MAT 001: ADVANCED PURE MATHEMATICS
1. (a) Let A = ( 2 1 1 1 ) \begin{pmatrix} 2 & 1 \ 1 & 1 \end{pmatrix} ( 2 1 1 1 ) [3 Marks]
(b) Compute the inverse of the matrix A = ( 1 5 2 3 0 − 2 − 1 0 3 ) A = \begin{pmatrix} 1 & 5 & 2 \ 3 & 0 & -2 \ -1 & 0 & 3 \end{pmatrix} A = ( 1 5 2 3 0 − 2 − 1 0 3 ) [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]
2. (a) Express ( 3 2 + 1 2 ) 17 \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)^{17} ( 2 3 + 2 1 ) 17 [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]
MAT 002: CALCULUS
3. (a) Prove that the area enclosed by the curve y 2 = x 2 4 x − 2 y^2 = \frac{x^2}{4x-2} y 2 = 4 x − 2 x 2 [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]
4. (a) Evaluate the following limits:
i. lim x → 3 ( 2 x 2 − 27 x + 3 ) \lim_{x \to 3} \left(\frac{2x^2 - 27}{x + 3}\right) lim x → 3 ( x + 3 2 x 2 − 27 ) [2 Marks]
ii. lim y → 0 5 y 3 − 3 y 2 + 6 y 4 y 2 + 3 y \lim_{y \to 0} \frac{5y^3 - 3y^2 + 6y}{4y^2 + 3y} lim y → 0 4 y 2 + 3 y 5 y 3 − 3 y 2 + 6 y [6 Marks]
(b) Differentiate y = x x 2 y = \frac{x}{x^2} y = x 2 x [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]
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
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]
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]
(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]
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]
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]
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]
© Maximize the function q = 12u + 156 subject to the constraints 4u + 3B ≤ 20. Where u ≥ 0, B ≥ 0. [4 Marks]
[TOTAL = 18 Marks]
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 = ( 2 1 1 1 ) \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} ( 2 1 1 1 )
f(A) = A² + 2A + 3I
A²:
A 2 = ( 2 1 1 1 ) ( 2 1 1 1 ) = ( 5 3 3 2 ) 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} A 2 = ( 2 1 1 1 ) ( 2 1 1 1 ) = ( 5 3 3 2 )
2A:
2 A = ( 4 2 2 2 ) 2A = \begin{pmatrix}4&2\\2&2\end{pmatrix} 2 A = ( 4 2 2 2 )
3I:
3 I = ( 3 0 0 3 ) 3I = \begin{pmatrix}3&0\\0&3\end{pmatrix} 3 I = ( 3 0 0 3 )
f ( A ) = ( 5 3 3 2 ) + ( 4 2 2 2 ) + ( 3 0 0 3 ) = ( 12 5 5 7 ) 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}} f ( A ) = ( 5 3 3 2 ) + ( 4 2 2 2 ) + ( 3 0 0 3 ) = ( 12 5 5 7 )
(b) Inverse of A = ( 1 5 2 3 0 − 2 − 1 0 3 ) A = \begin{pmatrix}1&5&2\\3&0&-2\\-1&0&3\end{pmatrix} A = ⎝ ⎛ 1 3 − 1 5 0 0 2 − 2 3 ⎠ ⎞
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 = ∣ 0 − 2 0 3 ∣ = 0 , C 12 = − ∣ 3 − 2 − 1 3 ∣ = − ( 9 − 2 ) = − 7 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 11 = ∣ ∣ 0 0 − 2 3 ∣ ∣ = 0 , C 12 = − ∣ ∣ 3 − 1 − 2 3 ∣ ∣ = − ( 9 − 2 ) = − 7
C 13 = ∣ 3 0 − 1 0 ∣ = 0 C_{13}=\begin{vmatrix}3&0\\-1&0\end{vmatrix}=0 C 13 = ∣ ∣ 3 − 1 0 0 ∣ ∣ = 0
C 21 = − ∣ 5 2 0 3 ∣ = − ( 15 ) = − 15 , C 22 = ∣ 1 2 − 1 3 ∣ = 3 + 2 = 5 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 21 = − ∣ ∣ 5 0 2 3 ∣ ∣ = − ( 15 ) = − 15 , C 22 = ∣ ∣ 1 − 1 2 3 ∣ ∣ = 3 + 2 = 5
C 23 = − ∣ 1 5 − 1 0 ∣ = − ( 0 + 5 ) = − 5 C_{23}=-\begin{vmatrix}1&5\\-1&0\end{vmatrix}=-(0+5)=-5 C 23 = − ∣ ∣ 1 − 1 5 0 ∣ ∣ = − ( 0 + 5 ) = − 5
C 31 = ∣ 5 2 0 − 2 ∣ = − 10 , C 32 = − ∣ 1 2 3 − 2 ∣ = − ( − 2 − 6 ) = 8 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 31 = ∣ ∣ 5 0 2 − 2 ∣ ∣ = − 10 , C 32 = − ∣ ∣ 1 3 2 − 2 ∣ ∣ = − ( − 2 − 6 ) = 8
C 33 = ∣ 1 5 3 0 ∣ = − 15 C_{33}=\begin{vmatrix}1&5\\3&0\end{vmatrix}=-15 C 33 = ∣ ∣ 1 3 5 0 ∣ ∣ = − 15
Adjugate (transpose of cofactor matrix):
adj ( A ) = ( 0 − 15 − 10 − 7 5 8 0 − 5 − 15 ) \text{adj}(A)=\begin{pmatrix}0&-15&-10\\-7&5&8\\0&-5&-15\end{pmatrix} adj ( A ) = ⎝ ⎛ 0 − 7 0 − 15 5 − 5 − 10 8 − 15 ⎠ ⎞
A − 1 = 1 − 35 ( 0 − 15 − 10 − 7 5 8 0 − 5 − 15 ) = ( 0 3 7 2 7 1 5 − 1 7 − 8 35 0 1 7 3 7 ) 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}} A − 1 = − 35 1 ⎝ ⎛ 0 − 7 0 − 15 5 − 5 − 10 8 − 15 ⎠ ⎞ = ⎝ ⎛ 0 5 1 0 7 3 − 7 1 7 1 7 2 − 35 8 7 3 ⎠ ⎞
©(i) Point dividing (3,−2) and (5,3) externally in ratio 1:3
External division formula: ( m x 2 − n x 1 m − n , m y 2 − n y 1 m − n ) \left(\frac{m x_2 - n x_1}{m-n},\ \frac{m y_2 - n y_1}{m-n}\right) ( m − n m x 2 − n x 1 , m − n m y 2 − n y 1 )
x = 1 ( 5 ) − 3 ( 3 ) 1 − 3 = 5 − 9 − 2 = − 4 − 2 = 2 x = \frac{1(5)-3(3)}{1-3}=\frac{5-9}{-2}=\frac{-4}{-2}=2 x = 1 − 3 1 ( 5 ) − 3 ( 3 ) = − 2 5 − 9 = − 2 − 4 = 2
y = 1 ( 3 ) − 3 ( − 2 ) 1 − 3 = 3 + 6 − 2 = 9 − 2 = − 4.5 y = \frac{1(3)-3(-2)}{1-3}=\frac{3+6}{-2}=\frac{9}{-2}=-4.5 y = 1 − 3 1 ( 3 ) − 3 ( − 2 ) = − 2 3 + 6 = − 2 9 = − 4.5
( 2 , − 4.5 ) \boxed{(2,\ -4.5)} ( 2 , − 4.5 )
©(ii) Ratio in which x + y − 6 = 0 divides (−2, 8) and (4, 5)
Let ratio = k:1. The dividing point:
x = 4 k − 2 k + 1 , y = 5 k + 8 k + 1 x=\frac{4k-2}{k+1},\quad y=\frac{5k+8}{k+1} x = k + 1 4 k − 2 , y = k + 1 5 k + 8
Substitute into x + y − 6 = 0:
4 k − 2 k + 1 + 5 k + 8 k + 1 − 6 = 0 \frac{4k-2}{k+1}+\frac{5k+8}{k+1}-6=0 k + 1 4 k − 2 + k + 1 5 k + 8 − 6 = 0
4 k − 2 + 5 k + 8 − 6 ( k + 1 ) = 0 4k-2+5k+8-6(k+1)=0 4 k − 2 + 5 k + 8 − 6 ( k + 1 ) = 0
9 k + 6 − 6 k − 6 = 0 9k+6-6k-6=0 9 k + 6 − 6 k − 6 = 0
3 k = 0 ⇒ k = 0 3k=0 \Rightarrow k=0 3 k = 0 ⇒ 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:
4 k − 2 k + 1 + 5 k + 3 k + 1 = 6 \frac{4k-2}{k+1}+\frac{5k+3}{k+1}=6 k + 1 4 k − 2 + k + 1 5 k + 3 = 6
9 k + 1 = 6 k + 6 ⇒ 3 k = 5 ⇒ k = 5 3 9k+1=6k+6 \Rightarrow 3k=5 \Rightarrow k=\frac{5}{3} 9 k + 1 = 6 k + 6 ⇒ 3 k = 5 ⇒ k = 3 5
Ratio = 5 : 3 internally \boxed{\text{Ratio} = 5:3 \text{ internally}} Ratio = 5 : 3 internally
Question 2
Note: 3 2 = cos 30 ° \frac{\sqrt{3}}{2}=\cos30° 2 3 = cos 30° 1 2 = sin 30 ° \frac{1}{2}=\sin30° 2 1 = sin 30°
So z = cos 30 ° + i sin 30 ° = e i π / 6 z = \cos30°+i\sin30° = e^{i\pi/6} z = cos 30° + i sin 30° = e iπ /6
By De Moivre’s theorem:
z 17 = cos ( 17 × 30 ° ) + i sin ( 17 × 30 ° ) = cos 510 ° + i sin 510 ° z^{17}=\cos(17\times30°)+i\sin(17\times30°)=\cos510°+i\sin510° z 17 = cos ( 17 × 30° ) + i sin ( 17 × 30° ) = cos 510° + i sin 510°
510 ° = 360 ° + 150 ° 510° = 360°+150° 510° = 360° + 150°
cos 510 ° = cos 150 ° = − 3 2 , sin 510 ° = sin 150 ° = 1 2 \cos510°=\cos150°=-\frac{\sqrt{3}}{2},\quad \sin510°=\sin150°=\frac{1}{2} cos 510° = cos 150° = − 2 3 , sin 510° = sin 150° = 2 1
z 17 = − 3 2 + 1 2 i \boxed{z^{17} = -\frac{\sqrt{3}}{2}+\frac{1}{2}i} z 17 = − 2 3 + 2 1 i
Midpoint of diagonal AC:
( 1 + 4 2 , 1 + 2 2 ) = ( 2.5 , 1.5 ) \left(\frac{1+4}{2},\frac{1+2}{2}\right)=\left(2.5,\ 1.5\right) ( 2 1 + 4 , 2 1 + 2 ) = ( 2.5 , 1.5 )
Midpoint of diagonal BD:
( 3 + 2 2 , 11 + 2 2 ) = ( 2.5 , 6.5 ) \left(\frac{3+2}{2},\frac{11+2}{2}\right)=\left(2.5,\ 6.5\right) ( 2 3 + 2 , 2 11 + 2 ) = ( 2.5 , 6.5 )
These midpoints are NOT equal — suggesting a possible misread. Let me use D(2,−2) (likely misread):
Midpoint BD: ( 3 + 2 2 , 11 − 2 2 ) = ( 2.5 , 4.5 ) \left(\frac{3+2}{2},\frac{11-2}{2}\right)=(2.5, 4.5) ( 2 3 + 2 , 2 11 − 2 ) = ( 2.5 , 4.5 )
Using the original points, let’s verify via vectors:
A B ⃗ = ( 2 , 10 ) \vec{AB} = (2, 10) A B = ( 2 , 10 )
D C ⃗ = ( 4 − 2 , 2 − 2 ) = ( 2 , 0 ) \vec{DC} = (4-2, 2-2) = (2, 0) D C = ( 4 − 2 , 2 − 2 ) = ( 2 , 0 )
Try B(3,1), C(4,2), D(2,2) … Image is unclear. Proceeding with given points and verifying via opposite sides:
A B ⃗ = ( 2 , 10 ) \vec{AB}=(2,10) A B = ( 2 , 10 ) D C ⃗ = ( 2 , 0 ) \vec{DC}=(2,0) D C = ( 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 A B = 11 − 1 3 − 1 = 10 2 = 5 m_{AB}=\frac{11-1}{3-1}=\frac{10}{2}=5 m A B = 3 − 1 11 − 1 = 2 10 = 5
y − 1 = 5 ( x − 1 ) ⇒ y = 5 x − 4 y-1=5(x-1) \Rightarrow \boxed{y=5x-4} y − 1 = 5 ( x − 1 ) ⇒ y = 5 x − 4
Equation of line AD through (1,1) and (2,2):
m A D = 2 − 1 2 − 1 = 1 m_{AD}=\frac{2-1}{2-1}=1 m A D = 2 − 1 2 − 1 = 1
y = x \boxed{y=x} y = x
Angle between AB and AD:
tan θ = ∣ 5 − 1 1 + 5 ( 1 ) ∣ = ∣ 4 6 ∣ = 2 3 \tan\theta=\left|\frac{5-1}{1+5(1)}\right|=\left|\frac{4}{6}\right|=\frac{2}{3} tan θ = ∣ ∣ 1 + 5 ( 1 ) 5 − 1 ∣ ∣ = ∣ ∣ 6 4 ∣ ∣ = 3 2
θ = tan − 1 ( 2 3 ) ≈ 33.69 ° \theta=\tan^{-1}\left(\frac{2}{3}\right)\approx\boxed{33.69°} θ = tan − 1 ( 3 2 ) ≈ 33.69°
© 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 ∪ C ∪ B ) = 0.7 + 0.5 + 0.4 − 0.3 − 0.3 − 0.2 + P ( P ∩ C ∩ B ) P(P\cup C\cup B)=0.7+0.5+0.4-0.3-0.3-0.2+P(P\cap C\cap B) P ( P ∪ C ∪ B ) = 0.7 + 0.5 + 0.4 − 0.3 − 0.3 − 0.2 + P ( P ∩ C ∩ B )
1 = 1.1 − 0.8 + P ( P ∩ C ∩ B ) 1=1.1-0.8+P(P\cap C\cap B) 1 = 1.1 − 0.8 + P ( P ∩ C ∩ B )
P ( P ∩ C ∩ B ) = 1 − 0.3 = 0.3 P(P\cap C\cap B)=1-0.3=\boxed{0.3} P ( P ∩ C ∩ B ) = 1 − 0.3 = 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 ∩ C ∩ B ) ⇒ P ( P ∩ C ∩ B ) = 0.2 1=0.8+P(P\cap C\cap B) \Rightarrow P(P\cap C\cap B)=0.2 1 = 0.8 + P ( P ∩ C ∩ B ) ⇒ P ( P ∩ C ∩ B ) = 0.2
(i) All three subjects:
P ( P ∩ C ∩ B ) = 0.2 \boxed{P(P\cap C\cap B)=0.2} P ( P ∩ C ∩ B ) = 0.2
(ii) Physics and Biology but NOT Chemistry:
P ( P ∩ B ∩ C ′ ) = P ( P ∩ B ) − P ( P ∩ B ∩ C ) P(P\cap B\cap C')=P(P\cap B)-P(P\cap B\cap C) P ( P ∩ B ∩ C ′ ) = P ( P ∩ B ) − P ( P ∩ B ∩ C )
= 0.2 − 0.2 = 0 =0.2-0.2=\boxed{0} = 0.2 − 0.2 = 0
(iii) P(P∩C) = 0.3 is given directly in the problem. ✓
MAT 002: CALCULUS
Question 3
(b) Maclaurin expansion of f(x) = sin x
f ( x ) = f ( 0 ) + x f ′ ( 0 ) + x 2 2 ! f ′ ′ ( 0 ) + x 3 3 ! f ′ ′ ′ ( 0 ) + ⋯ f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\cdots f ( x ) = f ( 0 ) + x f ′ ( 0 ) + 2 ! x 2 f ′′ ( 0 ) + 3 ! x 3 f ′′′ ( 0 ) + ⋯
| 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 |
sin x = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + ⋯ \boxed{\sin x = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots} sin x = x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + ⋯
© Show gradient of 3x² + 2xy − 5y² = 10 at (2, 0) is −3
Differentiating implicitly:
6 x + 2 y + 2 x d y d x − 10 y d y d x = 0 6x+2y+2x\frac{dy}{dx}-10y\frac{dy}{dx}=0 6 x + 2 y + 2 x d x d y − 10 y d x d y = 0
d y d x ( 2 x − 10 y ) = − ( 6 x + 2 y ) \frac{dy}{dx}(2x-10y)=-(6x+2y) d x d y ( 2 x − 10 y ) = − ( 6 x + 2 y )
d y d x = − ( 6 x + 2 y ) 2 x − 10 y \frac{dy}{dx}=\frac{-(6x+2y)}{2x-10y} d x d y = 2 x − 10 y − ( 6 x + 2 y )
At (2, 0):
d y d x = − ( 12 + 0 ) 4 − 0 = − 12 4 = − 3 ✓ \frac{dy}{dx}=\frac{-(12+0)}{4-0}=\frac{-12}{4}=\boxed{-3} \checkmark d x d y = 4 − 0 − ( 12 + 0 ) = 4 − 12 = − 3 ✓
Question 4
(a)(i) lim x → 3 2 x 2 − 27 x + 3 \lim_{x\to3}\frac{2x^2-27}{x+3} lim x → 3 x + 3 2 x 2 − 27
Direct substitution (no indeterminate form):
= 2 ( 9 ) − 27 3 + 3 = 18 − 27 6 = − 9 6 = − 3 2 =\frac{2(9)-27}{3+3}=\frac{18-27}{6}=\frac{-9}{6}=\boxed{-\frac{3}{2}} = 3 + 3 2 ( 9 ) − 27 = 6 18 − 27 = 6 − 9 = − 2 3
(a)(ii) lim y → 0 5 y 3 − 3 y 2 + 6 y 4 y 2 + 3 y \lim_{y\to0}\frac{5y^3-3y^2+6y}{4y^2+3y} lim y → 0 4 y 2 + 3 y 5 y 3 − 3 y 2 + 6 y
Factor y:
= lim y → 0 y ( 5 y 2 − 3 y + 6 ) y ( 4 y + 3 ) = lim y → 0 5 y 2 − 3 y + 6 4 y + 3 = 0 − 0 + 6 0 + 3 = 2 =\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} = y → 0 lim y ( 4 y + 3 ) y ( 5 y 2 − 3 y + 6 ) = y → 0 lim 4 y + 3 5 y 2 − 3 y + 6 = 0 + 3 0 − 0 + 6 = 2
(b) Differentiate y = x/x² = 1/x from first principles
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 1 x + h − 1 x h f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h} f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) = h → 0 lim h x + h 1 − x 1
= lim h → 0 x − ( x + h ) h ⋅ x ( x + h ) = lim h → 0 − h h x ( x + h ) = lim h → 0 − 1 x ( 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)} = h → 0 lim h ⋅ x ( x + h ) x − ( x + h ) = h → 0 lim h x ( x + h ) − h = h → 0 lim x ( x + h ) − 1
= − 1 x 2 =\boxed{-\frac{1}{x^2}} = − x 2 1
©(i) y = (2x³ − 4x² + 3x − 5)⁸
Let u = 2x³ − 4x² + 3x − 5
d u d x = 6 x 2 − 8 x + 3 \frac{du}{dx}=6x^2-8x+3 d x d u = 6 x 2 − 8 x + 3
d y d x = 8 u 7 ⋅ d u d x = 8 ( 2 x 3 − 4 x 2 + 3 x − 5 ) 7 ( 6 x 2 − 8 x + 3 ) \frac{dy}{dx}=8u^7\cdot\frac{du}{dx}=\boxed{8(2x^3-4x^2+3x-5)^7(6x^2-8x+3)} d x d y = 8 u 7 ⋅ d x d u = 8 ( 2 x 3 − 4 x 2 + 3 x − 5 ) 7 ( 6 x 2 − 8 x + 3 )
©(ii) y = 2x²eˣ + ln x
Using product rule on 2x²eˣ:
d d x ( 2 x 2 e x ) = 4 x e x + 2 x 2 e x = 2 x e x ( 2 + x ) \frac{d}{dx}(2x^2e^x)=4xe^x+2x^2e^x=2xe^x(2+x) d x d ( 2 x 2 e x ) = 4 x e x + 2 x 2 e x = 2 x e x ( 2 + x )
d d x ( ln x ) = 1 x \frac{d}{dx}(\ln x)=\frac{1}{x} d x d ( ln x ) = x 1
d y d x = 2 x e x ( x + 2 ) + 1 x \boxed{\frac{dy}{dx}=2xe^x(x+2)+\frac{1}{x}} d x d y = 2 x e x ( x + 2 ) + x 1
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 |
x ˉ = 409.5 45 = 9.1 \bar{x}=\frac{409.5}{45}=9.1 x ˉ = 45 409.5 = 9.1
s 2 = ∑ f x 2 n − x ˉ 2 = 3839.25 45 − 9. 1 2 = 85.317 − 82.81 = 2.507 s^2=\frac{\sum fx^2}{n}-\bar{x}^2=\frac{3839.25}{45}-9.1^2=85.317-82.81=2.507 s 2 = n ∑ f x 2 − x ˉ 2 = 45 3839.25 − 9. 1 2 = 85.317 − 82.81 = 2.507
s = 2.507 ≈ 1.583 s=\sqrt{2.507}\approx1.583 s = 2.507 ≈ 1.583
C V = 1.583 9.1 × 100 = 17.4 % CV=\frac{1.583}{9.1}\times100=\boxed{17.4\%} C V = 9.1 1.583 × 100 = 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.
(b) Find a and b added to {2, 3, 6, 9}
Original mean: x ˉ 1 = 2 + 3 + 6 + 9 4 = 5 \bar{x}_1=\frac{2+3+6+9}{4}=5 x ˉ 1 = 4 2 + 3 + 6 + 9 = 5
New mean (6 numbers) increased by 1: x ˉ 2 = 6 \bar{x}_2=6 x ˉ 2 = 6
2 + 3 + 6 + 9 + a + b 6 = 6 ⇒ 20 + a + b = 36 ⇒ a + b = 16 ⋯ ( 1 ) \frac{2+3+6+9+a+b}{6}=6 \Rightarrow 20+a+b=36 \Rightarrow a+b=16 \quad\cdots(1) 6 2 + 3 + 6 + 9 + a + b = 6 ⇒ 20 + a + b = 36 ⇒ a + b = 16 ⋯ ( 1 )
Original variance:
σ 1 2 = 4 + 9 + 36 + 81 4 − 25 = 130 4 − 25 = 32.5 − 25 = 7.5 \sigma_1^2=\frac{4+9+36+81}{4}-25=\frac{130}{4}-25=32.5-25=7.5 σ 1 2 = 4 4 + 9 + 36 + 81 − 25 = 4 130 − 25 = 32.5 − 25 = 7.5
New variance = 7.5 + 2.5 = 10
4 + 9 + 36 + 81 + a 2 + b 2 6 − 36 = 10 \frac{4+9+36+81+a^2+b^2}{6}-36=10 6 4 + 9 + 36 + 81 + a 2 + b 2 − 36 = 10
130 + a 2 + b 2 = 276 ⇒ a 2 + b 2 = 146 ⋯ ( 2 ) 130+a^2+b^2=276 \Rightarrow a^2+b^2=146 \quad\cdots(2) 130 + a 2 + b 2 = 276 ⇒ a 2 + b 2 = 146 ⋯ ( 2 )
From (1): ( a + b ) 2 = 256 ⇒ a 2 + 2 a b + b 2 = 256 (a+b)^2=256 \Rightarrow a^2+2ab+b^2=256 ( a + b ) 2 = 256 ⇒ a 2 + 2 ab + b 2 = 256
From (2): 2 a b = 256 − 146 = 110 ⇒ a b = 55 2ab=256-146=110 \Rightarrow ab=55 2 ab = 256 − 146 = 110 ⇒ ab = 55
So a and b are roots of: t 2 − 16 t + 55 = 0 t^2-16t+55=0 t 2 − 16 t + 55 = 0
( t − 5 ) ( t − 11 ) = 0 (t-5)(t-11)=0 ( t − 5 ) ( t − 11 ) = 0
a = 5 , b = 11 \boxed{a=5,\quad b=11} a = 5 , b = 11
© 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 ⇒ 2 r = 10 ⇒ r = 5 r = 10 - r \Rightarrow 2r = 10 \Rightarrow \boxed{r = 5} r = 10 − r ⇒ 2 r = 10 ⇒ r = 5
(Or the two values satisfying the complementary identity)
Question 6
(b) Normal Distribution — Bulbs (μ = 1570, σ = 85)
i. P(X > 1950):
z = 1950 − 1570 85 = 380 85 = 4.47 z=\frac{1950-1570}{85}=\frac{380}{85}=4.47 z = 85 1950 − 1570 = 85 380 = 4.47
P ( X > 1950 ) = 1 − Φ ( 4.47 ) ≈ 0.000004 ≈ 0 P(X>1950)=1-\Phi(4.47)\approx\boxed{0.000004 \approx 0} P ( X > 1950 ) = 1 − Φ ( 4.47 ) ≈ 0.000004 ≈ 0
ii. P(1730 < X < 1900):
z 1 = 1730 − 1570 85 = 160 85 = 1.88 z_1=\frac{1730-1570}{85}=\frac{160}{85}=1.88 z 1 = 85 1730 − 1570 = 85 160 = 1.88
z 2 = 1900 − 1570 85 = 330 85 = 3.88 z_2=\frac{1900-1570}{85}=\frac{330}{85}=3.88 z 2 = 85 1900 − 1570 = 85 330 = 3.88
P = Φ ( 3.88 ) − Φ ( 1.88 ) = 0.99995 − 0.9699 = 0.0300 P=\Phi(3.88)-\Phi(1.88)=0.99995-0.9699=\boxed{0.0300} P = Φ ( 3.88 ) − Φ ( 1.88 ) = 0.99995 − 0.9699 = 0.0300
iii. If tested (sample size needed — not given; express as probability):
Expected number = n × P(X > 1900)
z = 1900 − 1570 85 = 3.88 z=\frac{1900-1570}{85}=3.88 z = 85 1900 − 1570 = 3.88
P ( X > 1900 ) = 1 − Φ ( 3.88 ) = 1 − 0.99995 = 0.00005 P(X>1900)=1-\Phi(3.88)=1-0.99995=0.00005 P ( X > 1900 ) = 1 − Φ ( 3.88 ) = 1 − 0.99995 = 0.00005
Per 1000 bulbs: 0.00005 × 1000 ≈ 0.05 bulbs (essentially none)
© Show f(x) = c(⁴Cₓ)(¼)ˣ is a pdf for x = 0,1,2,3,4
For a pdf: ∑ x = 0 4 f ( x ) = 1 \sum_{x=0}^{4}f(x)=1 ∑ x = 0 4 f ( x ) = 1
∑ x = 0 4 c ( 4 x ) ( 1 4 ) x \sum_{x=0}^{4}c\binom{4}{x}\left(\frac{1}{4}\right)^x x = 0 ∑ 4 c ( x 4 ) ( 4 1 ) x
We recognize this relates to the binomial expansion of ( 1 + 1 4 ) 4 (1+\frac{1}{4})^4 ( 1 + 4 1 ) 4
∑ x = 0 4 ( 4 x ) ( 1 4 ) x ( 3 4 ) 4 − x = 1 \sum_{x=0}^{4}\binom{4}{x}\left(\frac{1}{4}\right)^x\left(\frac{3}{4}\right)^{4-x}=1 x = 0 ∑ 4 ( x 4 ) ( 4 1 ) x ( 4 3 ) 4 − x = 1
So likely f(x) = ⁴Cₓ(¼)ˣ(¾)⁴⁻ˣ (standard binomial). Given f(x) = c·⁴Cₓ·(¼)ˣ:
∑ x = 0 4 ( 4 x ) ( 1 4 ) x = ( 1 + 1 4 ) 4 = ( 5 4 ) 4 = 625 256 \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} x = 0 ∑ 4 ( x 4 ) ( 4 1 ) x = ( 1 + 4 1 ) 4 = ( 4 5 ) 4 = 256 625
c = 256 625 c=\frac{256}{625} c = 625 256
And since all f(x) ≥ 0 and Σf(x) = 1 (with this c), f(x) is a valid pdf. ✓
MAT 004A: APPLIED MATHEMATICS
Question 7
(a) Express forces as column vectors
F₁(25N, 050°): ( 25 sin 50 ° 25 cos 50 ° ) = ( 19.15 16.07 ) \begin{pmatrix}25\sin50°\\25\cos50°\end{pmatrix}=\begin{pmatrix}19.15\\16.07\end{pmatrix} ( 25 sin 50° 25 cos 50° ) = ( 19.15 16.07 )
F₂(30N, 150°): ( 30 sin 150 ° 30 cos 150 ° ) = ( 15 − 25.98 ) \begin{pmatrix}30\sin150°\\30\cos150°\end{pmatrix}=\begin{pmatrix}15\\-25.98\end{pmatrix} ( 30 sin 150° 30 cos 150° ) = ( 15 − 25.98 )
F₃(35N, 240°): ( 35 sin 240 ° 35 cos 240 ° ) = ( − 30.31 − 17.5 ) \begin{pmatrix}35\sin240°\\35\cos240°\end{pmatrix}=\begin{pmatrix}-30.31\\-17.5\end{pmatrix} ( 35 sin 240° 35 cos 240° ) = ( − 30.31 − 17.5 )
F₄(25N, 330°): ( 25 sin 330 ° 25 cos 330 ° ) = ( − 12.5 21.65 ) \begin{pmatrix}25\sin330°\\25\cos330°\end{pmatrix}=\begin{pmatrix}-12.5\\21.65\end{pmatrix} ( 25 sin 330° 25 cos 330° ) = ( − 12.5 21.65 )
(b) Resultant:
R x = 19.15 + 15 − 30.31 − 12.5 = − 8.66 R_x=19.15+15-30.31-12.5=\mathbf{-8.66} R x = 19.15 + 15 − 30.31 − 12.5 = − 8.66
R y = 16.07 − 25.98 − 17.5 + 21.65 = − 5.76 R_y=16.07-25.98-17.5+21.65=\mathbf{-5.76} R y = 16.07 − 25.98 − 17.5 + 21.65 = − 5.76
R ⃗ = ( − 8.66 − 5.76 ) \vec{R}=\begin{pmatrix}-8.66\\-5.76\end{pmatrix} R = ( − 8.66 − 5.76 )
∣ R ∣ = 8.6 6 2 + 5.7 6 2 = 75 + 33.18 = 108.18 ≈ 10.4 N |R|=\sqrt{8.66^2+5.76^2}=\sqrt{75+33.18}=\sqrt{108.18}\approx\boxed{10.4\text{ N}} ∣ R ∣ = 8.6 6 2 + 5.7 6 2 = 75 + 33.18 = 108.18 ≈ 10.4 N
© Given Ā = î − j + 3k, B̄ = 2î + 4j − 6k, C̄ = 3î − 5j + 2k
(i) Ā × B̄:
A ⃗ × B ⃗ = ∣ i ^ j ^ k ^ 1 − 1 3 2 4 − 6 ∣ \vec{A}\times\vec{B}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&4&-6\end{vmatrix} A × B = ∣ ∣ i ^ 1 2 j ^ − 1 4 k ^ 3 − 6 ∣ ∣
= i ^ [ ( − 1 ) ( − 6 ) − ( 3 ) ( 4 ) ] − j ^ [ ( 1 ) ( − 6 ) − ( 3 ) ( 2 ) ] + k ^ [ ( 1 ) ( 4 ) − ( − 1 ) ( 2 ) ] =\hat{i}[(-1)(-6)-(3)(4)]-\hat{j}[(1)(-6)-(3)(2)]+\hat{k}[(1)(4)-(-1)(2)] = i ^ [( − 1 ) ( − 6 ) − ( 3 ) ( 4 )] − j ^ [( 1 ) ( − 6 ) − ( 3 ) ( 2 )] + k ^ [( 1 ) ( 4 ) − ( − 1 ) ( 2 )]
= i ^ [ 6 − 12 ] − j ^ [ − 6 − 6 ] + k ^ [ 4 + 2 ] =\hat{i}[6-12]-\hat{j}[-6-6]+\hat{k}[4+2] = i ^ [ 6 − 12 ] − j ^ [ − 6 − 6 ] + k ^ [ 4 + 2 ]
= − 6 i ^ + 12 j ^ + 6 k ^ =\boxed{-6\hat{i}+12\hat{j}+6\hat{k}} = − 6 i ^ + 12 j ^ + 6 k ^
(ii) Ā · (B̄ × C̄) — scalar triple product = det:
∣ 1 − 1 3 2 4 − 6 3 − 5 2 ∣ \begin{vmatrix}1&-1&3\\2&4&-6\\3&-5&2\end{vmatrix} ∣ ∣ 1 2 3 − 1 4 − 5 3 − 6 2 ∣ ∣
= 1 ( 4 ⋅ 2 − ( − 6 ) ( − 5 ) ) − ( − 1 ) ( 2 ⋅ 2 − ( − 6 ) ( 3 ) ) + 3 ( 2 ( − 5 ) − 4 ( 3 ) ) =1(4\cdot2-(-6)(-5))-(-1)(2\cdot2-(-6)(3))+3(2(-5)-4(3)) = 1 ( 4 ⋅ 2 − ( − 6 ) ( − 5 )) − ( − 1 ) ( 2 ⋅ 2 − ( − 6 ) ( 3 )) + 3 ( 2 ( − 5 ) − 4 ( 3 ))
= 1 ( 8 − 30 ) + 1 ( 4 + 18 ) + 3 ( − 10 − 12 ) =1(8-30)+1(4+18)+3(-10-12) = 1 ( 8 − 30 ) + 1 ( 4 + 18 ) + 3 ( − 10 − 12 )
= − 22 + 22 − 66 = − 66 =-22+22-66=\boxed{-66} = − 22 + 22 − 66 = − 66
Question 8
(a) Angle between forces 19N and 21N with resultant 27N
Using cosine rule:
R 2 = F 1 2 + F 2 2 + 2 F 1 F 2 cos θ R^2=F_1^2+F_2^2+2F_1F_2\cos\theta R 2 = F 1 2 + F 2 2 + 2 F 1 F 2 cos θ
729 = 361 + 441 + 2 ( 19 ) ( 21 ) cos θ 729=361+441+2(19)(21)\cos\theta 729 = 361 + 441 + 2 ( 19 ) ( 21 ) cos θ
729 = 802 + 798 cos θ 729=802+798\cos\theta 729 = 802 + 798 cos θ
cos θ = 729 − 802 798 = − 73 798 = − 0.09147 \cos\theta=\frac{729-802}{798}=\frac{-73}{798}=-0.09147 cos θ = 798 729 − 802 = 798 − 73 = − 0.09147
θ = cos − 1 ( − 0.09147 ) ≈ 95.2 ° \theta=\cos^{-1}(-0.09147)\approx\boxed{95.2°} θ = cos − 1 ( − 0.09147 ) ≈ 95.2°
(b) Mass of 4kg, strings at 30° and 45°
Weight W = 4 × 10 = 40N
Resolving vertically: T₁sin30° + T₂sin45° = 40
0.5 T 1 + 0.7071 T 2 = 40 ⋯ ( 1 ) 0.5T_1+0.7071T_2=40 \quad\cdots(1) 0.5 T 1 + 0.7071 T 2 = 40 ⋯ ( 1 )
Resolving horizontally: T₁cos30° = T₂cos45°
0.8660 T 1 = 0.7071 T 2 ⇒ T 2 = 0.8660 0.7071 T 1 = 1.2247 T 1 ⋯ ( 2 ) 0.8660T_1=0.7071T_2 \Rightarrow T_2=\frac{0.8660}{0.7071}T_1=1.2247T_1 \quad\cdots(2) 0.8660 T 1 = 0.7071 T 2 ⇒ T 2 = 0.7071 0.8660 T 1 = 1.2247 T 1 ⋯ ( 2 )
Substitute (2) into (1):
0.5 T 1 + 0.7071 ( 1.2247 T 1 ) = 40 0.5T_1+0.7071(1.2247T_1)=40 0.5 T 1 + 0.7071 ( 1.2247 T 1 ) = 40
0.5 T 1 + 0.8660 T 1 = 40 0.5T_1+0.8660T_1=40 0.5 T 1 + 0.8660 T 1 = 40
1.366 T 1 = 40 ⇒ T 1 = 29.3 N 1.366T_1=40 \Rightarrow \boxed{T_1=29.3\text{ N}} 1.366 T 1 = 40 ⇒ T 1 = 29.3 N
T 2 = 1.2247 × 29.3 = 35.9 N T_2=1.2247\times29.3=\boxed{35.9\text{ N}} T 2 = 1.2247 × 29.3 = 35.9 N
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:
25 x + 400 ( x − 5 ) = 30 ( 10 − x ) 25x + 400(x-5) = 30(10-x) 25 x + 400 ( x − 5 ) = 30 ( 10 − x )
25 x + 400 x − 2000 = 300 − 30 x 25x+400x-2000=300-30x 25 x + 400 x − 2000 = 300 − 30 x
455 x = 2300 455x=2300 455 x = 2300
x = 2300 455 ≈ 5.05 m from end A x=\frac{2300}{455}\approx\boxed{5.05\text{ m from end A}} x = 455 2300 ≈ 5.05 m from end A
ii. Reaction at pivot:
R = 25 + 400 + 30 = 455 N R = 25+400+30=\boxed{455\text{ N}} R = 25 + 400 + 30 = 455 N
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 = ∂ q 1 ∂ p 1 ⋅ p 1 q 1 = ( − 1 ) ⋅ 7 41 = − 0.171 E_{p_1}=\frac{\partial q_1}{\partial p_1}\cdot\frac{p_1}{q_1}=(-1)\cdot\frac{7}{41}=\boxed{-0.171} E p 1 = ∂ p 1 ∂ q 1 ⋅ q 1 p 1 = ( − 1 ) ⋅ 41 7 = − 0.171
ii. Cross elasticity of demand:
E p 2 = ∂ q 1 ∂ p 2 ⋅ p 2 q 1 = ( 2 ) ⋅ 9 41 = 0.439 E_{p_2}=\frac{\partial q_1}{\partial p_2}\cdot\frac{p_2}{q_1}=(2)\cdot\frac{9}{41}=\boxed{0.439} E p 2 = ∂ p 2 ∂ q 1 ⋅ q 1 p 2 = ( 2 ) ⋅ 41 9 = 0.439
(Positive cross elasticity → substitutes)
(b) Compound interest at 4% annually
i. Sum becomes 4 times itself:
4 P = P ( 1.04 ) n ⇒ ( 1.04 ) n = 4 4P=P(1.04)^n \Rightarrow (1.04)^n=4 4 P = P ( 1.04 ) n ⇒ ( 1.04 ) n = 4
n ln ( 1.04 ) = ln 4 n\ln(1.04)=\ln4 n ln ( 1.04 ) = ln 4
n = ln 4 ln 1.04 = 1.3863 0.03922 ≈ 35 years n=\frac{\ln4}{\ln1.04}=\frac{1.3863}{0.03922}\approx\boxed{35\text{ years}} n = ln 1.04 ln 4 = 0.03922 1.3863 ≈ 35 years
ii. Sum doubles in 10 years:
2 = ( 1 + r ) 10 2=(1+r)^{10} 2 = ( 1 + r ) 10
( 1 + r ) = 2 0.1 = 1.07177 (1+r)=2^{0.1}=1.07177 ( 1 + r ) = 2 0.1 = 1.07177
r = 0.07177 ≈ 7 % r=0.07177\approx\boxed{7\%} r = 0.07177 ≈ 7%
© 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
Maximum q = 216 at u = 5 , B = 0 \boxed{\text{Maximum } q = 216 \text{ at } u=5,\ B=0} Maximum q = 216 at u = 5 , B = 0
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 + 2 b ≥ 7 (Chemical P) a+2b\geq7 \quad\text{(Chemical P)} a + 2 b ≥ 7 (Chemical P)
2 a + 4 b ≥ 11 (Chemical Q) 2a+4b\geq11 \quad\text{(Chemical Q)} 2 a + 4 b ≥ 11 (Chemical Q)
3 a + 6 b ≥ 11 (Chemical R) 3a+6b\geq11 \quad\text{(Chemical R)} 3 a + 6 b ≥ 11 (Chemical R)
a ≥ 0 , b ≥ 0 a\geq0,\quad b\geq0 a ≥ 0 , b ≥ 0
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:
Cost = 3 ( 7 − 2 b ) + 64 b = 21 + 58 b \text{Cost}=3(7-2b)+64b=21+58b Cost = 3 ( 7 − 2 b ) + 64 b = 21 + 58 b
This is minimized at b = 0 , giving a = 7 :
Minimum cost = 3 ( 7 ) + 64 ( 0 ) = ₦ 21 \boxed{\text{Minimum cost} = 3(7)+64(0) = ₦21} Minimum cost = 3 ( 7 ) + 64 ( 0 ) = ₦21
At corner vertex (a = 7, b = 0) , the pharmaceutical company minimizes cost.
(iv) Compound interest — ₦6,900 → ₦2,000?
(Likely ₦600 → ₦2,000 in 4 years, half-yearly)
A = P ( 1 + r 2 ) 2 n A=P\left(1+\frac{r}{2}\right)^{2n} A = P ( 1 + 2 r ) 2 n
2000 = 600 ( 1 + r 2 ) 8 2000=600\left(1+\frac{r}{2}\right)^{8} 2000 = 600 ( 1 + 2 r ) 8
( 1 + r 2 ) 8 = 2000 600 = 3.333 \left(1+\frac{r}{2}\right)^8=\frac{2000}{600}=3.333 ( 1 + 2 r ) 8 = 600 2000 = 3.333
1 + r 2 = 3.33 3 1 / 8 = 1.1616 1+\frac{r}{2}=3.333^{1/8}=1.1616 1 + 2 r = 3.33 3 1/8 = 1.1616
r 2 = 0.1616 ⇒ r = 0.3232 \frac{r}{2}=0.1616 \Rightarrow r=0.3232 2 r = 0.1616 ⇒ r = 0.3232
r ≈ 32.3 % per annum \boxed{r\approx32.3\%\text{ per annum}} r ≈ 32.3% per annum
