y − k):
- 4p = 4 → **p = 1**
- **Vertex = (0, 4)**
- Focus is p units below vertex = **(0, 3)**
$$\boxed{\text{Vertex} = (0,4), \quad \text{Focus} = (0,3)}$$
---
### SECTION B
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**Q6(a). Conic from x = 1 + 2tanθ, y = −1 + 3secθ**
From the parametric equations:
$$\frac{x-1}{2} = \tan\theta, \quad \frac{y+1}{3} = \sec\theta$$
Using the identity sec²θ − tan²θ = 1:
$$\left(\frac{y+1}{3}\right)^2 - \left(\frac{x-1}{2}\right)^2 = 1$$
$$\boxed{\frac{(y+1)^2}{9} - \frac{(x-1)^2}{4} = 1}$$
This is a **Hyperbola** with:
- Center: (1, −1)
- Transverse axis vertical
- a² = 9, b² = 4
- c² = 9 + 4 = 13, so c = √13
- Foci: (1, −1 ± √13)
- Asymptotes: y + 1 = ±(3/2)(x − 1)
---
**Q6(b). Sketch r = 1 + sinθ (Cardioid)**
Key values:
| θ | sinθ | r |
|---|------|---|
| 0° | 0 | 1 |
| 90° | 1 | 2 |
| 180° | 0 | 1 |
| 270° | −1 | 0 |
| 360° | 0 | 1 |
This is a **cardioid** opening upward, symmetric about the vertical axis, with maximum r = 2 at θ = 90° and passing through the pole at θ = 270°.
---
**Q6(c). Show r = 1/(1 + cosθ) is a parabola**
Converting to Cartesian: r(1 + cosθ) = 1
$$r + r\cos\theta = 1 \Rightarrow r = 1 - x$$
Since r = √(x² + y²):
$$\sqrt{x^2+y^2} = 1 - x$$
Squaring: $x^2 + y^2 = 1 - 2x + x^2$
$$y^2 = 1 - 2x$$
$$\boxed{y^2 = -2(x - \tfrac{1}{2})}$$
This is a **parabola** (second-degree equation with only y², no x² term). ✓
Also recognizable as the standard polar conic $r = \frac{l}{1 + e\cos\theta}$ with **e = 1** → parabola. ✓
---
**Q7(a). 9x² + 25y² − 54x − 100y − 44 = 0 → Ellipse**
Complete the square:
$$9(x^2 - 6x) + 25(y^2 - 4y) = 44$$
$$9(x-3)^2 - 81 + 25(y-2)^2 - 100 = 44$$
$$9(x-3)^2 + 25(y-2)^2 = 225$$
$$\frac{(x-3)^2}{25} + \frac{(y-2)^2}{9} = 1$$
Since this is of the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ with **two different positive denominators**, it is an **Ellipse**.
$$\boxed{\text{Center} = (3, 2), \quad a=5, \quad b=3}$$
---
**Q7(b). Line parallel to x − 3y = 5 through mid-point of (3,−2) and (7,8)**
Mid-point = $\left(\frac{3+7}{2}, \frac{-2+8}{2}\right) = (5, 3)$
Slope of x − 3y = 5 is **1/3** (parallel lines share same slope)
Equation: $y - 3 = \frac{1}{3}(x - 5)$
$$3y - 9 = x - 5$$
$$\boxed{x - 3y + 4 = 0}$$
---
**Q7(c). Find 2x + y if (2,3) is midpoint of (x,3) and (−3,y)**
$$\frac{x + (-3)}{2} = 2 \Rightarrow x - 3 = 4 \Rightarrow x = 7$$
$$\frac{3 + y}{2} = 3 \Rightarrow 3 + y = 6 \Rightarrow y = 3$$
$$2x + y = 2(7) + 3 = \boxed{17}$$
---
**Q8(a). Condition for y = mx to be tangent to y² = 4a(x − 5)**
Substitute y = mx into y² = 4a(x − 5):
$$m^2x^2 = 4ax - 20a$$
$$m^2x^2 - 4ax + 20a = 0$$
For tangency, discriminant = 0:
$$(-4a)^2 - 4(m^2)(20a) = 0$$
$$16a^2 = 80am^2$$
$$\boxed{a = 5m^2}$$
General tangent to y² = 4a(x − 5):
Substitute x' = x − 5, the parabola is y² = 4ax'. Standard tangent: y = m·x' + a/m → restore x:
$$\boxed{y = m(x-5) + \frac{a}{m}}$$
---
**Q8(b). Midpoints A, B, C of △RST; intersection of RA and SB**
Given: R(2,−7), S(4,−5), T(6,3)
- A = midpoint of ST = $\left(\frac{4+6}{2}, \frac{-5+3}{2}\right) = (5, -1)$
- B = midpoint of RT = $\left(\frac{2+6}{2}, \frac{-7+3}{2}\right) = (4, -2)$
- C = midpoint of SR = $\left(\frac{4+2}{2}, \frac{-5-7}{2}\right) = (3, -6)$
**Line RA** through R(2,−7) and A(5,−1):
$$m = \frac{-1-(-7)}{5-2} = \frac{6}{3} = 2$$
$$y + 7 = 2(x-2) \Rightarrow y = 2x - 11 \quad \cdots (1)$$
**Line SB** through S(4,−5) and B(4,−2):
Both points have x = 4, so **x = 4** ⋯ (2)
**Intersection P**: x = 4, y = 2(4) − 11 = −3
$$\boxed{P = (4, -3)}$$
**Verify TC passes through P:**
Line TC through T(6,3) and C(3,−6):
$$m = \frac{-6-3}{3-6} = \frac{-9}{-3} = 3$$
$$y - 3 = 3(x-6) \Rightarrow y = 3x - 15$$
At x = 4: y = 12 − 15 = **−3** ✓
TC passes through P(4,−3). ✓ (**P is the centroid**)
---
**Q8(c). Slope of line parallel to line joining (3,7) and (−2,−5)**
$$m = \frac{-5-7}{-2-3} = \frac{-12}{-5} = \boxed{\frac{12}{5}}$$
Parallel line has same slope = **12/5**
---
### SECTION C
---
**Q9(a). Biased coin: P(H) = 2/3, P(T) = 1/3, n = 6 tosses**
Using Binomial: P(X = k) = C(6,k)·(2/3)^k·(1/3)^(6−k)
**(i) All six tails [k=0 heads]:**
$$P = \left(\frac{1}{3}\right)^6 = \boxed{\frac{1}{729}}$$
**(ii) None of the six results in a tail [all heads, k=6]:**
$$P = \left(\frac{2}{3}\right)^6 = \boxed{\frac{64}{729}}$$
**(iii) At least three heads [k ≥ 3]:**
$$P = \sum_{k=3}^{6}\binom{6}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{6-k}$$
- k=3: C(6,3)·(8/27)·(1/27) = 20·8/729 = 160/729
- k=4: C(6,4)·(16/81)·(1/9) = 15·16/729 = 240/729
- k=5: C(6,5)·(32/243)·(1/3) = 6·32/729 = 192/729
- k=6: 64/729
$$P = \frac{160+240+192+64}{729} = \boxed{\frac{656}{729}}$$
**(iv) Not more than three tails [T ≤ 3, i.e., H ≥ 3]:**
This is the same as (iii): **P = 656/729**
**(v) First two heads, remaining four tails:**
$$P = \left(\frac{2}{3}\right)^2 \cdot \left(\frac{1}{3}\right)^4 = \frac{4}{9} \cdot \frac{1}{81} = \boxed{\frac{4}{729}}$$
**(vi) Heads and tails appear alternately:**
Two patterns: HTHTHT or THTHTH
$$P(HTHTHT) = \left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^3 = \frac{8}{729}$$
$$P(THTHTH) = \left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^3 = \frac{8}{729}$$
$$P = \frac{8}{729} + \frac{8}{729} = \boxed{\frac{16}{729}}$$
---
**Q9(b). Harmonic Mean and Geometric Mean of 3, 5, 7**
**Harmonic Mean:**
$$HM = \frac{n}{\sum \frac{1}{x_i}} = \frac{3}{\frac{1}{3}+\frac{1}{5}+\frac{1}{7}} = \frac{3}{\frac{35+21+15}{105}} = \frac{3 \times 105}{71} = \boxed{\frac{315}{71} \approx 4.437}$$
**Geometric Mean:**
$$GM = \sqrt[3]{3 \times 5 \times 7} = \sqrt[3]{105} \approx \boxed{4.718}$$
---
**Q10(a). Two dice tossed; total outcomes = 36**
**(i) Sum < 7:**
Favourable: (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**
$$P = \boxed{\frac{15}{36} = \frac{5}{12}}$$
**(ii) Same number (doubles):**
(1,1),(2,2),(3,3),(4,4),(5,5),(6,6) = **6**
$$P = \boxed{\frac{6}{36} = \frac{1}{6}}$$
**(iii) Difference is 2:**
(1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4) = **8**
$$P = \boxed{\frac{8}{36} = \frac{2}{9}}$$
**(iv) Product > 18:**
(4,5),(5,4),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6),(3,6),(6,3) → products: 20,20,24,24,25,30,30,36,18✗,18✗ = **8**
$$P = \boxed{\frac{8}{36} = \frac{2}{9}}$$
**(v) First number is even:**
First die is 2, 4, or 6 → **18 outcomes**
$$P = \boxed{\frac{18}{36} = \frac{1}{2}}$$
---
**Q10(b). Normal distribution: μ = 60.5, σ = 2.5; P(58.5 ≤ X ≤ 63.5)**
$$z_1 = \frac{58.5 - 60.5}{2.5} = \frac{-2}{2.5} = -0.8$$
$$z_2 = \frac{63.5 - 60.5}{2.5} = \frac{3}{2.5} = 1.2$$
$$P(-0.8 \leq Z \leq 1.2) = \Phi(1.2) - \Phi(-0.8)$$
$$= 0.8849 - 0.2119 = \boxed{0.6730}$$
About **67.3%** of ball bearings fall in this range.
---
**Q11(a). f(x) = mx², 0 ≤ x ≤ 2**
**(i) Find m:**
$$\int_0^2 mx^2 dx = 1 \Rightarrow m\left[\frac{x^3}{3}\right]_0^2 = 1 \Rightarrow m \cdot \frac{8}{3} = 1 \Rightarrow \boxed{m = \frac{3}{8}}$$
**(ii) P(0 ≤ x ≤ 1):**
$$\int_0^1 \frac{3}{8}x^2 dx = \frac{3}{8}\cdot\frac{1}{3} = \boxed{\frac{1}{8}}$$
**(iii) Mean E(X):**
$$E(X) = \int_0^2 x \cdot \frac{3}{8}x^2 dx = \frac{3}{8}\int_0^2 x^3 dx = \frac{3}{8}\cdot 4 = \boxed{\frac{3}{2}}$$
**(iv) Variance:**
$$E(X^2) = \int_0^2 \frac{3}{8}x^4 dx = \frac{3}{8}\cdot\frac{32}{5} = \frac{12}{5}$$
$$Var(X) = E(X^2) - [E(X)]^2 = \frac{12}{5} - \frac{9}{4} = \frac{48-45}{20} = \boxed{\frac{3}{20}}$$
---
**Q11(b). Poisson distribution: λ = 2, e⁻² = 1/7**
Formula: $P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}$
**(i) P(X = 2):**
$$P = \frac{e^{-2} \cdot 4}{2} = \frac{(1/7)\cdot 4}{2} = \frac{4}{14} = \boxed{\frac{2}{7}}$$
**(ii) P(X = 0):**
$$P = e^{-2} = \boxed{\frac{1}{7}}$$
**(iii) Least n such that P(X = n) < 0.1:**
$$P(X=n) = \frac{e^{-2}\cdot 2^n}{n!} = \frac{2^n}{7 \cdot n!} < 0.1$$
Test values:
- n=0: 1/7 ≈ 0.143 ✗
- n=1: 2/7 ≈ 0.286 ✗
- n=2: 4/14 ≈ 0.286 ✗
- n=3: 8/42 ≈ 0.190 ✗
- n=4: 16/168 ≈ 0.095 ✓
$$\boxed{n = 4}$$