## SECTION A
**1.** Find the equation of a straight line which is perpendicular to the line x − 3y + 2 = 0 and passes through the mid-point of the line joining the points (3, −2) and (−1, 5). [05 marks]
**2.** Show that f(x) = 5e^(−5x), 0 ≤ x ≤ ∞ is a probability density function, hence find the probability for x ≤ 3. [04 marks]
**3.** A bag contains 15 balls out of which 7 are yellow balls, 3 white balls and the rest are black. A ball is picked from the bag at random. Find the probability that the ball is black. [03 marks]
**4.** Find the equation of the circle whose centre is the mid-point of (−1, 2) and (2, 3) and radius 5. [04 marks]
**5.** Find the focus of the parabola y² = 3(x + 1). [04 marks]
---
## SECTION B: COORDINATE GEOMETRY
**6. (a)** Obtain the equation of the conic whose parametric equations are y = 5 + 4 tan θ, x = 7 + 3 sec θ and describe it completely. [12 marks]
**(b)** Write the following in polar form:
- (i) 3x + 2y = 8
- (ii) x² = √(x² + y²) [08 marks]
**7. (a)** Obtain the equation of the conic whose parametric equations are y = k + 2t, x = h + t² and obtain the condition for which the line y = mx + c touches it. [12 marks]
**(b)** Describe the locus of the points P(x, y) equidistant from the points A(3, −2) and B(−3, 5). [08 marks]
**8. (a)** Find the equation of the tangent to the ellipse (x−h)²/a² + (y−k)²/b² = 1, at the point (x₁, y₁). Hence, obtain the equation of the tangent to the ellipse (x+2)²/16 + y²/4 = 1, at the point (−2, 2). [10 marks]
**(b)** Sketch the curve r = 2(1 + sin θ) in polar form for 0 ≤ θ ≤ 360°. [10 marks]
---
## SECTION C: STATISTICS
**9. (a)** Two numbers are selected at random from integers 10 to 40, both inclusive, repetition being allowed. Find the probability that:
- (i) Both are multiples of 5
- (ii) Both are powers of 2
- (iii) One is a power of 2 and the other is a multiple of 5. What are the corresponding probabilities if repetition is not allowed? [12 marks]
**(b)** If X ~ N(74, 9), find:
- (i) P(x ≤ 73)
- (ii) P(70 ≤ X ≤ 78) [08 marks]
**10. (a)** For a Binomial distribution with 5 trials and P = 1/4, find the probability that:
- (i) Exactly 3 trials are successful
- (ii) At least 2 trials are successful
- (iii) Less than 4 trials are successful
- (iv) All 5 trials are successful
- (v) None of the trials is successful [12 marks]
**(b)** If a card is randomly selected from a deck of cards, find the probability that it is:
- (i) A spade or a diamond
- (ii) An ace or a heart [08 marks]
**11. (a)** Three balls are drawn at random without replacement from a bag containing 10 black, 13 white and 11 red balls. Find the probability that:
- (i) The balls are of the same color
- (ii) 2 of the balls are black and one is white [10 marks]
**(b)** If X ~ P₀(2.5), find:
- (i) P(x = 1)
- (ii) P(x ≥ 2)
- (iii) P(x ≤ 2) [10 marks]
---
---
# SOLUTIONS
## SECTION A
---
### Question 1
**Given line:** x − 3y + 2 = 0 → slope = 1/3
**Perpendicular slope:** m = −3
**Midpoint of (3, −2) and (−1, 5):**
$$M = \left(\frac{3+(-1)}{2}, \frac{-2+5}{2}\right) = \left(1, \frac{3}{2}\right)$$
**Equation of line:**
$$y - \frac{3}{2} = -3(x - 1)$$
$$y - \frac{3}{2} = -3x + 3$$
$$y = -3x + \frac{9}{2}$$
$$\boxed{6x + 2y - 9 = 0}$$
---
### Question 2
**To show f(x) = 5e^(−5x) is a PDF:**
A PDF requires: ∫₀^∞ f(x) dx = 1
$$\int_0^\infty 5e^{-5x}\,dx = 5\left[\frac{e^{-5x}}{-5}\right]_0^\infty = \left[-e^{-5x}\right]_0^\infty = 0 - (-1) = 1 ✓$$
**P(x ≤ 3):**
$$\int_0^3 5e^{-5x}\,dx = \left[-e^{-5x}\right]_0^3 = -e^{-15} + e^0 = 1 - e^{-15}$$
$$\boxed{P(x \leq 3) = 1 - e^{-15} \approx 0.9999997}$$
---
### Question 3
**Total balls:** 15
**Yellow:** 7, **White:** 3, **Black:** 15 − 7 − 3 = **5**
$$P(\text{black}) = \frac{5}{15} = \boxed{\frac{1}{3}}$$
---
### Question 4
**Centre = midpoint of (−1, 2) and (2, 3):**
$$C = \left(\frac{-1+2}{2}, \frac{2+3}{2}\right) = \left(\frac{1}{2}, \frac{5}{2}\right)$$
**Equation of circle** with centre (h, k) and radius r = 5:
$$(x - \tfrac{1}{2})^2 + (y - \tfrac{5}{2})^2 = 25$$
Expanding:
$$x^2 - x + \frac{1}{4} + y^2 - 5y + \frac{25}{4} = 25$$
$$x^2 + y^2 - x - 5y + \frac{26}{4} - 25 = 0$$
$$\boxed{x^2 + y^2 - x - 5y - \frac{37}{2} = 0}$$
---
### Question 5
**Rewrite:** y² = 3(x + 1)
This is of the form y² = 4a(x − h) with vertex (−1, 0):
$$4a = 3 \Rightarrow a = \frac{3}{4}$$
**Focus:** (h + a, k) = (−1 + 3/4, 0)
$$\boxed{\text{Focus} = \left(-\frac{1}{4},\ 0\right)}$$
---
## SECTION B: COORDINATE GEOMETRY
---
### Question 6(a)
**Given:** y = 5 + 4 tan θ, x = 7 + 3 sec θ
$$\tan\theta = \frac{y-5}{4}, \quad \sec\theta = \frac{x-7}{3}$$
Using identity: sec²θ − tan²θ = 1:
$$\frac{(x-7)^2}{9} - \frac{(y-5)^2}{16} = 1$$
**This is a hyperbola** with:
- Centre: (7, 5)
- a² = 9, b² = 16, so a = 3, b = 4
- Transverse axis: horizontal
- c = √(9 + 16) = √25 = 5
- Foci: (7 ± 5, 5) = **(2, 5) and (12, 5)**
- Eccentricity: e = 5/3
---
### Question 6(b)
Using x = r cos θ, y = r sin θ, x² + y² = r²
**(i) 3x + 2y = 8:**
$$3r\cos\theta + 2r\sin\theta = 8$$
$$\boxed{r(3\cos\theta + 2\sin\theta) = 8}$$
**(ii) x² = √(x² + y²):**
$$x^2 = \sqrt{x^2 + y^2}$$
$$r^2\cos^2\theta = r$$
$$\boxed{r = \sec^2\theta}$$
---
### Question 7(a)
**Given:** y = k + 2t, x = h + t²
From the first equation: t = (y − k)/2
Substituting into x = h + t²:
$$x = h + \left(\frac{y-k}{2}\right)^2$$
$$(y-k)^2 = 4(x-h)$$
**This is a parabola** with vertex (h, k), opening rightward.
**Condition for y = mx + c to touch the parabola:**
Substitute y = mx + c into (y − k)² = 4(x − h):
$$(mx + c - k)^2 = 4(x - h)$$
$$m^2x^2 + 2m(c-k)x + (c-k)^2 = 4x - 4h$$
$$m^2x^2 + [2m(c-k) - 4]x + [(c-k)^2 + 4h] = 0$$
For tangency, discriminant = 0:
$$[2m(c-k) - 4]^2 - 4m^2[(c-k)^2 + 4h] = 0$$
$$4m^2(c-k)^2 - 16m(c-k) + 16 - 4m^2(c-k)^2 - 16m^2h = 0$$
$$-16m(c-k) + 16 - 16m^2h = 0$$
$$\boxed{c - k = \frac{1 - m^2h}{m}}$$
---
### Question 7(b)
**Locus equidistant from A(3, −2) and B(−3, 5):**
PA = PB:
$$(x-3)^2 + (y+2)^2 = (x+3)^2 + (y-5)^2$$
Expanding:
$$x^2 - 6x + 9 + y^2 + 4y + 4 = x^2 + 6x + 9 + y^2 - 10y + 25$$
$$-6x + 4y + 13 = 6x - 10y + 34$$
$$-12x + 14y - 21 = 0$$
$$\boxed{12x - 14y + 21 = 0}$$
This is the **perpendicular bisector** of AB.
---
### Question 8(a)
**General tangent to ellipse** (x−h)²/a² + (y−k)²/b² = 1 at (x₁, y₁):
$$\frac{(x-h)(x_1-h)}{a^2} + \frac{(y-k)(y_1-k)}{b^2} = 1$$
**For ellipse** (x+2)²/16 + y²/4 = 1 at (−2, 2):
Here h = −2, k = 0, a² = 16, b² = 4, x₁ = −2, y₁ = 2:
$$\frac{(x+2)(-2+2)}{16} + \frac{(y)(2)}{4} = 1$$
$$0 + \frac{2y}{4} = 1$$
$$\frac{y}{2} = 1$$
$$\boxed{y = 2}$$
---
### Question 8(b): Sketch r = 2(1 + sin θ)
This is a **cardioid**. Key values:
| θ | sin θ | r = 2(1 + sin θ) |
|---|-------|-----------------|
| 0° | 0 | 2 |
| 90° | 1 | 4 |
| 180° | 0 | 2 |
| 270° | −1 | 0 |
| 360° | 0 | 2 |
The curve is a **cardioid** symmetric about the y-axis, with maximum r = 4 at θ = 90° (top), and a cusp at the origin when θ = 270°.
---
## SECTION C: STATISTICS
---
### Question 9(a)
**Integers 10 to 40 inclusive:** Total = 31 numbers
**Multiples of 5 in range:** {10, 15, 20, 25, 30, 35, 40} → **7 numbers**
**Powers of 2 in range:** {16, 32} → **2 numbers**
**WITH repetition (sample space = 31 × 31 = 961):**
**(i) Both multiples of 5:**
$$P = \frac{7}{31} \times \frac{7}{31} = \frac{49}{961}$$
**(ii) Both powers of 2:**
$$P = \frac{2}{31} \times \frac{2}{31} = \frac{4}{961}$$
**(iii) One power of 2, other multiple of 5:**
$$P = 2 \times \frac{2}{31} \times \frac{7}{31} = \frac{28}{961}$$
**WITHOUT repetition (sample space = 31 × 30 = 930):**
**(i) Both multiples of 5:**
$$P = \frac{7 \times 6}{930} = \frac{42}{930} = \frac{7}{155}$$
**(ii) Both powers of 2:**
$$P = \frac{2 \times 1}{930} = \frac{2}{930} = \frac{1}{465}$$
**(iii) One power of 2, other multiple of 5:**
$$P = \frac{2 \times 2 \times 7}{930} = \frac{28}{930} = \frac{14}{465}$$
---
### Question 9(b)
**X ~ N(74, 9)**, so μ = 74, σ² = 9, σ = 3
**(i) P(X ≤ 73):**
$$z = \frac{73 - 74}{3} = -\frac{1}{3} = -0.333$$
$$P(X \leq 73) = P(Z \leq -0.333) = 1 - P(Z \leq 0.333) = 1 - 0.6306 = \boxed{0.3694}$$
**(ii) P(70 ≤ X ≤ 78):**
$$z_1 = \frac{70-74}{3} = -1.333, \quad z_2 = \frac{78-74}{3} = 1.333$$
$$P = P(Z \leq 1.333) - P(Z \leq -1.333) = 0.9088 - 0.0912 = \boxed{0.8176}$$
---
### Question 10(a)
**X ~ B(5, 1/4)**, so p = 1/4, q = 3/4, n = 5
P(X = r) = C(5,r)(1/4)^r(3/4)^(5−r)
**(i) P(X = 3):**
$$= \binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2 = 10 \times \frac{1}{64} \times \frac{9}{16} = \frac{90}{1024} \approx \boxed{0.0879}$$
**(ii) P(X ≥ 2) = 1 − P(X = 0) − P(X = 1):**
$$P(X=0) = \left(\frac{3}{4}\right)^5 = \frac{243}{1024}$$
$$P(X=1) = 5 \times \frac{1}{4} \times \left(\frac{3}{4}\right)^4 = 5 \times \frac{81}{1024} = \frac{405}{1024}$$
$$P(X \geq 2) = 1 - \frac{243+405}{1024} = 1 - \frac{648}{1024} = \frac{376}{1024} \approx \boxed{0.3672}$$
**(iii) P(X < 4) = 1 − P(X = 4) − P(X = 5):**
$$P(X=4) = \binom{5}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right) = 5 \times \frac{3}{1024} = \frac{15}{1024}$$
$$P(X=5) = \left(\frac{1}{4}\right)^5 = \frac{1}{1024}$$
$$P(X < 4) = 1 - \frac{16}{1024} = \frac{1008}{1024} \approx \boxed{0.9844}$$
**(iv) P(X = 5):**
$$= \left(\frac{1}{4}\right)^5 = \frac{1}{1024} \approx \boxed{0.000977}$$
**(v) P(X = 0):**
$$= \left(\frac{3}{4}\right)^5 = \frac{243}{1024} \approx \boxed{0.2373}$$
---
### Question 10(b)
Standard deck: 52 cards
**(i) P(Spade or Diamond):**
$$= \frac{13 + 13}{52} = \frac{26}{52} = \boxed{\frac{1}{2}}$$
**(ii) P(Ace or Heart):**
$$= \frac{4 + 13 - 1}{52} = \frac{16}{52} = \boxed{\frac{4}{13}}$$
(subtract 1 for ace of hearts counted twice)
---
### Question 11(a)
**Bag:** 10 black, 13 white, 11 red → Total = **34 balls**
Draw 3 without replacement. Total ways = C(34, 3) = **5984**
**(i) All same colour:**
$$P = \frac{\binom{10}{3} + \binom{13}{3} + \binom{11}{3}}{5984}$$
$$= \frac{120 + 286 + 165}{5984} = \frac{571}{5984} \approx \boxed{0.0954}$$
**(ii) 2 black, 1 white:**
$$P = \frac{\binom{10}{2} \times \binom{13}{1}}{5984} = \frac{45 \times 13}{5984} = \frac{585}{5984} \approx \boxed{0.0977}$$
---
### Question 11(b)
**X ~ Poisson(λ = 2.5)**
$$P(X = x) = \frac{e^{-2.5}(2.5)^x}{x!}, \quad e^{-2.5} \approx 0.0821$$
**(i) P(X = 1):**
$$= \frac{e^{-2.5}(2.5)^1}{1!} = 0.0821 \times 2.5 = \boxed{0.2052}$$
**(ii) P(X ≥ 2) = 1 − P(X = 0) − P(X = 1):**
$$P(X=0) = e^{-2.5} = 0.0821$$
$$P(X \geq 2) = 1 - 0.0821 - 0.2052 = \boxed{0.7127}$$
**(iii) P(X ≤ 2) = P(0) + P(1) + P(2):**
$$P(X=2) = \frac{e^{-2.5}(2.5)^2}{2!} = \frac{0.0821 \times 6.25}{2} = 0.2565$$
$$P(X \leq 2) = 0.0821 + 0.2052 + 0.2565 = \boxed{0.5438}$$