COURSE DESCRIPTION
This course gives the basic mathematical skills needed for entry level into sciences, engineering and social sciences. The course covered topics such as;
- Set Theory
- Algebra
- Simultaneous Equations
- Quadratic Equations
- Sequence and series
- Complex Numbers
- Trigonometry
Course Materials
Recommendably, we advise you to download Summaries and Textbooks related to this course.
Find the centre and radius of the circle (x -1)2 + (y + 2)2 = 16
General equations of a circle:
(x – a)2 + (y – b)2 = r2
Where (a, b) are centre coordinates and r = radius
Or
Taking g = -a and f = -b
x2 + y2 + 2gx + 2by + g2 + f2 = r2
x2 + y2 + 2gx + 2fy + c = 0
c = g2 + f2 – r2
r = √(g2 + f2 – c)
Given: (x – 1)2 + (y + 2)2 = 16
a = -(-1) b = -2
Centre = (a, b) = (1, -2)
Radiu = r2 = √16 = 4
If one root of the equation
X2 – 3x + p = 0 is 2 then p is ?
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Find the equation of the circle which is concentric (shares the same centre) with the circle x2 + y2 + 2x – 4y = 0 and which has a radius of 5 units
From x2 + y2 + 2x – 4y = 0, 2g = 2 and 2f = -4
g = 1 and f = -2; a = -1 and b = 2
Centre = (a, b) = (-1, 2)
With radius 5 units, r2 = 25
Equation of circle: (x + 1)2 + (x – 2)2 = 25
x2 + y2 + 2x – 4y + 1 + 4 = 25
x2 + y2 + 2x – 4y = 20
Trigonometry
1. Give the possible values of x if sin x = 6 sin 2x.
2. Find the possible values of y, if 3cosy+2siny = 2.5
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Given that:
U = {Natural numbers less than 10}
Subsets: X = {Even numbers} , Y = {Perfect squares} and Z = {Numbers divisible by 4}
Find the following:
a) (X’ U Y’) U Z
b) Y’ n Z’
c) (X n Y)’
d) (X U Z) n Z
e) (Y n Z)’ n X
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Tutor: Fatimah Garba Yusuf»
The sum of the first 3 terms of A.P is 21. If the difference between and 3rd term and the 1st term is 4, find the next four (4) terms of the sequence
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Tutor: Fatimah Garba Yusuf»
Find the centre and radius of the circle (x -1)2 + (y + 2)2 = 16
General equations of a circle:
(x – a)2 + (y – b)2 = r2
Where (a, b) are centre coordinates and r = radius
Or
Taking g = -a and f = -b
x2 + y2 + 2gx + 2by + g2 + f2 = r2
x2 + y2 + 2gx + 2fy + c = 0
c = g2 + f2 – r2
r = √(g2 + f2 – c)
Given: (x – 1)2 + (y + 2)2 = 16
a = -(-1) b = -2
Centre = (a, b) = (1, -2)
r2 = √16 = 4
Trigonometry
1. Given that sin x = 5/13, find cot x.
2. Prove that Sin 85 – Cos 55 = Sin 25.
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Tutor: Fatimah Garba Yusuf»
Find the centre and radius of the circle x2 + y2 – x + y = 0
From x2 + y2 – x + y = 0, 2g = -1 and 2f = 1
g = – ½ and f = ½, a = ½ and b = – ½
centre = (a, b) = (½, -½)
r = √(g2 + f2 – c)
r = √((- ½)2 + (½)2 – 0)
r = √1/4+1/4 = √1/2 = 1/√2 = √2/2
Find the equation of the circle which passes through the origin and through points (3,0) and (0,4)
Circle passes through the origin (0, 0) and points (3, 0), (0, 4)
A circle passing through all three points will satisfy equation of circle for all values of x and y.
Solving simultaneously:
02 + 02 + g(2 x 0) + f(2 x 0) + c = 0
33 + 02 + g(2 x 3) + f(2 x 0) + c = 0
02 + 42 + g(2 x 0) + f(2 x 4) + c = 0
These are simplified to:
c = 0
6g + c = -9
8f + c = -16
The values are c = 0, g = -9/6 = -3/2 and f = -16/8 = -2
Equation of circle = x2 + y2 + 2(-3/2)x + 2(-2)y + 0 = 0
This gives x2 + y2 – 3x – 4y = 0
Find the equation of the circle passing through the points (0,0) (3,1) (5,5)
Given points: (0,0) (3,1) (5,5)
Solving simultaneously:
02 + 02 + g(2 x 0) + f(2 x 0) + c = 0
32 + 12 + g(2 x 3) + f(2 x 1) + c = 0
52 + 52 + g(2 x 5) + f(2 x 5) + c = 0
These are simplified to:
c = 0
6g + 2f = -10
10g + 10f = -50
f = (−10 – 6g)/2
substituting in the second equation gives 10g + 5(-10 – 6g) = – 50
10g – 50 – 30g = – 50
-20g = 50 – 50 = 0
Hence g = 0, f = (−10 – 0)/2 = -5
The values are c = 0, g = 0 and f = -5
Equation: x2 + y2 + 2(0)x + 2(-5)y + 0 = 0
This gives x2 + y2 – 10y = 0
Find the equation of the circle passing through the points (5,0) (6,0) (8,1)
Given points (5,0) (6,0) (8,1)
Solving simultaneously:
52 + 02 + g(2 x 5) + f(2 x 0) + c = 0
62 + 02 + g(2 x 6) + f(2 x 0) + c = 0
82 + 12 + g(2 x 8) + f(2 x 1) + c = 0
These are simplified to:
25 + 10g + c = 0 → 10g + c = -25
36 + 12g + c = 0 → 12g + c = 36
64 + 1 + 16g + 2f + c = 0 → 16g + 2f + c = -65
The values gotten are: g = -11/2 , f = -7/2 and c = 30
Equation: x2 + y2 + 2(-11/2)x + 2(-7/2)y + 30 = 0
This gives x2 + y2 – 11x – 7y + 30 = 0
Find the equation of the tangent to the circle x2 + y2 – 6x – 4y – 12 = 0 at the point (7,5)
Formula for equation of the tangent to a circle is: xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
Where (x1, y1) is the point at which the tangent is.
Given x1 = 7 and y1 = 5 on the circle x2 + y2 – 6x – 4y – 12 = 0
2g = -6 and 2f = -4 → g = -3 and f = -2
Equation of tangent: 7x + 5y – 3(x + 7) – 2(y + 5) – 12 = 0
This gives 7x + 5y – 3x – 21 – 2y – 10 – 12 = 0
Simplified to 4x + 3y – 43 = 0
Find the equation of the tangent to the circle x2 + y2 + 16x – 4y – 101 = 0 at the point (-8,-11)
To find equation of the tangent to the circle x2 + y2 + 16x – 4y – 101 = 0 at the point (-8,-11)
g = 16/2 = 8, f = -4/2 = -2, x1 = -8 and y1 = -11
Equation of tangent = -8x – 11y + 8(x – 8) – 2(y – 11) – 101 = 0
Simplified to: -8x -11y +8x – 64 – 2y + 22 – 101 = 0
This gives -13y – 143 = 0
Dividing through by 13 gives –y – 11 = 0 which is the same as y + 11 = 0
Find the equation of the tangent to the circle x2 + y2 – 6x – 4y – 12 = 0 at the point (8,2)
Finding equation of the tangent to the circle x2 + y2 – 6x – 4y – 12 = 0 at the point (8,2)
g = -6/2 = -3, f = -4/2 = -2, x1 = 8 and y1 = 2
Equation of tangent = 8x + 2y – 3(x + 8) – 2(y + 2) – 12 = 0
Simplified to: 8x + 2y – 3x – 24 – 2y – 4 – 12 = 0
This gives 5x – 40 = 0
Dividing through by 5 gives x – 8 = 0
Find the length of the tangent from the point (5,-2) to the circle x2 + y2 + 2x – 3y = 0
To find length of tangent from a point outside a given circle
IMAGE
Distance PT = Distance from point to tangent
PT2 = PO2 – OT2
Centre = (a, b) = (-g, -f)
P has coordinates (x, y)
PO2 = Distance between P and centre
(x – (-g))2 + (y – (-f))2 = PO2
OT = radius = √(g2 + f2 – c)
PT2 = [(x + g)2 + (y + f)2] – (g2 + f2 – c)
PT2 = x2 +2gx + g2 + y2 + 2fy + f2 – g2 – f2 + c
PT2 = x2 + y2 +2gx + 2fy + c where (x, y) refers to coordinates of the point given.
Given: Circle x2 + y2 + 2x – 3y = 0 and the point (5, -2)
PT2 = 52 + (-2)2 + 2(5) – 3(-2) = 25 + 4 + 10 + 6 = 45
PT = √45 = 3√5
Cross-check
From the given equation of circle, it has centre (-1, 3/2)
Radius r = √(12+(32)2−0)= √134= √132
D = Shortest distance between centre and given point: √[5− −1)2+(−2−32)2] = √[62+(−72)2]= √(36+494)= √1934=√1932 𝑈𝑠𝑖𝑛𝑔 𝑝𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎′𝑠 𝑟𝑢𝑙𝑒,
D2 – r2 = PT2 𝑃𝑇2= 1934− 134=1804=45 𝑢𝑛𝑖𝑡𝑠 𝑃𝑇= √45=3√5 𝑢𝑛𝑖𝑡𝑠
Which of this is NOT true about 43042’ and 0.243 radians
43042’ = 434260 degrees = 43.70 = 43.7π/180 𝑟𝑎𝑑𝑖𝑎𝑛𝑠=0.763 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
Therefore, 43042’ > 0.243 radians
0.243 radians =( 0.243 x 180)/𝜋 = 13.920
Simplify the trigonometry function sec((1/2)𝜋−𝑥)
In simplifying sec(𝜋2−𝑥), recall that sec = 1𝑐𝑜𝑠, 𝜋= 1800 , 𝜋2 = 900
sec((𝜋/2)−𝑥) = 1/𝑐𝑜𝑠(90−𝑥)
cos(90 – x) = sin(x)
Therefore, 1/𝑐𝑜𝑠(90−𝑥) = 1/ sin (𝑥) = cosec (𝑥) 15. cosec(−3𝜋4) = cosec(-3 x 180/4) = cosec(-135)
Find the exact values of cosec(−3𝜋/4)
cosec(−3𝜋/4) = cosec(-3 x 180/4) = cosec(-135) = cosec(-135 + 3600) = cosec225
2250 = 1800 + 450
Recall cosec = 1/sin
sin(A + B) = sinAcosB + sinBcosA
sin(180 + 45) = (sin180 x cos45) + (sin45 x cos180)
sin180 = 0, cos180 = -1, sin45 = cos45 = √22
sin(180 + 45) = (0 x (√2/2) ) + (-1 x (√2/2))= -√22
cosec(225) = – 1/(√2/2) = -2/√2 = -√2
Simplify the trigonometry function 𝑡𝑎𝑛∅/(sec2∅−1 )
In simplifying 𝑡𝑎𝑛∅/(sec2∅ − 1) recall that sec∅= 1/𝑐𝑜𝑠∅; sec2∅= 1/cos2∅
Simplifying the denominator gives (1/cos2∅) – 1 =( 1−cos2∅)/ cos2∅
𝑡𝑎𝑛∅/(sec2∅ − 1) = 𝑡𝑎𝑛∅/(1−cos2∅)/ cos2∅ = 𝑡𝑎𝑛∅cos2∅/ sin2∅
But 𝑠𝑖𝑛∅/𝑐𝑜𝑠∅=𝑡𝑎𝑛∅ , 𝑐𝑜𝑠∅/𝑠𝑖𝑛∅ = 1/𝑡𝑎𝑛∅, cos2∅/ sin2∅ =1/tan2∅
𝑡𝑎𝑛∅cos2∅/ sin2∅= 𝑡𝑎𝑛∅/tan2∅ = 1/𝑡𝑎𝑛∅ = 𝑐𝑜𝑡∅
Given that sin A = 3/5 where A is acute and cos B = -1/2 where B is obtuse, find the exact value of cot B
sinA = 3/5 acute, A = sin-1(3/5) = 36.90
sinB = -1/2 obtuse, B = 1800 – sin-1(1/2) = 1800 – 600 = 1200 [Angle in second quadrant]
cotB = 1/tanB = 1/tan1200 = 1/-√3 = – √3/3
Given that sin A = 8/17 and sin B = 12/13 and 0 < B < π/2 < A < π. Find the exact value of tan (A + B)
sinA = 8/17, sinB = 12/13 and 0 < B <𝜋/2 < A < 𝜋
A is obtuse, B is acute.
A = 180 – sin-1(8/17) = 180 – 28.07 = 151.930
B = sin-1(12/13) = 67.380
tan(A + B) = tan(151.93 + 67.38)
tan(A + B) =( tanA + tanB)/(1− tanAtanB )
tan(151.93 + 67.38) =( tan151.93 + tan67.38)/(1− tan151.93tan67.38 )= 1.867/2.28 = 0.82 = 140/171
What is the equation of the circle with centre (0,-2) and radius 5
Given: Centre = (0, -2) , radius = 5
Equation of circle: (x – 0)2 + (y + 2)2 = 52
This gives x2 + y2 + 4y + 4 = 25
x2 + y2 + 4y – 21 =0
Find the equation of the tangent to the circle x2 + y2 – 10x + 4y + 24 = 0
Equation of tangent to the circle x2 + y2 – 10x + 4y + 24 = 0 at point (4, -3)
g = -10/2 = -5, f = 4/2 = 2, x1 = 4, y1 = -3
Eq’n: x(4) + y(-3) – 5(x +4) + 2(y – 3) + 24 + 0
4x – 5x – 3y + 2y -20 – 6 + 24 = 0
This gives – x – y – 2 = 0 which is the same as x + y + 2 = 0
Find the radius of the circle passing through the points (3,2) (1,1) (1,0)
Given points (3,2) (1,1) (1,0) on a circle
Substituting into the general equation of a circle and solving simultaneously:
32 + 22 + g(2 x 3) + f(2 x 2) + c = 0
12 + 12 + g(2 x 1) + f(2 x 1) + c = 0
12 + 02 + g(2 x 1) + f(2 x 0) + c = 0
These are simplified to:
9 + 4 + 6g + 4f + c = 0 → 6g + 4f + c = -13
1 + 1 + 2g + 2f + c = 0 → 2g + 2f + c = -2
1 + 0 + 2g + c = 0 → 2g + c = -1
The values gotten are: g = -5/2 , f = -1/2 and c = 4
Centre: (-g, -f) = (5/2, ½)
Radius = √(5/2)2+(1/2)2−4= √25/4+1/4−4=√26/4−4=√5/2=√10/2
Find the centre of the circle passing through the points (2,1) (-2,5) (-3,2)
Given points (2,1) (-2,5) (-3,2) on a circle
Substituting into the general equation of a circle and solving simultaneously:
22 + 12 + g(2 x 2) + f(2 x 1) + c = 0
(-2)2 + 52 + g(2 x -2) + f(2 x 5) + c = 0
(-3)2 + 22 + g(2 x -3) + f(2 x 2) + c = 0
These are simplified to:
4 + 1 + 4g + 2f + c = 0 → 4g + 2f + c = -5
4 + 25 – 4g + 10f + c = 0 → -4g + 10f + c = -29
9 + 4 – 6g + 4f + c = 0 → -6g + 4f + c = -13
The values gotten are: g = 1/4 , f = -11/4 and c = -1/2
Centre: (-g, -f) = (-1/4, 11/4)
Find the equation of the tangent to the circle 9x2 + 9y2 – 12x + 42y – 236 = 0 at the point (-2,8/3)
Equation of the tangent of a circle 9x2 + 9y2 – 12x + 42y – 236 = 0 at the point (-2,8/3)
g = -12/2 = -6, f = 42/2 = 21
Eq’n: 9x(-2) + 9y(8/3) – 6(x -2) + 21(y + 8/3) – 236 = 0
-18x + 24y – 6x + 12 + 21y + 56 – 236 = 0
This gives -24x + 45y – 168 = 0 which is the same as 24x – 45y + 168
Is the point (8,6) common to the circles x2 + y2 – 11x – 7y + 30 = 0 and x2 + y2 – x + 3y – 110 = 0
A point is on a circle if it satisfies its equation.
Substitute (8, 6) as values for x and y in the equations x2 + y2 – 11x – 7y + 30 = 0 and x2 + y2 – x + 3y – 110 = 0
i. 82 + 62 – 11(8) – 7(6) + 30 = 64 + 36 – 88 – 42 + 30 = 100 + 30 – 130 = 0
ii. 82 + 62 – 8 + 3(6) – 110 = 64 + 36 – 8 + 18 – 110 = 118 – 8 – 110 = 0
The point satisfies both equations, therefore it is common to both circles.
In the complex number expression i is the number such that i3 is
i3 = (i2)i
i2 = -1, therefore i3 = -i
Use the information to answer Questions 29 to 32
Given that Z1 = 3 + 2i, Z2 = 4 + 3i
Z1 + Z2
Z1 + Z2 = (3 + 4) + (2 + 3)i = 7 + 5i
Z1 – Z2
Z1 – Z2 = (3 + 2i) – (4 + 3i) = -1 – i
Z1 Z2
Z1.Z2 = (3 + 2i)(4 + 3i) = 3(4) + 3(3i) + 4(2i) + 2i(3i)
i2 = -1
12 + 9i + 8i + (6i2) = 12 + 17i – 6 = 6 + 17i
Z1/ Z2
Z1/Z2
= (3+2𝑖)/4+3𝑖)
= (3+2𝑖)(4−3𝑖) / (4-3𝑖)(4−3𝑖) (multiplying by conjugate of the denominator)
This gives (12 – 9i + 8i + 6)/16+9
=( 18 − i)/25
If Z = a + bi, then its conjugate is denoted by
Conjugate of Z = a + bi is Z’ = a – bi
Reference: John Bird Engineering Mathematics, page 316
If the complex number Z = 2 + 3i, find Z2
Z = 2 + 3i
Z2 = (2 + 3i)( 2 + 3i) = 4 + 6i + 6i + 9i2 = 4 – 9 + 12i = – 5 + 12i
Simplify i5
i5 = i4(i)
i4 = (i2)2 = (-1 x -1) = 1
i5 = 1 x i =i
Simplify (3i)(4i)
(3i)(4i) = (3 x 4)i2 = 12 x – 1 = -12
_______ is a formula useful for finding powers and roots of complex numbers
De Moivre’s formula is a formula useful for finding powers and roots of complex numbers.
Simplify the expression 3/(1+ i)
3/(1+ i)= 3/(1+ i )𝑥 (1−𝑖)/(1− i)=( 3−3𝑖)/(1−𝑖2 ) =( 3−3𝑖)/(1+ i)= (3−3𝑖)/2= 3/2− (3/2)𝑖
In quadratic equation ax2 + bx + c = 0, then the expression that determines the type of solution is
In quadratic equation ax2 + bx + c = 0, the expression that determines the type of solution is Dis criminant.
In the division of complex numbers, conjugate of the denominator is multiplied to both numerator and denominator
In the division of complex numbers, conjugate of the denominator is truly multiplied to both numerator and denominator.
If Z = 3 + 2i, find 8(Z)
Z = 3 + 2i
8(Z) = 8(3 + 2i) = 24 + 16i
Simplify (i)(2i)(-3i)
(i)(2i)(-3i) = 2i2(-3i) = -2(-3i) = 6i
Determine the polar coordinates (r,∅) of complex number (3 + 4i)
Given complex number (3 + 4i)
r = √32+42= √9+16 = √25 = 5
∅ = tan-1(4/3) =( 53.10 x 𝜋)/180= 0.927 radians
When a plotted complex number is multiplied by i, its position will rotate in an anti-clockwise direction by 900
When a plotted complex number is multiplied by i, its position will truly rotate in an anti-clockwise direction by 900
Given that A = 3 + 2i, B = 4 + 3i, find 2A + B
Given that A = 3 + 2i, B = 4 + 3i, 2A + B = 2(3 + 2i) + (4 + 3i) = (6 + 4i) + (4 + 3i) = 10 + 7i
The set P of prime numbers less than 20 can be expressed by __
The set P of prime numbers less than 20 can be expressed by {2, 3, 5, 7, 11, 13, 17, 19}
Let the set A be {a, b, c, d, e} and B {a, d, i, o, u}. Then the set of elements in A U B is
If set A = {a, b, c, d, e} and B = {a, d, i, o, u}
The set of elements in A U B is a merged set of A and B without repetition of elements = {a, b, c, d, e, i, o, u}.
Let the set X be {1, 2, 3, 4, 6, 7, 9} and Y be {9, 2, 6, 8, 10, 12}. Elements in Y ∩ X is
From set X = {1, 2, 3, 4, 6, 7, 9} and Y = {9, 2, 6, 8, 10, 12}, elements in X ∩ Y are the elements present in both X and Y: {2, 6, 9}
Let the set A be {2, 4, 6, 9, 10} and B is {3, 5, 7, 9, 10}. Then the set A – B is
From set A = {2, 4, 6, 9, 10} and B = {3, 5, 7, 9, 10}, the set A – B contains the elements in A which are not in B: {2, 4, 6}
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