2024 JUPEB Physics

MAT 001: ADVANCED PURE MATHEMATICS

1. (a) Let A = $\begin{pmatrix} 2 & 1 \ 1 & 1 \end{pmatrix}$ and f(x) = x² + 2x + 3, find f(A). [3 Marks]

(b) Compute the inverse of the matrix $A = \begin{pmatrix} 1 & 5 & 2 \ 3 & 0 & -2 \ -1 & 0 & 3 \end{pmatrix}$ [6 Marks]

©

• i. Find the coordinates of the point which divides the line segment joining (3, -2) and (5, 3) externally in the ratio 1 : 3. [3 Marks]
• ii. In what ratio at the line joining (-2, 8) and (4, 5) divided by the line x + y − 6 = 0? [3 Marks]

[TOTAL = 15 Marks]

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2. (a) Express $\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)^{17}$ in the form z = x + iy. [4 Marks]

(b) Show that the points A(1,1), B(3,11), C(4,2), D(2,2) are vertices of a Parallelogram ABCD. Find the equations of the lines AB and AD and the angle between them. [5 Marks]

© In a certain Faculty of Science, 70% of students studied Physics, 50% studied Chemistry, 40% studied Biology, 30% studied Physics and Chemistry, 30% studied Chemistry and Biology, while 20% studied Physics and Biology. If a student is to be selected at random, what is the probability that the student studied:

• i. all the three subjects, and [2 Marks]
• ii. Physics and Biology but not Chemistry? [2 Marks]
• iii. Show that the probability of students that studied Physics and Chemistry is 30%. [2 Marks]

[TOTAL = 15 Marks]

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MAT 002: CALCULUS

3. (a) Prove that the area enclosed by the curve $y^2 = \frac{x^2}{4x-2}$ and the line x = a is (n − 2)a². [7 Marks]

(b) Find the Maclaurin expansion of f(x) = sin x. [6 Marks]

© A curve has equation 3x² + 2xy − 5y² = 10. Show that the gradient of the tangent at point (2, 0) is -3. [2 Marks]

[TOTAL = 15 Marks]

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4. (a) Evaluate the following limits:

• i. $\lim_{x \to 3} \left(\frac{2x^2 - 27}{x + 3}\right)$ [2 Marks]
• ii. $\lim_{y \to 0} \frac{5y^3 - 3y^2 + 6y}{4y^2 + 3y}$ [6 Marks]

(b) Differentiate $y = \frac{x}{x^2}$ with respect to x from first principle. [2 Marks]

© Find the derivatives of the following functions with respect to x:

• i. y = (2x³ − 4x² + 3x − 5)⁸ [3 Marks]
• ii. y = 2x²·e^(2x) + ln x [TOTAL = 15 Marks]

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MAT 003: STATISTICS

5. (a) The top 45 stocks of the NSE market, ranked by percentage of outstanding shares traded on one day last year are as follows:

8.9, 12.4, 9.6, 11.3, 9.2, 8.8, 5.1, 6.2, 7.0, 7.1, 11.8, 10.7, 7.6, 9.1, 9.2
8.7, 9.1, 10.9, 10.3, 9.6, 7.8, 11.5, 9.3, 7.9, 8.8, 8.8, 12.7, 8.4, 7.8, 5.7
9.6, 8.9, 10.2, 10.3, 7.7, 10.6, 8.3, 8.8, 9.5, 8.8, 9.4, 9.0, 10.5, 8.2, 10.5

By using 4 class 5.0–5.9, 6.0–6.9, …

• (i) prepare the frequency distribution table; [3 Marks]
• (ii) find the coefficient of variation of the distribution; [3 Marks]
• (iii) Does the data represent a sample or a population? [3 Marks]

(b) Two numbers a and b is to be added to set of four numbers: 2, 3, 6, 9 such that the mean is increased by 1 and the variance is increased by 2.5. Find a and b. [4 Marks]

© Given that ¹⁰Cᵣ = ¹⁰Cᵣ, find the value of r. [2 Marks]

[TOTAL = 15 Marks]

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6. The table shows heights x and y of a sample of 12 mothers and their oldest daughters.

Height x of Mothers (inches)

65

63

67

64

68

62

70

66

68

67

69

71

Height y of Daughters (inches)

68

66

68

65

69

66

68

65

71

67

68

70

(a) Construct a scatter diagram. [4 Marks]

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(b) The mean life span of bulbs manufactured by a company is 1570 hours with standard deviation of 85 hours. If the life-span of the bulbs is normally distributed, calculate (with the extract of the Normal distribution table below) the probability that a bulb will cease to function:

• i. in more than 1950 hours, [2 Marks]
• ii. between 1730 hours and 1900 hours, [2 Marks]
• iii. How many bulbs would be expected to last beyond 1900 hours, if tested? [2 Marks]

© If X is a discrete random variable with sample space S = {x: x = 0, 1, 2, 3, 4} and f(x) = c(⁴Cₓ)(¼)ˣ. Show that f(x) defines a probability density function. [5 Marks]

[TOTAL = 15 Marks]

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MAT 004A: APPLIED MATHEMATICS

7. Given that four coplanar forces F₁(25N, 050°), F₂(30N, 150°), F₃(35N, 240°), and F₄(25N, 330°) act on a particle P.

(a) express each of the four coplanar forces as a column vector. [8 Marks]

(b) find the resultant of these coplanar forces as a column vector. [3 Marks]

© If Ā = î − j + 3k, B̄ = 2î + 4j − 6k and C̄ = 3î − 5j + 2k. Evaluate:

• (i) Ā × B̄ × C̄ [2 Marks]
• (ii) Ā · (B̄ × C̄) [2 Marks]

[TOTAL = 15 Marks]

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8. (a) Calculate, correct to the nearest degree, the angle between two forces of magnitude 19N and 21N, if the resultant of the two forces has a magnitude of 27N. [6 Marks]

(b) A mass of 4kg hangs on a light inextensible string, fixed at point A and B, such that the object rest in equilibrium with the strings inclined at 30° and 45° at A and B respectively. Find the tensions in the strings. (g = 10 m/s²) [4 Marks]

© A uniform bar of mass 40kg is 10m long and has weights 25N and 30N suspended from its ends.

• i. At what point must the bar be pivoted for it to rest in equilibrium horizontally? [3 Marks]
• ii. What is the reaction on this pivot? (g = 10 m/s²) [2 Marks]

[TOTAL = 15 Marks]

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MAT 004B: APPLIED BUSINESS MATHEMATICS

9. (a) The demand function q₁ for Beans and yams is given in terms of their prices p₁ and p₂ respectively as q₁ = 30 + 2p₂ − p₁. Given that p₁ = N7 and N9, determine:

• i. the price elasticity of demand for beans; [3 Marks]
• ii. the cross elasticity of demand for beans. [2 Marks]

(b) If an interest on a sum of money compounded at rate of 4% annually, find:

• i. how many years that the sum will be 4 times itself, correct to nearest year; [3 Marks]
• ii. the rate to the nearest whole number if the sum is doubled within 10 years. [3 Marks]

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© Maximize the function q = 12u + 156 subject to the constraints 4u + 3B ≤ 20. Where u ≥ 0, B ≥ 0. [4 Marks]

[TOTAL = 18 Marks]

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10. A pharmaceutical company is formulating a drug which contains three chemicals in the following proportions: chemical P at least 7 units, chemical Q at least 11 units, and Chemical R at least 11 units. The pharmaceutical company has three chemical suppliers: company A supplies chemical P with 1 unit, 2 units of Q, 3 units of R and costs N3 per kg; chemical B: company B supplies chemical Y which contains 2 units of P, 4 units of Q, 6 units of R at N64 per kg. While Y sells for N30/kg.

• (i) formulate the problem and determine the constraints. [25 Marks]
• (ii) solve the problem graphically. [5 Marks]
• (iii) at what corner vertex should the pharmaceutical company buy to minimize cost? [1 Mark]
• (iv) What is the minimum annual compound interest until the sum of N6,900 amounts to N2,000 in 4 years, if reckoned half-yearly? [7 Marks]

#Solution

MAT 001: ADVANCED PURE MATHEMATICS

Question 1

(a) Find f(A) where A = $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, f(x) = x² + 2x + 3

f(A) = A² + 2A + 3I

A²:

$$A^2 = \begin{pmatrix}2&1\\1&1\end{pmatrix}\begin{pmatrix}2&1\\1&1\end{pmatrix} = \begin{pmatrix}5&3\\3&2\end{pmatrix}$$

2A:

$$2A = \begin{pmatrix}4&2\\2&2\end{pmatrix}$$

3I:

$$3I = \begin{pmatrix}3&0\\0&3\end{pmatrix}$$

$$f(A) = \begin{pmatrix}5&3\\3&2\end{pmatrix}+\begin{pmatrix}4&2\\2&2\end{pmatrix}+\begin{pmatrix}3&0\\0&3\end{pmatrix} = \boxed{\begin{pmatrix}12&5\\5&7\end{pmatrix}}$$

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(b) Inverse of $A = \begin{pmatrix}1&5&2\\3&0&-2\\-1&0&3\end{pmatrix}$

det(A):

= 1(0·3 − (−2)·0) − 5(3·3 − (−2)(−1)) + 2(3·0 − 0·(−1))

= 1(0) − 5(9 − 2) + 2(0)

= 0 − 35 + 0 = −35

Matrix of Cofactors:

$$C_{11}=\begin{vmatrix}0&-2\\0&3\end{vmatrix}=0, \quad C_{12}=-\begin{vmatrix}3&-2\\-1&3\end{vmatrix}=-(9-2)=-7$$

$$C_{13}=\begin{vmatrix}3&0\\-1&0\end{vmatrix}=0$$

$$C_{21}=-\begin{vmatrix}5&2\\0&3\end{vmatrix}=-(15)=-15, \quad C_{22}=\begin{vmatrix}1&2\\-1&3\end{vmatrix}=3+2=5$$

$$C_{23}=-\begin{vmatrix}1&5\\-1&0\end{vmatrix}=-(0+5)=-5$$

$$C_{31}=\begin{vmatrix}5&2\\0&-2\end{vmatrix}=-10, \quad C_{32}=-\begin{vmatrix}1&2\\3&-2\end{vmatrix}=-(-2-6)=8$$

$$C_{33}=\begin{vmatrix}1&5\\3&0\end{vmatrix}=-15$$

Adjugate (transpose of cofactor matrix):

$$\text{adj}(A)=\begin{pmatrix}0&-15&-10\\-7&5&8\\0&-5&-15\end{pmatrix}$$

$$A^{-1}=\frac{1}{-35}\begin{pmatrix}0&-15&-10\\-7&5&8\\0&-5&-15\end{pmatrix} = \boxed{\begin{pmatrix}0&\frac{3}{7}&\frac{2}{7}\\\frac{1}{5}&-\frac{1}{7}&-\frac{8}{35}\\0&\frac{1}{7}&\frac{3}{7}\end{pmatrix}}$$

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©(i) Point dividing (3,−2) and (5,3) externally in ratio 1:3

External division formula: $\left(\frac{m x_2 - n x_1}{m-n},\ \frac{m y_2 - n y_1}{m-n}\right)$

$$x = \frac{1(5)-3(3)}{1-3}=\frac{5-9}{-2}=\frac{-4}{-2}=2$$

$$y = \frac{1(3)-3(-2)}{1-3}=\frac{3+6}{-2}=\frac{9}{-2}=-4.5$$

$$\boxed{(2,\ -4.5)}$$

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©(ii) Ratio in which x + y − 6 = 0 divides (−2, 8) and (4, 5)

Let ratio = k:1. The dividing point:

$$x=\frac{4k-2}{k+1},\quad y=\frac{5k+8}{k+1}$$

Substitute into x + y − 6 = 0:

$$\frac{4k-2}{k+1}+\frac{5k+8}{k+1}-6=0$$

$$4k-2+5k+8-6(k+1)=0$$

$$9k+6-6k-6=0$$

$$3k=0 \Rightarrow k=0$$

Hmm, let me recheck using the section formula approach with the line values:

Substituting A(−2,8): −2+8−6 = 0 ← A lies ON the line!

Let me recheck the question — it likely involves line joining (−2, 8) and (4, 5) divided by x + y − 6 = 0… but (−2+8−6=0), meaning A is on the line.

Likely the points are (−2, 3) and (4, 5) (possible misread due to image quality).

Using (−2, 3) and (4, 5), ratio k:1:

$$\frac{4k-2}{k+1}+\frac{5k+3}{k+1}=6$$

$$9k+1=6k+6 \Rightarrow 3k=5 \Rightarrow k=\frac{5}{3}$$

$$\boxed{\text{Ratio} = 5:3 \text{ internally}}$$

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Question 2

(a) Express $\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)^{17}$ in the form z = x + iy

Note: $\frac{\sqrt{3}}{2}=\cos30°$, $\frac{1}{2}=\sin30°$

So $z = \cos30°+i\sin30° = e^{i\pi/6}$

By De Moivre’s theorem:

$$z^{17}=\cos(17\times30°)+i\sin(17\times30°)=\cos510°+i\sin510°$$

$$510° = 360°+150°$$

$$\cos510°=\cos150°=-\frac{\sqrt{3}}{2},\quad \sin510°=\sin150°=\frac{1}{2}$$

$$\boxed{z^{17} = -\frac{\sqrt{3}}{2}+\frac{1}{2}i}$$

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(b) Show A(1,1), B(3,11), C(4,2), D(2,2) form a Parallelogram; find equations of AB, AD and angle between them

Midpoint of diagonal AC:

$$\left(\frac{1+4}{2},\frac{1+2}{2}\right)=\left(2.5,\ 1.5\right)$$

Midpoint of diagonal BD:

$$\left(\frac{3+2}{2},\frac{11+2}{2}\right)=\left(2.5,\ 6.5\right)$$

These midpoints are NOT equal — suggesting a possible misread. Let me use D(2,−2) (likely misread):

Midpoint BD: $\left(\frac{3+2}{2},\frac{11-2}{2}\right)=(2.5, 4.5)$ — still not equal.

Using the original points, let’s verify via vectors:

• $\vec{AB} = (2, 10)$

• $\vec{DC} = (4-2, 2-2) = (2, 0)$ ← not equal

Try B(3,1), C(4,2), D(2,2)… Image is unclear. Proceeding with given points and verifying via opposite sides:

• $\vec{AB}=(2,10)$, $\vec{DC}=(2,0)$ — not parallel with given points

The coordinates appear affected by image quality. Using what’s clearly readable:

Equation of line AB through (1,1) and (3,11):

$$m_{AB}=\frac{11-1}{3-1}=\frac{10}{2}=5$$

$$y-1=5(x-1) \Rightarrow \boxed{y=5x-4}$$

Equation of line AD through (1,1) and (2,2):

$$m_{AD}=\frac{2-1}{2-1}=1$$

$$\boxed{y=x}$$

Angle between AB and AD:

$$\tan\theta=\left|\frac{5-1}{1+5(1)}\right|=\left|\frac{4}{6}\right|=\frac{2}{3}$$

$$\theta=\tan^{-1}\left(\frac{2}{3}\right)\approx\boxed{33.69°}$$

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© Probability questions

Let P=Physics, C=Chemistry, B=Biology

Given:

• P§=0.7, P©=0.5, P(B)=0.4

• P(P∩C)=0.3, P(C∩B)=0.3, P(P∩B)=0.2

• P(P∪C∪B)=1 (faculty of science students)

Using inclusion-exclusion:

$$P(P\cup C\cup B)=0.7+0.5+0.4-0.3-0.3-0.2+P(P\cap C\cap B)$$

$$1=1.1-0.8+P(P\cap C\cap B)$$

$$P(P\cap C\cap B)=1-0.3=\boxed{0.3}$$

Wait: 0.7+0.5+0.4 = 1.6; 1.6−0.3−0.3−0.2 = 0.8

$$1=0.8+P(P\cap C\cap B) \Rightarrow P(P\cap C\cap B)=0.2$$

(i) All three subjects:

$$\boxed{P(P\cap C\cap B)=0.2}$$

(ii) Physics and Biology but NOT Chemistry:

$$P(P\cap B\cap C')=P(P\cap B)-P(P\cap B\cap C)$$

$$=0.2-0.2=\boxed{0}$$

(iii) P(P∩C) = 0.3 is given directly in the problem. ✓

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MAT 002: CALCULUS

Question 3

(b) Maclaurin expansion of f(x) = sin x

$$f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\cdots$$

| n | f⁽ⁿ⁾(x) | f⁽ⁿ⁾(0) |

|—|---------|---------|

| 0 | sin x | 0 |

| 1 | cos x | 1 |

| 2 | −sin x | 0 |

| 3 | −cos x | −1 |

| 4 | sin x | 0 |

$$\boxed{\sin x = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots}$$

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© Show gradient of 3x² + 2xy − 5y² = 10 at (2, 0) is −3

Differentiating implicitly:

$$6x+2y+2x\frac{dy}{dx}-10y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}(2x-10y)=-(6x+2y)$$

$$\frac{dy}{dx}=\frac{-(6x+2y)}{2x-10y}$$

At (2, 0):

$$\frac{dy}{dx}=\frac{-(12+0)}{4-0}=\frac{-12}{4}=\boxed{-3} \checkmark$$

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Question 4

(a)(i) $\lim_{x\to3}\frac{2x^2-27}{x+3}$

Direct substitution (no indeterminate form):

$$=\frac{2(9)-27}{3+3}=\frac{18-27}{6}=\frac{-9}{6}=\boxed{-\frac{3}{2}}$$

(a)(ii) $\lim_{y\to0}\frac{5y^3-3y^2+6y}{4y^2+3y}$

Factor y:

$$=\lim_{y\to0}\frac{y(5y^2-3y+6)}{y(4y+3)}=\lim_{y\to0}\frac{5y^2-3y+6}{4y+3}=\frac{0-0+6}{0+3}=\boxed{2}$$

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(b) Differentiate y = x/x² = 1/x from first principles

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$

$$=\lim_{h\to0}\frac{x-(x+h)}{h\cdot x(x+h)}=\lim_{h\to0}\frac{-h}{hx(x+h)}=\lim_{h\to0}\frac{-1}{x(x+h)}$$

$$=\boxed{-\frac{1}{x^2}}$$

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©(i) y = (2x³ − 4x² + 3x − 5)⁸

Let u = 2x³ − 4x² + 3x − 5

$$\frac{du}{dx}=6x^2-8x+3$$

$$\frac{dy}{dx}=8u^7\cdot\frac{du}{dx}=\boxed{8(2x^3-4x^2+3x-5)^7(6x^2-8x+3)}$$

©(ii) y = 2x²eˣ + ln x

Using product rule on 2x²eˣ:

$$\frac{d}{dx}(2x^2e^x)=4xe^x+2x^2e^x=2xe^x(2+x)$$

$$\frac{d}{dx}(\ln x)=\frac{1}{x}$$

$$\boxed{\frac{dy}{dx}=2xe^x(x+2)+\frac{1}{x}}$$

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MAT 003: STATISTICS

Question 5

(a) NSE Stocks Data — Frequency Distribution

Data range: 5.1 to 12.7, using class width 1.0

| Class | Tally | Frequency |

|-------|-------|-----------|

| 5.0–5.9 | II | 2 |

| 6.0–6.9 | I | 1 |

| 7.0–7.9 | IIII II | 7 |

| 8.0–8.9 | IIII IIII III | 13 |

| 9.0–9.9 | IIII IIII | 9 |

| 10.0–10.9 | IIII III | 8 |

| 11.0–11.9 | III | 3 |

| 12.0–12.9 | II | 2 |

| Total | | 45 |

(ii) Coefficient of Variation = (SD/Mean) × 100

Using midpoints (x): 5.5, 6.5, 7.5, 8.5, 9.5, 10.5, 11.5, 12.5

| Class | f | x | fx | fx² |

|-------|—|---|----|-----|

| 5.0–5.9 | 2 | 5.5 | 11 | 60.5 |

| 6.0–6.9 | 1 | 6.5 | 6.5 | 42.25 |

| 7.0–7.9 | 7 | 7.5 | 52.5 | 393.75 |

| 8.0–8.9 | 13 | 8.5 | 110.5 | 939.25 |

| 9.0–9.9 | 9 | 9.5 | 85.5 | 812.25 |

| 10.0–10.9 | 8 | 10.5 | 84 | 882 |

| 11.0–11.9 | 3 | 11.5 | 34.5 | 396.75 |

| 12.0–12.9 | 2 | 12.5 | 25 | 312.5 |

| Σ | 45 | | 409.5 | 3839.25 |

$$\bar{x}=\frac{409.5}{45}=9.1$$

$$s^2=\frac{\sum fx^2}{n}-\bar{x}^2=\frac{3839.25}{45}-9.1^2=85.317-82.81=2.507$$

$$s=\sqrt{2.507}\approx1.583$$

$$CV=\frac{1.583}{9.1}\times100=\boxed{17.4\%}$$

(iii) Since it’s the top 45 stocks selected from the NSE market, it represents a sample (not a population), as it is a subset chosen from all stocks.

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(b) Find a and b added to {2, 3, 6, 9}

Original mean: $\bar{x}_1=\frac{2+3+6+9}{4}=5$

New mean (6 numbers) increased by 1: $\bar{x}_2=6$

$$\frac{2+3+6+9+a+b}{6}=6 \Rightarrow 20+a+b=36 \Rightarrow a+b=16 \quad\cdots(1)$$

Original variance:

$$\sigma_1^2=\frac{4+9+36+81}{4}-25=\frac{130}{4}-25=32.5-25=7.5$$

New variance = 7.5 + 2.5 = 10

$$\frac{4+9+36+81+a^2+b^2}{6}-36=10$$

$$130+a^2+b^2=276 \Rightarrow a^2+b^2=146 \quad\cdots(2)$$

From (1): $(a+b)^2=256 \Rightarrow a^2+2ab+b^2=256$

From (2): $2ab=256-146=110 \Rightarrow ab=55$

So a and b are roots of: $t^2-16t+55=0$

$$(t-5)(t-11)=0$$

$$\boxed{a=5,\quad b=11}$$

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© ¹⁰Cᵣ = ¹⁰C₍ᵣ₎ (likely ¹⁰Cᵣ = ¹⁰C₍ₗₒ₋ᵣ₎ type problem)

If ¹⁰C₃ = ¹⁰Cᵣ, then either r = 3 or r = 10−3 = 7.

Since the original reads ¹⁰Cᵣ = ¹⁰Cᵣ (likely ¹⁰C₄ = ¹⁰Cᵣ or similar), using the identity nCₓ = nC₍ₙ₋ₓ₎:

$$r = 10 - r \Rightarrow 2r = 10 \Rightarrow \boxed{r = 5}$$

(Or the two values satisfying the complementary identity)

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Question 6

(b) Normal Distribution — Bulbs (μ = 1570, σ = 85)

i. P(X > 1950):

$$z=\frac{1950-1570}{85}=\frac{380}{85}=4.47$$

$$P(X>1950)=1-\Phi(4.47)\approx\boxed{0.000004 \approx 0}$$

ii. P(1730 < X < 1900):

$$z_1=\frac{1730-1570}{85}=\frac{160}{85}=1.88$$

$$z_2=\frac{1900-1570}{85}=\frac{330}{85}=3.88$$

$$P=\Phi(3.88)-\Phi(1.88)=0.99995-0.9699=\boxed{0.0300}$$

iii. If tested (sample size needed — not given; express as probability):

Expected number = n × P(X > 1900)

$$z=\frac{1900-1570}{85}=3.88$$

$$P(X>1900)=1-\Phi(3.88)=1-0.99995=0.00005$$

Per 1000 bulbs: 0.00005 × 1000 ≈ 0.05 bulbs (essentially none)

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© Show f(x) = c(⁴Cₓ)(¼)ˣ is a pdf for x = 0,1,2,3,4

For a pdf: $\sum_{x=0}^{4}f(x)=1$

$$\sum_{x=0}^{4}c\binom{4}{x}\left(\frac{1}{4}\right)^x$$

We recognize this relates to the binomial expansion of $(1+\frac{1}{4})^4$… but for a proper pdf we need:

$$\sum_{x=0}^{4}\binom{4}{x}\left(\frac{1}{4}\right)^x\left(\frac{3}{4}\right)^{4-x}=1$$

So likely f(x) = ⁴Cₓ(¼)ˣ(¾)⁴⁻ˣ (standard binomial). Given f(x) = c·⁴Cₓ·(¼)ˣ:

$$\sum_{x=0}^{4}\binom{4}{x}\left(\frac{1}{4}\right)^x = \left(1+\frac{1}{4}\right)^4=\left(\frac{5}{4}\right)^4=\frac{625}{256}$$

$$c=\frac{256}{625}$$

And since all f(x) ≥ 0 and Σf(x) = 1 (with this c), f(x) is a valid pdf. ✓

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MAT 004A: APPLIED MATHEMATICS

Question 7

(a) Express forces as column vectors

F₁(25N, 050°): $\begin{pmatrix}25\sin50°\\25\cos50°\end{pmatrix}=\begin{pmatrix}19.15\\16.07\end{pmatrix}$

F₂(30N, 150°): $\begin{pmatrix}30\sin150°\\30\cos150°\end{pmatrix}=\begin{pmatrix}15\\-25.98\end{pmatrix}$

F₃(35N, 240°): $\begin{pmatrix}35\sin240°\\35\cos240°\end{pmatrix}=\begin{pmatrix}-30.31\\-17.5\end{pmatrix}$

F₄(25N, 330°): $\begin{pmatrix}25\sin330°\\25\cos330°\end{pmatrix}=\begin{pmatrix}-12.5\\21.65\end{pmatrix}$

(b) Resultant:

$$R_x=19.15+15-30.31-12.5=\mathbf{-8.66}$$

$$R_y=16.07-25.98-17.5+21.65=\mathbf{-5.76}$$

$$\vec{R}=\begin{pmatrix}-8.66\\-5.76\end{pmatrix}$$

$$|R|=\sqrt{8.66^2+5.76^2}=\sqrt{75+33.18}=\sqrt{108.18}\approx\boxed{10.4\text{ N}}$$

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© Given Ā = î − j + 3k, B̄ = 2î + 4j − 6k, C̄ = 3î − 5j + 2k

(i) Ā × B̄:

$$\vec{A}\times\vec{B}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&4&-6\end{vmatrix}$$

$$=\hat{i}[(-1)(-6)-(3)(4)]-\hat{j}[(1)(-6)-(3)(2)]+\hat{k}[(1)(4)-(-1)(2)]$$

$$=\hat{i}[6-12]-\hat{j}[-6-6]+\hat{k}[4+2]$$

$$=\boxed{-6\hat{i}+12\hat{j}+6\hat{k}}$$

(ii) Ā · (B̄ × C̄) — scalar triple product = det:

$$\begin{vmatrix}1&-1&3\\2&4&-6\\3&-5&2\end{vmatrix}$$

$$=1(4\cdot2-(-6)(-5))-(-1)(2\cdot2-(-6)(3))+3(2(-5)-4(3))$$

$$=1(8-30)+1(4+18)+3(-10-12)$$

$$=-22+22-66=\boxed{-66}$$

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Question 8

(a) Angle between forces 19N and 21N with resultant 27N

Using cosine rule:

$$R^2=F_1^2+F_2^2+2F_1F_2\cos\theta$$

$$729=361+441+2(19)(21)\cos\theta$$

$$729=802+798\cos\theta$$

$$\cos\theta=\frac{729-802}{798}=\frac{-73}{798}=-0.09147$$

$$\theta=\cos^{-1}(-0.09147)\approx\boxed{95.2°}$$

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(b) Mass of 4kg, strings at 30° and 45°

Weight W = 4 × 10 = 40N

Resolving vertically: T₁sin30° + T₂sin45° = 40

$$0.5T_1+0.7071T_2=40 \quad\cdots(1)$$

Resolving horizontally: T₁cos30° = T₂cos45°

$$0.8660T_1=0.7071T_2 \Rightarrow T_2=\frac{0.8660}{0.7071}T_1=1.2247T_1 \quad\cdots(2)$$

Substitute (2) into (1):

$$0.5T_1+0.7071(1.2247T_1)=40$$

$$0.5T_1+0.8660T_1=40$$

$$1.366T_1=40 \Rightarrow \boxed{T_1=29.3\text{ N}}$$

$$T_2=1.2247\times29.3=\boxed{35.9\text{ N}}$$

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© Uniform bar 40kg, 10m long, weights 25N at one end, 30N at other

Bar weight = 40×10 = 400N acting at centre (5m from each end).

Let pivot be at distance x from end A (where 25N hangs).

Taking moments about pivot:

$$25x + 400(x-5) = 30(10-x)$$

$$25x+400x-2000=300-30x$$

$$455x=2300$$

$$x=\frac{2300}{455}\approx\boxed{5.05\text{ m from end A}}$$

ii. Reaction at pivot:

$$R = 25+400+30=\boxed{455\text{ N}}$$

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MAT 004B: APPLIED BUSINESS MATHEMATICS

Question 9

(a) q₁ = 30 + 2p₂ − p₁, p₁ = 7, p₂ = 9

At these prices: q₁ = 30 + 2(9) − 7 = 30 + 18 − 7 = 41

i. Price elasticity of demand for beans:

$$E_{p_1}=\frac{\partial q_1}{\partial p_1}\cdot\frac{p_1}{q_1}=(-1)\cdot\frac{7}{41}=\boxed{-0.171}$$

ii. Cross elasticity of demand:

$$E_{p_2}=\frac{\partial q_1}{\partial p_2}\cdot\frac{p_2}{q_1}=(2)\cdot\frac{9}{41}=\boxed{0.439}$$

(Positive cross elasticity → substitutes)

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(b) Compound interest at 4% annually

i. Sum becomes 4 times itself:

$$4P=P(1.04)^n \Rightarrow (1.04)^n=4$$

$$n\ln(1.04)=\ln4$$

$$n=\frac{\ln4}{\ln1.04}=\frac{1.3863}{0.03922}\approx\boxed{35\text{ years}}$$

ii. Sum doubles in 10 years:

$$2=(1+r)^{10}$$

$$(1+r)=2^{0.1}=1.07177$$

$$r=0.07177\approx\boxed{7\%}$$

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© Maximize q = 12u + 156 subject to 4u + 3B ≤ 20, u ≥ 0, B ≥ 0

Corner points:

• (0, 0): q = 156

• (5, 0): q = 12(5)+156 = 216

• (0, 6.67): q = 12(0)+156 = 156

$$\boxed{\text{Maximum } q = 216 \text{ at } u=5,\ B=0}$$

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Question 10

Pharmaceutical Company LP Problem

Let x = kg of Chemical P, y = kg of Chemical Y

Constraints from reading:

• Chemical component constraints (units required ≥ minimum)

• P ≥ 7 units, Q ≥ 11 units, R ≥ 11 units

Based on supplies:

• Company A per kg: 1P, 2Q, 3R at cost ₦3

• Company B per kg: 2P, 4Q, 6R at ₦64

Let a = kg from A, b = kg from B:

Constraints:

$$a+2b\geq7 \quad\text{(Chemical P)}$$

$$2a+4b\geq11 \quad\text{(Chemical Q)}$$

$$3a+6b\geq11 \quad\text{(Chemical R)}$$

$$a\geq0,\quad b\geq0$$

Objective: Minimize Cost = 3a + 64b

Graphical solution:

From constraint 1: a ≥ 7 − 2b

Corner points (checking intersections):

• Set a + 2b = 7 and 2a + 4b = 11:

• 2(7−2b)+4b = 11 → 14 = 11 (inconsistent → parallel lines)

• Use a + 2b = 7 and 3a + 6b = 11:

• 3(7−2b)+6b = 11 → 21 = 11 (also inconsistent)

So constraint 1 is the binding one. Minimize along a + 2b = 7:

$$\text{Cost}=3(7-2b)+64b=21+58b$$

This is minimized at b = 0, giving a = 7:

$$\boxed{\text{Minimum cost} = 3(7)+64(0) = ₦21}$$

At corner vertex (a = 7, b = 0), the pharmaceutical company minimizes cost.

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(iv) Compound interest — ₦6,900 → ₦2,000?

(Likely ₦600 → ₦2,000 in 4 years, half-yearly)

$$A=P\left(1+\frac{r}{2}\right)^{2n}$$

$$2000=600\left(1+\frac{r}{2}\right)^{8}$$

$$\left(1+\frac{r}{2}\right)^8=\frac{2000}{600}=3.333$$

$$1+\frac{r}{2}=3.333^{1/8}=1.1616$$

$$\frac{r}{2}=0.1616 \Rightarrow r=0.3232$$

$$\boxed{r\approx32.3\%\text{ per annum}}$$