QUESTIONS
SECTION A
1. Express y² − 4x = 4y in polar coordinates.
2. Find the eccentricity of the ellipse x²/25 + y²/16 = 1, hence its foci.
3. Given that the coordinates of the centre and the length of the radius of a circle are (−2,−5) and 5 respectively, obtain its equation.
4. Calculate P(A∩B) where P(A) = 0.4, P(B) = 0.6, P(A∪B) = 0.8.
5. What is the remainder when the polynomial x³ + 3x² + 6x + 1 is divided by (x − 2)?
SECTION B: CO-ORDINATE GEOMETRY
6a. Find the condition for the line y = ax + b to be tangent to the circle x² + y² + x + y = 0.
6b. Show that r = 1/(2+2cosθ) represents a parabola in polar form.
7a. Find the condition for the line y = ax + b to touch the parabola y² − y + 3x = 0.
7b. Find the centre, eccentricity and foci of the ellipse (2x−2)²/9a + (x+4)²/16 = 1.
8a. Sketch the curve in polar form r = 3(1 − sin3θ).
8b. Obtain the equation of the tangent to the ellipse x²/a² + y²/b² = 1 at the point (x₁,y₁). Hence, write down the equation of the tangent at (a sinθ, b cosθ).
SECTION C: STATISTICS
9a. Show that f(x) = 2e⁻²ˣ, 0 ≤ x < ∞ represents probability density function for a random variable x. Calculate:
(i) P(x ≤ 3); (ii) P(x ≥ 2).
9b. For a Binomial distribution with 6 trials and p = 0.125, find the probability that:
(i) all six trials are successful; (ii) at most two trials are successful; (iii) none of the trials is successful.
9c. A card is selected from a pack of playing cards, what is the probability that it is a queen or a diamond?
10a. A random variable has the distribution given by the probability density function f(x) = mx², 1 ≤ x ≤ 4. Calculate the value of m. For this value of m, find the mean and standard deviation of the distribution.
10b. The average birth in a certain community is 5 per day; find the probability that in a certain day:
(i) there will be 2 births; (ii) at least 3 births.
10c. In a single toss of a fair die, what is the probability of obtaining a prime or even number?
11. Two dice are tossed once. What is the probability that the outcomes are:
(a) prime and
(b) the product of the outcome is (i) less than 7, (ii) even, (iii) odd?
(c) Find the Harmonic and Geometric means of 7, 9 and 11.
ANSWERS
Question 1
y² − 4x = 4y
In polar coordinates: x = r cosθ, y = r sinθ
(r sinθ)² − 4r cosθ = 4r sinθ
r² sin²θ − 4r cosθ − 4r sinθ = 0
r² sin²θ = 4r cosθ + 4r sinθ
r sin²θ = 4cosθ + 4sinθ
r = 4(cosθ + sinθ)/sin²θ
Question 2
x²/25 + y²/16 = 1
a² = 25, b² = 16 → a = 5, b = 4
b² = a²(1−e²)
16 = 25(1−e²)
1−e² = 16/25
e² = 9/25
e = 3/5 = 0.6
Foci: (±ae, 0) = (±5×0.6, 0) = (±3, 0)
Question 3
Centre (−2,−5), radius 5
Equation: (x+2)² + (y+5)² = 5²
x² + 4x + 4 + y² + 10y + 25 = 25
x² + y² + 4x + 10y + 4 = 0
Question 4
P(A∪B) = P(A) + P(B) − P(A∩B)
0.8 = 0.4 + 0.6 − P(A∩B)
0.8 = 1.0 − P(A∩B)
P(A∩B) = 0.2
Question 5
By the Remainder Theorem, for divisor (x−2), remainder = f(2):
f(x) = x³ + 3x² + 6x + 1
f(2) = 8 + 12 + 12 + 1 = 33
Question 6
a) Condition for y = ax + b to be tangent to x² + y² + x + y = 0
Circle: x² + y² + x + y = 0 → centre (−1/2, −1/2), radius² = 1/4+1/4 = 1/2, radius = 1/√2
Substitute y = ax+b:
x² + (ax+b)² + x + ax + b = 0
(1+a²)x² + (2ab+1+a)x + (b²+b) = 0
For tangency, discriminant = 0:
(2ab+1+a)² − 4(1+a²)(b²+b) = 0
Condition: (2ab+1+a)² = 4(1+a²)(b²+b)
b) Show r = 1/(2+2cosθ) represents a parabola
r = 1/(2+2cosθ) = (1/2)/(1+cosθ)
This is of the standard polar conic form: r = l/(1+e cosθ), with l = 1/2 and e = 1.
Since eccentricity e = 1, the curve represents a parabola (by definition, a conic with e=1 is a parabola).
Question 7
a) Condition for y = ax+b to touch y² − y + 3x = 0
From the parabola equation: 3x = y − y² → x = (y−y²)/3
Substitute x = (y−b)/a (from y=ax+b, x=(y-b)/a):
(y−b)/a = (y−y²)/3
3(y−b) = a(y−y²)
3y − 3b = ay − ay²
ay² + 3y − ay − 3b = 0
ay² + (3−a)y − 3b = 0
For tangency, discriminant = 0:
(3−a)² − 4(a)(−3b) = 0
(3−a)² + 12ab = 0
Condition: (3−a)² + 12ab = 0
b) (The equation as transcribed contains inconsistent terms; based on standard ellipse form, assuming the equation is (x−1)²/9 + (y+4)²/16 = 1):
Centre: (1, −4)
a² = 16, b² = 9 (major axis along y since 16>9) → a=4, b=3
b² = a²(1−e²) → 9 = 16(1−e²) → e² = 7/16 → e = √7/4 ≈ 0.66
Foci: (1, −4±ae) = (1, −4±4×√7/4) = (1, −4±√7)
Question 8
a) Sketch r = 3(1 − sin3θ)
This is a limaçon-type polar curve with a period of 2π/3 due to the sin3θ term, producing a rose-like pattern with three "petals" or lobes of varying size, since the coefficient of sinθ (1) equals the constant term (1) in the (1−sin) form — giving cardioid-like loops repeated three times around the origin, with maximum r = 6 (at sin3θ=−1) and minimum r = 0 (at sin3θ=1).
b) Equation of tangent to ellipse at (x₁,y₁), and at (a sinθ, b cosθ)
General equation of tangent to x²/a² + y²/b² = 1 at point (x₁,y₁):
xx₁/a² + yy₁/b² = 1
At point (a sinθ, b cosθ):
x(a sinθ)/a² + y(b cosθ)/b² = 1
(x sinθ)/a + (y cosθ)/b = 1
Question 9
a) Show f(x) = 2e⁻²ˣ is a valid pdf
∫₀^∞ 2e⁻²ˣ dx = [−e⁻²ˣ]₀^∞ = 0−(−1) = 1 ✓ (integrates to 1, and f(x) ≥ 0 for all x ≥ 0, so it is a valid pdf — this is an exponential distribution with λ=2)
(i) P(x≤3) = ∫₀³ 2e⁻²ˣ dx = [−e⁻²ˣ]₀³ = −e⁻⁶−(−1) = 1−e⁻⁶ = 1 − 0.00248 ≈ 0.9975
(ii) P(x≥2) = ∫₂^∞ 2e⁻²ˣ dx = [−e⁻²ˣ]₂^∞ = 0−(−e⁻⁴) = e⁻⁴ ≈ 0.0183
b) Binomial distribution, n=6, p=0.125, q=0.875
(i) P(all 6 successful) = p⁶ = 0.125⁶ = 3.815×10⁻⁶
(ii) P(at most 2 successful) = P(0)+P(1)+P(2)
P(0) = q⁶ = 0.875⁶ = 0.4488
P(1) = 6p q⁵ = 6×0.125×0.875⁵ = 6×0.125×0.5129 = 0.3847
P(2) = C(6,2)p²q⁴ = 15×0.015625×0.5862 = 0.1373
Sum = 0.4488+0.3847+0.1373 = 0.9708
(iii) P(none successful) = q⁶ = 0.875⁶ ≈ 0.4488
c) P(queen or diamond)
P(queen) = 4/52; P(diamond) = 13/52; P(queen and diamond) = 1/52 (queen of diamonds)
P(queen or diamond) = 4/52+13/52−1/52 = 16/52 = 4/13 ≈ 0.3077
Question 10
a) f(x) = mx², 1≤x≤4
∫₁⁴ mx² dx = 1
m[x³/3]₁⁴ = 1
m(64/3 − 1/3) = 1
m(63/3) = 1
21m = 1
m = 1/21
Mean: E(x) = ∫₁⁴ x·mx² dx = m∫₁⁴x³dx = (1/21)[x⁴/4]₁⁴ = (1/21)(256/4−1/4) = (1/21)(63.75) = 3.036
E(x²) = ∫₁⁴ x²·mx² dx = m∫₁⁴x⁴dx = (1/21)[x⁵/5]₁⁴ = (1/21)(1024/5−1/5) = (1/21)(204.6) = 9.743
Variance = E(x²)−[E(x)]² = 9.743 − 9.217 = 0.526
Standard deviation = √0.526 ≈ 0.725
b) Poisson distribution, λ=5
(i) P(x=2) = e⁻⁵×5²/2! = 0.006738×25/2 = 0.0842
(ii) P(x≥3) = 1 − [P(0)+P(1)+P(2)]
P(0) = e⁻⁵ = 0.006738
P(1) = e⁻⁵×5 = 0.03369
P(2) = 0.0842
Sum = 0.1246
P(x≥3) = 1−0.1246 = 0.8754
c) P(prime or even number) on a die
Numbers 1–6; Prime = {2,3,5}; Even = {2,4,6}; Prime∩Even = {2}
P(prime or even) = P(prime)+P(even)−P(both) = 3/6+3/6−1/6 = 5/6 = 0.833
Question 11: Two dice tossed once
a) P(outcomes are prime)
Primes on a die: {2,3,5}. P(single die prime) = 3/6 = 1/2
P(both prime) = 1/2×1/2 = 1/4
b) Product of outcomes:
Total outcomes = 36
(i) P(product < 7):
Count pairs (a,b) with a×b<7: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1),(5,1),(6,1) = 14 pairs
P = 14/36 = 7/18 ≈ 0.389
(ii) P(product even):
Product is even unless both numbers are odd. P(both odd) = (3/6)×(3/6) = 9/36
P(product even) = 1 − 9/36 = 27/36 = 3/4 = 0.75
(iii) P(product odd):
P(product odd) = P(both odd) = 9/36 = 1/4 = 0.25
c) Harmonic and Geometric means of 7, 9, 11
Geometric mean = ∛(7×9×11) = ∛693 = 8.847
Harmonic mean = 3/(1/7+1/9+1/11) = 3/(0.1429+0.1111+0.0909) = 3/0.3449 = 8.699