2023 IJMB mathematics paper III

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:

$$\frac{\sqrt{3}}{2}x - \frac{1}{2}y = 1$$

$$\sqrt{3}\,x - y = 2$$

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Q2. Eccentricity and foci of x²/36 − y²/16 = 1

This is a hyperbola with a² = 36, b² = 16 → a = 6, b = 4

$$c^2 = a^2 + b^2 = 36 + 16 = 52 \Rightarrow c = 2\sqrt{13}$$

$$\text{Eccentricity: } e = \frac{c}{a} = \frac{2\sqrt{13}}{6} = \frac{\sqrt{13}}{3} \approx 1.202$$

$$\text{Foci: } (\pm 2\sqrt{13},\ 0) \approx (\pm 7.21,\ 0)$$

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Q3. Circle with centre (−2, −5), radius 5

$$(x+2)^2 + (y+5)^2 = 25$$

Expanding:

$$x^2 + y^2 + 4x + 10y + 4 = 0$$

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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. ✓

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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):

$$y - 5 = 3(x - 1)$$

$$y = 3x + 2$$

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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:

$$x = -2 + 3\left(\frac{y+1}{5}\right)^2$$

$$x + 2 = \frac{3(y+1)^2}{25}$$

$$(y+1)^2 = \frac{25(x+2)}{3}$$

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

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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°.

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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:

$$\frac{(x-3)^2}{25} + \frac{(y-2)^2}{9} = 1$$

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)

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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):

$$y - 1 = \frac{3}{2}(x-1)$$

$$2y - 2 = 3x - 3$$

$$3x - 2y - 1 = 0$$

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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:

$$m = \pm\frac{b}{a}$$

These are the asymptotes of the hyperbola. Any other line y = mx + c is tangent when:

$$c^2 = a^2m^2 - b^2$$

General tangent equation:

$$y = mx \pm\sqrt{a^2m^2 - b^2}$$

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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). ✓

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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):

$$= \int_0^{\pi/4}\frac{1}{2}\sin x\,dx = \frac{1}{2}[-\cos x]_0^{\pi/4} = \frac{1}{2}\left(-\frac{\sqrt{2}}{2}+1\right) = \frac{1}{2}\left(1-\frac{\sqrt{2}}{2}\right)$$

$$= \frac{2-\sqrt{2}}{4} \approx 0.1464$$

(ii) P(x ≥ π/4):

$$= 1 - P(x \leq \pi/4) = 1 - \frac{2-\sqrt{2}}{4} = \frac{2+\sqrt{2}}{4} \approx 0.8536$$

(iii) Find c such that P(x ≤ c) = 1/2:

$$\int_0^c\frac{1}{2}\sin x\,dx = \frac{1}{2}[-\cos x]_0^c = \frac{1}{2}(1-\cos c) = \frac{1}{2}$$

$$1 - \cos c = 1 \Rightarrow \cos c = 0 \Rightarrow c = \frac{\pi}{2}$$

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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):

$$= (0.25)^5 = 0.000977$$

(ii) P(exactly 4) = P(X=4):

$$= \binom{5}{4}(0.25)^4(0.75)^1 = 5 \times 0.003906 \times 0.75 = 0.01465$$

(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)

$$P(X\geq3) = 0.08789 + 0.01465 + 0.000977 = 0.10352$$

(iv) P(none successful) = P(X=0):

$$= (0.75)^5 = 0.2373$$

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Q10(a). f(x) = kx², x = 1, 2, 3, 4

Find k: Σf(x) = 1

$$k(1 + 4 + 9 + 16) = 1 \Rightarrow 30k = 1 \Rightarrow k = \frac{1}{30}$$

Mean E(X):

$$E(X) = \sum x \cdot f(x) = \frac{1}{30}\left[1(1)+2(4)+3(9)+4(16)\right] = \frac{1+8+27+64}{30} = \frac{100}{30} = \frac{10}{3}$$

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:

$$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{59}{5} - \frac{100}{9} = \frac{531-500}{45} = \frac{31}{45} \approx 0.689$$

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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

$$P(X\geq2) = 1 - 0.04979 - 0.14936 = 0.8009$$

(ii) P(X = 2):

$$= \frac{e^{-3}\cdot9}{2} = \frac{9\times0.04979}{2} = 0.2240$$

(iii) P(X ≤ 2):

$$= P(0)+P(1)+P(2) = 0.04979+0.14936+0.22404 = 0.4232$$

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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.

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(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

$$\text{Median} = L + \frac{\frac{N}{2}-F}{f}\times h = 17.5 + \frac{20-16}{5}\times 4 = 17.5 + 3.2 = 20.7$$

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(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

$$\text{Mode} = 29.5 + \frac{4}{4+10}\times 4 = 29.5 + \frac{16}{14} = 29.5 + 1.14 = 30.64$$

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(iii) Mean:

$$\bar{x} = \frac{\sum fx}{N} = \frac{868}{40} = 21.7$$

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(iv) Standard Deviation:

$$\sigma = \sqrt{\frac{\sum fx^2}{N} - \bar{x}^2} = \sqrt{\frac{21106}{40} - (21.7)^2}$$

$$= \sqrt{527.65 - 470.89} = \sqrt{56.76} = 7.534$$