### SECTION A
1. Express r cos(θ + π/6) = 1 in rectangular coordinates. [04marks]
2. Find the eccentricity and foci of the hyperbola x²/36 − y²/16 = 1. [04marks]
3. Find the equation of the circle whose centre is at (−2, −5) and radius 5. [04marks]
4. Show that f(x) = 2e^(−2x), 0 ≤ x < ∞ is a probability density function. [04marks]
5. Find the equation of the perpendicular bisector of the line joining the points (7, 3) and (−5, 7). [04marks]
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### SECTION B: COORDINATE GEOMETRY
6. (a) Obtain the equation of the conic whose parametric equations are given by x = −2 + 3t² and y = −1 + 5t, and describe it completely. [10marks]
(b) Sketch the curve r = 5(1 − cos 2θ) in polar form for 0 ≤ θ ≤ 360°. [10marks]
7. (a) Obtain the equation of the conic whose parametric equations are given by x = 3 + 5 sin θ and y = 2 + 3 cos θ, and describe it completely. [10marks]
(b) Find the equation of a straight line perpendicular to the line 2x + 3y = 3 which passes through the point of intersection of the lines 3x − 5y = −2 and 2x − 3y = −1. [10marks]
8. (a) Determine the condition for the line y = mx to be tangent to the hyperbola x²/a² − y²/b² = 1. Write down the general equation of the tangent to the hyperbola. [10marks]
(b) Let D be the midpoint of side BC of a triangle A(−1, 5), B(3, −5), C(7, 3). Find the coordinates of the point G on AD which divides AD in the ratio 2:1. Show that the medians of the triangle are concurrent. [10marks]
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### SECTION C: STATISTICS
9. (a) Show that f(x) = (1/2)sin x, 0 ≤ x ≤ π represents a probability density function for a random variable x. Calculate:
- (i) P(x ≤ π/4)
- (ii) P(x ≥ π/4)
- (iii) find c such that P(x ≤ c) = 1/2
[10marks]
(b) For a Binomial distribution with 5 trials and p = 0.25, find the probability that:
- (i) all five trials are successful
- (ii) only four trials are successful
- (iii) at least three trials are successful
- (iv) none of the trials is successful
[10marks]
10. (a) A random variable has the distribution given by the probability function f(x) = kx², x = 1, 2, 3, 4. Calculate the value of k. For this value of k, find the mean and variance of the distribution. [10marks]
(b) If X ≈ P₀(3), find:
- (i) P(x ≥ 2)
- (ii) P(x = 2)
- (iii) P(x ≤ 2)
[10marks]
11. Given the distribution below:
| Marks | 10-13 | 14-17 | 18-21 | 22-25 | 26-29 | 30-33 |
|-------|-------|-------|-------|-------|-------|-------|
| Frequency | 7 | 9 | 5 | 3 | 6 | 10 |
Calculate:
- (i) The median
- (ii) The mode
- (iii) The mean
- (iv) The standard deviation of the distribution
[20marks]
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# FULL SOLUTIONS
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## SECTION A
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### Q1. Express r cos(θ + π/6) = 1 in rectangular coordinates
Expand using cos(A+B):
$$r\left[\cos\theta\cos\frac{\pi}{6} - \sin\theta\sin\frac{\pi}{6}\right] = 1$$
$$r\cos\theta\cdot\frac{\sqrt{3}}{2} - r\sin\theta\cdot\frac{1}{2} = 1$$
Since x = r cosθ, y = r sinθ:
$$\frac{\sqrt{3}}{2}x - \frac{1}{2}y = 1$$
$$\boxed{\sqrt{3}\,x - y = 2}$$
---
### Q2. Eccentricity and foci of x²/36 − y²/16 = 1
This is a hyperbola with a² = 36, b² = 16 → a = 6, b = 4
$$c^2 = a^2 + b^2 = 36 + 16 = 52 \Rightarrow c = 2\sqrt{13}$$
$$\text{Eccentricity: } e = \frac{c}{a} = \frac{2\sqrt{13}}{6} = \boxed{\frac{\sqrt{13}}{3} \approx 1.202}$$
$$\text{Foci: } (\pm 2\sqrt{13},\ 0) \approx (\pm 7.21,\ 0)$$
---
### Q3. Circle with centre (−2, −5), radius 5
$$(x+2)^2 + (y+5)^2 = 25$$
Expanding:
$$\boxed{x^2 + y^2 + 4x + 10y + 4 = 0}$$
---
### Q4. Show f(x) = 2e^(−2x), 0 ≤ x < ∞ is a PDF
Two conditions required: f(x) ≥ 0 ✓ (exponential is always positive)
$$\int_0^{\infty} 2e^{-2x}\,dx = 2\left[-\frac{1}{2}e^{-2x}\right]_0^{\infty} = \left[-e^{-2x}\right]_0^{\infty}$$
$$= 0 - (-1) = \mathbf{1}\ ✓$$
Since f(x) ≥ 0 and the total integral = 1, **f(x) is a valid PDF.** ✓
---
### Q5. Perpendicular bisector of line joining (7, 3) and (−5, 7)
**Midpoint:**
$$M = \left(\frac{7-5}{2},\ \frac{3+7}{2}\right) = (1,\ 5)$$
**Slope of joining line:**
$$m = \frac{7-3}{-5-7} = \frac{4}{-12} = -\frac{1}{3}$$
**Slope of perpendicular bisector** = 3
**Equation through M(1, 5):**
$$y - 5 = 3(x - 1)$$
$$\boxed{y = 3x + 2}$$
---
## SECTION B: COORDINATE GEOMETRY
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### Q6(a). Parametric: x = −2 + 3t², y = −1 + 5t
From y equation: t = (y+1)/5
Substitute into x:
$$x = -2 + 3\left(\frac{y+1}{5}\right)^2$$
$$x + 2 = \frac{3(y+1)^2}{25}$$
$$\boxed{(y+1)^2 = \frac{25(x+2)}{3}}$$
This is a **parabola** with vertex at (−2, −1), opening to the right, axis parallel to the x-axis.
- Comparing with Y² = 4aX: 4a = 25/3 → a = 25/12
- Focus at (−2 + 25/12, −1) = (1/12, −1)
- Directrix: x = −2 − 25/12 = −49/12
---
### Q6(b). Sketch r = 5(1 − cos 2θ) for 0 ≤ θ ≤ 360°
Key values table:
| θ | cos 2θ | r = 5(1−cos2θ) |
|---|--------|----------------|
| 0° | 1 | 0 |
| 45° | 0 | 5 |
| 90° | −1 | 10 |
| 135° | 0 | 5 |
| 180° | 1 | 0 |
| 225° | 0 | 5 |
| 270° | −1 | 10 |
| 315° | 0 | 5 |
| 360° | 1 | 0 |
This traces a **four-petalled rose-like curve** (lemniscate variant) — r ranges from 0 to 10 and is symmetric about both axes. Maximum r = 10 at θ = 90° and 270°; r = 0 at θ = 0°, 180°, 360°.
---
### Q7(a). Parametric: x = 3 + 5 sinθ, y = 2 + 3 cosθ
$$\frac{x-3}{5} = \sin\theta, \quad \frac{y-2}{3} = \cos\theta$$
Using sin²θ + cos²θ = 1:
$$\boxed{\frac{(x-3)^2}{25} + \frac{(y-2)^2}{9} = 1}$$
This is an **ellipse** with:
- Centre: (3, 2)
- Semi-major axis: a = 5 (along x-direction)
- Semi-minor axis: b = 3 (along y-direction)
- c = √(25−9) = √16 = 4
- Foci: (3±4, 2) = **(7, 2) and (−1, 2)**
- Eccentricity: e = 4/5 = 0.8
- Vertices: (8, 2), (−2, 2), (3, 5), (3, −1)
---
### Q7(b). Line ⊥ to 2x + 3y = 3 through intersection of 3x − 5y = −2 and 2x − 3y = −1
**Step 1: Find intersection**
3x − 5y = −2 ...(i)
2x − 3y = −1 ...(ii)
(i)×3: 9x − 15y = −6
(ii)×5: 10x − 15y = −5
Subtract: −x = −1 → **x = 1**
From (ii): 2 − 3y = −1 → y = 1
Intersection: **(1, 1)**
**Step 2: Slope of required line**
2x + 3y = 3 has slope −2/3
Perpendicular slope = **3/2**
**Step 3: Equation through (1,1):**
$$y - 1 = \frac{3}{2}(x-1)$$
$$2y - 2 = 3x - 3$$
$$\boxed{3x - 2y - 1 = 0}$$
---
### Q8(a). Condition for y = mx to be tangent to x²/a² − y²/b² = 1
Substitute y = mx into the hyperbola:
$$\frac{x^2}{a^2} - \frac{m^2x^2}{b^2} = 1$$
$$x^2\left(\frac{b^2 - a^2m^2}{a^2b^2}\right) = 1$$
For tangency, this must have exactly one solution, which requires the coefficient to produce a repeated root. For real tangency we need:
$$b^2 - a^2m^2 \neq 0 \text{ (for intersection to exist)}$$
The line y = mx + c is tangent when c² = a²m² − b².
For y = mx (i.e., c = 0): **0 = a²m² − b²**, giving:
$$\boxed{m = \pm\frac{b}{a}}$$
These are the **asymptotes** of the hyperbola. Any other line y = mx + c is tangent when:
$$\boxed{c^2 = a^2m^2 - b^2}$$
**General tangent equation:**
$$y = mx \pm\sqrt{a^2m^2 - b^2}$$
---
### Q8(b). Triangle A(−1,5), B(3,−5), C(7,3); D = midpoint BC; G divides AD in 2:1
**D = midpoint of BC:**
$$D = \left(\frac{3+7}{2},\ \frac{-5+3}{2}\right) = (5,\ -1)$$
**G divides AD in ratio 2:1** (from A):
$$G = \left(\frac{2(5)+1(-1)}{3},\ \frac{2(-1)+1(5)}{3}\right) = \left(\frac{9}{3},\ \frac{3}{3}\right) = (3,\ 1)$$
**Show medians are concurrent (all pass through G):**
Median from B: midpoint of AC = $\left(\frac{-1+7}{2}, \frac{5+3}{2}\right) = (3, 4)$
Line from B(3,−5) to midpoint E(3, 4): this is the vertical line x = 3. G = (3,1) lies on x = 3 ✓
Median from C: midpoint of AB = $\left(\frac{-1+3}{2}, \frac{5-5}{2}\right) = (1, 0)$
Line from C(7,3) to F(1,0): slope = (3−0)/(7−1) = 1/2
Equation: y − 0 = (1/2)(x−1) → y = (x−1)/2
At x = 3: y = 1 ✓ G(3,1) lies on this median.
**All three medians pass through G(3,1) — they are concurrent (centroid).** ✓
---
## SECTION C: STATISTICS
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### Q9(a). f(x) = (1/2)sinx, 0 ≤ x ≤ π is a PDF
**Verify:**
$$\int_0^{\pi}\frac{1}{2}\sin x\,dx = \frac{1}{2}[-\cos x]_0^{\pi} = \frac{1}{2}[-\cos\pi + \cos 0] = \frac{1}{2}[1+1] = 1\ ✓$$
f(x) = (1/2)sinx ≥ 0 on [0,π] ✓ — **confirmed PDF.**
**(i) P(x ≤ π/4):**
$$= \int_0^{\pi/4}\frac{1}{2}\sin x\,dx = \frac{1}{2}[-\cos x]_0^{\pi/4} = \frac{1}{2}\left(-\frac{\sqrt{2}}{2}+1\right) = \frac{1}{2}\left(1-\frac{\sqrt{2}}{2}\right)$$
$$\boxed{= \frac{2-\sqrt{2}}{4} \approx 0.1464}$$
**(ii) P(x ≥ π/4):**
$$= 1 - P(x \leq \pi/4) = 1 - \frac{2-\sqrt{2}}{4} = \frac{2+\sqrt{2}}{4} \approx \boxed{0.8536}$$
**(iii) Find c such that P(x ≤ c) = 1/2:**
$$\int_0^c\frac{1}{2}\sin x\,dx = \frac{1}{2}[-\cos x]_0^c = \frac{1}{2}(1-\cos c) = \frac{1}{2}$$
$$1 - \cos c = 1 \Rightarrow \cos c = 0 \Rightarrow \boxed{c = \frac{\pi}{2}}$$
---
### Q9(b). Binomial: n = 5, p = 0.25, q = 0.75
P(X = r) = C(5,r)(0.25)^r(0.75)^(5−r)
**(i) P(all 5 successful) = P(X=5):**
$$= (0.25)^5 = \boxed{0.000977}$$
**(ii) P(exactly 4) = P(X=4):**
$$= \binom{5}{4}(0.25)^4(0.75)^1 = 5 \times 0.003906 \times 0.75 = \boxed{0.01465}$$
**(iii) P(at least 3) = P(X≥3):**
P(X=3) = C(5,3)(0.25)³(0.75)² = 10 × 0.015625 × 0.5625 = 0.08789
P(X=4) = 0.01465 (above)
P(X=5) = 0.000977 (above)
$$P(X\geq3) = 0.08789 + 0.01465 + 0.000977 = \boxed{0.10352}$$
**(iv) P(none successful) = P(X=0):**
$$= (0.75)^5 = \boxed{0.2373}$$
---
### Q10(a). f(x) = kx², x = 1, 2, 3, 4
**Find k:** Σf(x) = 1
$$k(1 + 4 + 9 + 16) = 1 \Rightarrow 30k = 1 \Rightarrow \boxed{k = \frac{1}{30}}$$
**Mean E(X):**
$$E(X) = \sum x \cdot f(x) = \frac{1}{30}\left[1(1)+2(4)+3(9)+4(16)\right] = \frac{1+8+27+64}{30} = \frac{100}{30} = \boxed{\frac{10}{3}}$$
**E(X²):**
$$E(X^2) = \frac{1}{30}\left[1(1)+4(4)+9(9)+16(16)\right] = \frac{1+16+81+256}{30} = \frac{354}{30} = \frac{59}{5}$$
**Variance:**
$$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{59}{5} - \frac{100}{9} = \frac{531-500}{45} = \boxed{\frac{31}{45} \approx 0.689}$$
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### Q10(b). X ~ Poisson(λ = 3)
$$P(X=x) = \frac{e^{-3}\cdot 3^x}{x!}, \quad e^{-3} \approx 0.04979$$
**(i) P(X ≥ 2):**
$$= 1 - P(X=0) - P(X=1)$$
P(X=0) = e⁻³ = 0.04979
P(X=1) = 3e⁻³ = 0.14936
$$P(X\geq2) = 1 - 0.04979 - 0.14936 = \boxed{0.8009}$$
**(ii) P(X = 2):**
$$= \frac{e^{-3}\cdot9}{2} = \frac{9\times0.04979}{2} = \boxed{0.2240}$$
**(iii) P(X ≤ 2):**
$$= P(0)+P(1)+P(2) = 0.04979+0.14936+0.22404 = \boxed{0.4232}$$
---
### Q11. Grouped frequency distribution
| Marks | f | Midpoint x | fx | fx² |
|-------|---|------------|-----|------|
| 10-13 | 7 | 11.5 | 80.5 | 925.75 |
| 14-17 | 9 | 15.5 | 139.5 | 2162.25 |
| 18-21 | 5 | 19.5 | 97.5 | 1901.25 |
| 22-25 | 3 | 23.5 | 70.5 | 1656.75 |
| 26-29 | 6 | 27.5 | 165.0 | 4537.50 |
| 30-33 | 10 | 31.5 | 315.0 | 9922.50 |
| **Total** | **40** | | **868** | **21106** |
**Note:** Class width = 4 throughout.
---
**(i) Median:**
N = 40, N/2 = 20
Cumulative frequencies:
- 10-13: 7
- 14-17: 16
- 18-21: 21 ← 20th value falls here
Median class = **18-21**
L = 17.5 (lower boundary), f = 5, F = 16, h = 4
$$\text{Median} = L + \frac{\frac{N}{2}-F}{f}\times h = 17.5 + \frac{20-16}{5}\times 4 = 17.5 + 3.2 = \boxed{20.7}$$
---
**(ii) Mode:**
Modal class = **30-33** (highest frequency = 10)
Using the formula:
L = 29.5, Δ₁ = 10−6 = 4, Δ₂ = 10−0 = 10, h = 4
$$\text{Mode} = 29.5 + \frac{4}{4+10}\times 4 = 29.5 + \frac{16}{14} = 29.5 + 1.14 = \boxed{30.64}$$
---
**(iii) Mean:**
$$\bar{x} = \frac{\sum fx}{N} = \frac{868}{40} = \boxed{21.7}$$
---
**(iv) Standard Deviation:**
$$\sigma = \sqrt{\frac{\sum fx^2}{N} - \bar{x}^2} = \sqrt{\frac{21106}{40} - (21.7)^2}$$
$$= \sqrt{527.65 - 470.89} = \sqrt{56.76} = \boxed{7.534}$$