### SECTION A
**1.** Express x² − 10y = 10x in polar coordinates. [04 marks]
**2.** If the eccentricity of the hyperbola x²/a² − y²/16 = 1 is ±√13/3, find the value of a and hence its foci. [04 marks]
**3.** Calculate the coordinates of center and the length of the radius of a circle whose equation is given by x² + y² + 4x + 10y + 4 = 0. [04 marks]
**4.** Calculate the mean of the distribution f(x) = 3x², 0 ≤ x ≤ 1. [04 marks]
**5.** Find the equation of the line that passes through the origin and perpendicular to the line joining the points (5,3) and (−5,6). [04 marks]
---
### SECTION B: COORDINATE GEOMETRY
**6. (a)** Find the condition for the line y = mx + c to be tangent to the ellipse 4x² + 9y² + 8x + 36y + 4 = 0. [12 marks]
**(b)** Show that r = 1/(1 + sinθ) represents a parabola in polar form. [08 marks]
**7. (a)** Find the condition for the line y = mx to touch the parabola y² − 4y − 4x + 8 = 0. [10 marks]
**(b)** Find the center, eccentricity and foci of the hyperbola (x+3)²/49 − (x−1)²/25 = 1. [10 marks]
**8. (a)** Sketch the curve in polar form r = 5(1 + 2sin 2θ). [10 marks]
**(b)** Obtain the equation of the tangent to the parabola y² = 4ax at the point (x₁, y₁). Hence, write down the equation of the tangent at (a, 2a). [10 marks]
---
### SECTION C: STATISTICS
**9. (a)** A random variable x has a distribution given by the probability density function f(x) = kx where x = 1, 3, 4, 5. Calculate the constant k, the mean and the variance of the distribution. [08 marks]
**(b)** If for two events Q and R, P(Q) = 0.4, P(Q ∪ R) = 0.7, for what value(s) of P(R) are Q and R mutually exclusive? [06 marks]
**(c)** The probability distribution of a random variable x is given by:
| x | 0 | 2 | 4 | 7 | k |
|---|---|---|---|---|---|
| p(x) | 0.3 | 0.2 | 0.2 | l | 0.15 |
Find the value of k and l given that the expected value of x is 3.9. [06 marks]
**10. (a)** The normal distribution is given by the probability density function:
f(x) = (1/√2πσ) · e^(−½((x−μ)/σ)²)
(i) What is the range of x?
(ii) Interpret μ and σ.
(iii) Is the distribution symmetrical about any axis?
(iv) A Binomial distribution for a large number of trials n each with success p is approximated by a normal distribution, what is the mean and the standard deviation of the normal distribution? [10 marks]
**(b)** Two numbers are selected from the integers 1 to 10 (both inclusive). What is the probability that:
(i) their sum is 12,
(ii) their difference is 7,
(iii) one is double the other,
(iv) the first one is even and the second one is odd,
(v) both are multiples of 3,
(vi) one is a multiple of 2 and the other is a multiple of 3. [10 marks]
**11. (a)** A continuous random variable has the probability density function f(x) = 4e^(−4x), x > 0. Calculate:
(i) P(x ≤ 1)
(ii) P(x ≥ 2)
(iii) the value of k such that P(x ≤ k) = ½. [10 marks]
**(b)** The marks in a science subject (x) and in an art subject (y) for eight (8) students are tabulated below:
| x | 6 | 8 | 7 | 6 | 9 | 5 | 8 | 7 |
|---|---|---|---|---|---|---|---|---|
| y | 8 | 7 | 9 | 9 | 4 | 10 | 8 | 9 |
(i) Compute x̄, ȳ for the above data, the mean of x and y respectively.
(ii) Obtain the equation of the line of regression of y on x.
(iii) Exhibit the points on a scatter diagram.
(iv) Comment on the scatter diagram. [10 marks]
---
## ANSWERS
---
### SECTION A
---
**Q1. Express x² − 10y = 10x in polar coordinates.**
Substitute x = r cosθ, y = r sinθ:
r²cos²θ − 10r sinθ = 10r cosθ
r²cos²θ = 10r cosθ + 10r sinθ
r²cos²θ = 10r(cosθ + sinθ)
**r = 10(cosθ + sinθ)/cos²θ**
or equivalently: **r = 10(secθ tanθ + sec²θ · sinθ/cosθ)**
Simplified: **r cos²θ = 10(cosθ + sinθ)**
---
**Q2. Eccentricity of x²/a² − y²/16 = 1 is √13/3. Find a and foci.**
For a hyperbola: e² = 1 + b²/a²
(√13/3)² = 1 + 16/a²
13/9 = 1 + 16/a²
13/9 − 1 = 16/a²
4/9 = 16/a²
a² = 36 → **a = 6**
c = ae = 6 × (√13/3) = 2√13
**Foci: (±2√13, 0)**
---
**Q3. Center and radius of x² + y² + 4x + 10y + 4 = 0.**
Complete the square:
(x² + 4x + 4) + (y² + 10y + 25) = −4 + 4 + 25
(x + 2)² + (y + 5)² = 25
**Center: (−2, −5), Radius: 5**
---
**Q4. Mean of f(x) = 3x², 0 ≤ x ≤ 1.**
E(x) = ∫₀¹ x · 3x² dx = 3∫₀¹ x³ dx = 3[x⁴/4]₀¹ = 3/4
**Mean = 3/4 = 0.75**
---
**Q5. Line through origin perpendicular to line joining (5,3) and (−5,6).**
Slope of line joining (5,3) and (−5,6):
m₁ = (6−3)/(−5−5) = 3/(−10) = −3/10
Perpendicular slope: m₂ = 10/3
Line through origin: y = (10/3)x
**10x − 3y = 0**
---
### SECTION B: COORDINATE GEOMETRY
---
**Q6(a). Condition for y = mx + c to be tangent to 4x² + 9y² + 8x + 36y + 4 = 0.**
Rewrite in standard form by completing the square:
4(x² + 2x + 1) + 9(y² + 4y + 4) = −4 + 4 + 36
4(x+1)² + 9(y+2)² = 36
(x+1)²/9 + (y+2)²/4 = 1
This is an ellipse with center (−1, −2), a² = 9, b² = 4.
Substitute the translated coordinates: X = x+1, Y = y+2
Line becomes: Y + 2 = m(X − 1) + c → Y = mX + (c − m + 2 − 2) → Y = mX + (c − m)
Wait, let X = x+1, Y = y+2, so x = X−1, y = Y−2:
y = mx + c → Y − 2 = m(X−1) + c → Y = mX + (c − m + 2)
Let C' = c − m + 2
For tangency to X²/9 + Y²/4 = 1:
**Condition: c − m + 2 = ±√(9m² + 4)**
i.e., **(c − m + 2)² = 9m² + 4**
---
**Q6(b). Show r = 1/(1 + sinθ) is a parabola.**
r(1 + sinθ) = 1
r + r sinθ = 1
Since r sinθ = y and r = √(x²+y²):
√(x²+y²) + y = 1
√(x²+y²) = 1 − y
Square both sides: x² + y² = 1 − 2y + y²
x² = 1 − 2y
**x² = −2(y − ½)**
This is a parabola with vertex at (0, ½), opening downward. ∎
---
**Q7(a). Condition for y = mx to touch y² − 4y − 4x + 8 = 0.**
Substitute x = y/m into the parabola equation:
y² − 4y − 4(y/m) + 8 = 0
y² − (4 + 4/m)y + 8 = 0
For tangency, discriminant = 0:
(4 + 4/m)² − 32 = 0
16(1 + 1/m)² = 32
(1 + 1/m)² = 2
1 + 1/m = ±√2
1/m = ±√2 − 1
**m = 1/(√2 − 1) = √2 + 1 or m = −1/(√2 + 1) = 1 − √2**
---
**Q7(b). Center, eccentricity and foci of (x+3)²/49 − (y−1)²/25 = 1.**
*(Note: The image shows (x−1)² in denominator 25 — interpreted as (y−1)² for a standard hyperbola)*
**Center: (−3, 1)**
a² = 49, b² = 25
e = √(1 + b²/a²) = √(1 + 25/49) = √(74/49) = √74/7
**Eccentricity: e = √74/7**
c = ae = 7 × √74/7 = √74
**Foci: (−3 ± √74, 1)**
---
**Q8(a). Sketch r = 5(1 + 2sin 2θ).**
| θ | 0° | 45° | 90° | 135° | 180° | 270° |
|---|---|---|---|---|---|---|
| sin2θ | 0 | 1 | 0 | −1 | 0 | 0 |
| r | 5 | 15 | 5 | −5 | 5 | 5 |
This is a limaçon-type curve with inner loops due to |2| > 1 in the sin 2θ term. The curve has maximum r = 15 at θ = 45°, minimum r = −5 at θ = 135° (inner loop). The curve is periodic with period π due to sin 2θ, producing a 4-petal rose-like limaçon. Plot symmetrically about θ = 45° and θ = 225°.
---
**Q8(b). Tangent to y² = 4ax at (x₁, y₁); tangent at (a, 2a).**
Differentiating y² = 4ax implicitly:
2y · dy/dx = 4a → dy/dx = 2a/y
At (x₁, y₁): slope = 2a/y₁
Equation of tangent:
y − y₁ = (2a/y₁)(x − x₁)
y·y₁ − y₁² = 2ax − 2ax₁
Since y₁² = 4ax₁:
**yy₁ = 2a(x + x₁)** ← equation of tangent at (x₁, y₁)
At (a, 2a): x₁ = a, y₁ = 2a:
y(2a) = 2a(x + a)
2ay = 2ax + 2a²
**y = x + a**
---
### SECTION C: STATISTICS
---
**Q9(a). f(x) = kx, x = 1, 3, 4, 5. Find k, mean, variance.**
Sum of probabilities = 1:
k(1) + k(3) + k(4) + k(5) = 1
13k = 1 → **k = 1/13**
| x | p(x) = x/13 | xp(x) | x²p(x) |
|---|---|---|---|
| 1 | 1/13 | 1/13 | 1/13 |
| 3 | 3/13 | 9/13 | 27/13 |
| 4 | 4/13 | 16/13 | 64/13 |
| 5 | 5/13 | 25/13 | 125/13 |
**Mean = E(X) = (1+9+16+25)/13 = 51/13 ≈ 3.92**
E(X²) = (1+27+64+125)/13 = 217/13
Var(X) = E(X²) − [E(X)]² = 217/13 − (51/13)²
= 217/13 − 2601/169 = 2821/169 − 2601/169 = **220/169 ≈ 1.30**
---
**Q9(b). P(Q) = 0.4, P(Q ∪ R) = 0.7. Find P(R) if Q and R are mutually exclusive.**
If mutually exclusive: P(Q ∪ R) = P(Q) + P(R)
0.7 = 0.4 + P(R)
**P(R) = 0.3**
---
**Q9(c). Find k and l given E(x) = 3.9.**
Sum of probabilities = 1:
0.3 + 0.2 + 0.2 + l + 0.15 = 1
0.85 + l = 1 → **l = 0.15**
E(x) = 0(0.3) + 2(0.2) + 4(0.2) + 7(0.15) + k(0.15) = 3.9
0 + 0.4 + 0.8 + 1.05 + 0.15k = 3.9
2.25 + 0.15k = 3.9
0.15k = 1.65 → **k = 11**
---
**Q10(a). Normal Distribution questions.**
**(i) Range of x:** −∞ < x < +∞
**(ii) μ** is the mean (also median and mode); **σ** is the standard deviation measuring spread.
**(iii) Yes**, the distribution is symmetrical about x = μ (the vertical axis x = μ).
**(iv)** For Binomial approximated by Normal:
- **Mean = np**
- **Standard deviation = √(npq)** where q = 1 − p
---
**Q10(b). Two numbers selected from 1–10. Total outcomes = 10 × 10 = 100** *(with replacement, ordered pairs)*
**(i) Sum = 12:**
Pairs: (2,10),(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3),(10,2) = 9 pairs
**P = 9/100**
**(ii) Difference = 7:**
Pairs: (8,1),(9,2),(10,3),(1,8),(2,9),(3,10) = 6 pairs
**P = 6/100 = 3/50**
**(iii) One is double the other:**
Pairs: (1,2),(2,4),(3,6),(4,8),(5,10),(2,1),(4,2),(6,3),(8,4),(10,5) = 10 pairs
**P = 10/100 = 1/10**
**(iv) First is even, second is odd:**
Even in 1–10: {2,4,6,8,10} = 5; Odd: {1,3,5,7,9} = 5
**P = 5×5/100 = 25/100 = 1/4**
**(v) Both multiples of 3:**
Multiples of 3 in 1–10: {3,6,9} = 3
**P = 3×3/100 = 9/100**
**(vi) One multiple of 2, other multiple of 3:**
Multiples of 2: {2,4,6,8,10} = 5; Multiples of 3: {3,6,9} = 3
Both ways: 5×3 + 3×5 = 30
**P = 30/100 = 3/10**
---
**Q11(a). f(x) = 4e^(−4x), x > 0.**
**(i) P(x ≤ 1):**
P(x ≤ 1) = ∫₀¹ 4e^(−4x) dx = [−e^(−4x)]₀¹ = 1 − e^(−4)
**= 1 − e^(−4) ≈ 0.9817**
**(ii) P(x ≥ 2):**
P(x ≥ 2) = ∫₂^∞ 4e^(−4x) dx = [−e^(−4x)]₂^∞ = e^(−8)
**= e^(−8) ≈ 0.000335**
**(iii) P(x ≤ k) = ½:**
1 − e^(−4k) = ½
e^(−4k) = ½
−4k = ln(½) = −ln 2
**k = ln2/4 ≈ 0.1733**
---
**Q11(b). Regression of y on x.**
| x | y |
|---|---|
| 6 | 8 |
| 8 | 7 |
| 7 | 9 |
| 6 | 9 |
| 9 | 4 |
| 5 | 10 |
| 8 | 8 |
| 7 | 9 |
**(i) Means:**
Σx = 6+8+7+6+9+5+8+7 = 56 → **x̄ = 56/8 = 7**
Σy = 8+7+9+9+4+10+8+9 = 64 → **ȳ = 64/8 = 8**
**(ii) Regression of y on x: y = a + bx**
Compute Σxy and Σx²:
| x | y | xy | x² |
|---|---|---|---|
| 6 | 8 | 48 | 36 |
| 8 | 7 | 56 | 64 |
| 7 | 9 | 63 | 49 |
| 6 | 9 | 54 | 36 |
| 9 | 4 | 36 | 81 |
| 5 | 10 | 50 | 25 |
| 8 | 8 | 64 | 64 |
| 7 | 9 | 63 | 49 |
| **56** | **64** | **434** | **404** |
b = (Σxy − nx̄ȳ)/(Σx² − nx̄²)
= (434 − 8×7×8)/(404 − 8×49)
= (434 − 448)/(404 − 392)
= −14/12 = **−7/6 ≈ −1.167**
a = ȳ − bx̄ = 8 − (−7/6)(7) = 8 + 49/6 = 48/6 + 49/6 = **97/6 ≈ 16.17**
**Regression line: y = 97/6 − (7/6)x** or **y ≈ 16.17 − 1.167x**
**(iii)** Plot the 8 points on axes with x (science) horizontal, y (art) vertical.
**(iv)** The scatter diagram shows a **negative correlation** — as science marks increase, art marks tend to decrease, suggesting students who perform better in science tend to score lower in art, and vice versa.