Further Mathematics Lesson Note SS2 First Term
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WEEK 1
SCHEME OF WORK FOR SS2 FIRST TERM FURTHER MATHEMATICS
WEEKS TOPICS
- Revision of SS 1 work
- Nature of roots of quadratic equations and co-efficient
- Intersection of a line and a given curve
- Polynomials
- Factorization of Polynomials
- Sum of roots of Polynomials
- Revision of first half and test
- Legal reasoning
- Trigonometric function
- Relationship between graphs of trig ratios
- Graph of inverse by ration
- Revision and first term examination
REFERENCE MATERIAL
- NEW FURTHER MATHEMATICS PROJECT BY M.R TULTUH – ADEGUN
D. GODSPOWER ADEGOKE
WEEK ONE
REVISION OF SS ONE WORK
WEEK TWO
Finding quadratic equations given sum and product of roots
Recall that if ax2 + bx + c = 0 where a, b, c are constants such that a 0, suppose are represent the distinct roots as
then sum of roots = ———— (1)
Product of roots = —————– (2)
In general, if a quadratic equation has roots then ( which gives —————– (3)
That is X2 – (sum of roots) x + product of roots = o
Note: The above consideration gives rise to two results
- Given a quadratic equation can find the sum and product of the roots
- Given the roots, we can formulate the corresponding quadratic equation.
Examples: Find the sum and product of the roots of each of the following quadratic equations.
Solution
a = 2, b = 3, c = -1
= =
= =
a=3, b = -5, c=-2
= =
= =
a=1, b = -4, c=-3
= = ,
= =
a = 1, b = -6, c = -2
= = ,
= =
Example 2
Find the quadratic equation whose roots are (a) 3 (b) and 5 (c) -1 and 8 (d) and
Solution
The quadratic equation whose roots are is
- = 3 + -2 = 1
= 3x – 2 = -6
- = + 5 =
= x 5 =
- = 1 + 8 = 7
= -1 x 8 = -8
- = + =
= =
– 10x + 3 = 0
Exercise: if are the roots of the equation Find the values of , , ,
Hint =
- The roots of the quadratic equation are , with greater than . Find the values of (i) (ii)
Nature of roots
We recall that the solution of is where D = – 4ac called the discriminant.
Three restrictions can be place in the value of D
- If D = 0, The roots are real and equal
- If D > 0 The roots of the equation are real and district
- If D<0 The roots are not real (imaginary)
The graph of y= is related to the x-axis as follows
- Y y= f(x) (b) y y = f(x) (c) y y=(x)
X x
Real and equal roots real and distinct roots no real roots
Solution
Example 1
Determine the nature of roots of the following quadratic equations.
- 2
Solution
- a = 1, b = 3, c = -2
D = – 4a c = 9+8=17
D = 17 > 0
Roots are real and distinct.
a = 1, b = -6, c =9
D = – 4ac = 36 – 36 = 0
Roots are real and equal
A = 2, b = -2, c = 5
D = – 4ac = 4 – 40 = -36
The roots are imaginary
Example 2: Find the value of k for which the roots of the quadratic equation are equal.
Solution
a = 1, b = k-1, c = -k
For equal roots – 4ac = 0
– 2k + 1 + 4k = 0
= 0
K + 1 = 0
K = -1
Exercise: Find the possible values of the constant k if the roots of the quadratic equation
Assignment: Ex 1 D Question 22, 27, 28, 29.
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WEEK 3
Conditions for given line to intersect a curve
Simultaneous equations involving one linear and one quadratic can be solved as follows:
Example: Solve simultaneously
- y – 2x = -2
X2 + y2 + 2x – 3y = 19
- 2y – y =4
2x2 + 3y2 – x + 4y = 17
Solution
- x2 + (2x-2)2 + 2x – 3(2x – 2) = 19
x2 + 4x2 – 8x + 4 +2x – 6x = 19
5x2 – 12x – 9 = 0
(5x + 3) (x+3) = 0
X = 3 or x =
And y = 4 or y =
- 2(2y -4)2 + 3y2 – (2y – 4) + 4y = 17
2(4y2 – 16y + 16) + 3y2 – 2y + 4 +4y = 17
11y2 – 30y + 19 = 0
(11y – 9) (y – 1) = 0
Either y = 1 or y =
And x = -2 or x =
Condition for a tangent to a curve
A tangent to a curve touches the curve of one point
Example: Find the value of c if the line y = x + c is to be a tangent to the curve x2+y2=1
Solution
y = x + c ———– (1)
x2+y2=1 ———— (2)
x2 + (x + c)2 = 1
x2+ x2 + 2cx + c2 = 1
2x2 + 2cx + c2 – 1 = 0
The quadratic equation has equal roots since y = x + c is a tangent to x2 + y2 = 1
a = 2, b = x, c = c2 – 1
D = b2 – 4ac = 0
4c2 – 8(c2 – 1) = 0
4c2 – 8c2 + 8 = 0
C2 = 2
C =
Assignment: Find the equation of the tangent to the curve x2 + y2 – 8x – 8y + 28 = 0 which is parallel to the line y = 3x + 6
Condition for a line not to intersect a curve
a. Prove that the line y=3x-1 neither cuts nor touches the curve x2 +y2 – 8x – 2y + 8 = 0
b. prove that the line y=2x – 3 is a tangent to the curve x2 + y2 – 10x – 4y + 24 = 0
Solution
y = 3x – 1 ————- (1)
x2 + y2 – 8x – 2y + 8 = 0 ———— (2)
x2 + (3x – 1)2 – 8x – 2 (3x – 1) + 8 = 0
10x2 – 20x + 11 = 0
D = b2 – 4ac
400 – 440
-40 < 0
Y = 3x – 1 does not meet the curve x2 + y2-8x-2y +8 =0
b. x2 + (2x +3) – 10x -4 (2x – 3) + 24 = 0
5x2 – 30x + 45 = 0
X2 – 6x + 9 = 0
D=b2 – 4ac
36– 36 = 0
The line y=2x-3 is a tangent to the curve
Assignment
- Show that the line y = x does not meet the curve x2 + y2 – 4x – 2cy + 86 = 0
- Prove that the line 3x – y + 1 = 0 is a tangent to the curve x2 + y2 – 14x -4y +13 = 0
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WEEK 4
POLYNOMIALS
A polynomial p(x) is in the form p(x) = anxn + an-1 xn-1 + _________ + a2x2 + a1x + a0
The numerical constants an, an-1– a2 a1 are called coefficients of xn, xn-1…X2, x respectively while a0 is called the constant term of the polynomial. The highest power n is called the degree of the polynomial. Examples
P(x) = 3x2 + 2x + 5, degree 2
H(x) = x4 + 3x2 – 6x degree 4
Example: Given that P(x) = 7x3 – 4x2 + 3x + 4
P1(x) = 5x2 + 6x + 1 Find P(x) + P1 (x) (b) P(x) – P1(x)
Solution
- P(x) + P1(x) = (7x3 – 4x2 + 3x + 4)
(+) 5x2 + 6x + 1
= 7x3 + x2 + 9x + 5
- P(x) – P1(x) = 7x3 – 4x2+ 3x + 4
(-) 5x2 + 6x + 1
7x3 – 9x2 – 3x + 3
2. Given that P1(x) = 4x3 – 2x2 + 3x – 1 and P2(x) = 3x2 – 4. Find P1(x) x P2(x)
Solution
P1(x) x P2(x) = (3x2 – 4) (4x3 – 2x2 + 3x – 1)
= 3x2(4x3 – 2x2 + 3x – 1)-4(4x3 – 2x2 + 3x – 1)
= 12x5 – 6x4 + 9x3 – 3x2 – 16x3 + 8x2 – 12x + 4
= 12x5 – 6x4 – 7x3 + 5x2 – 12x + 4
Division of Polynomials
If a polynomial P(x) is divided by another Polynomial D(x) to obtain Q(x) Then we have P(x) = D(x) x Q(x) + R where R is the remainder
P (x) is the dividend
D(x) is the divisor
Q(x) is the Quotient
Example: Divide the polynomial P1 (x) = 3x2 – 2x + 4 by the polynomial P(x) = x + 2
Solution
Step 1: divide the first term of the poly by the first term of the divisor to get the first term of quotient
Step 2: Multiply each term of the division by the quotient
3x
Step 3: Subtract the product obtain in step 2 from the first two terms of the dividend and
3x
add the next term of the dividend.
(-)-8x + 4
Step 4: Using -8x + 4 as a new dividend repeat
3x – 8
Steps 1, 2 and 3.
(-) – 8x + 4
-8x – 16
20
Note x + 2 is the divisor, 3x2 – 4x + 4 the dividend 3x – 8 is the quotient and x the remainder.
2) Divide 4x3 + 6x2– 2x + 7 by 2x – 3 and hence find the quotient and the remainder.
2x2 + 6x + 8
12x2 – 2x Quotient = 2x2 + 6x + 8
12x2 – 18x Remainder = 31
16x + 7
16x – 24
31
Exercise: Find the quotient and remainder when
- 2x4 – 3x3 + x2 – 4x + 5 is divided by x2 + 3x + 1
- X3 + 8 is divided by x2 – 2x + 4
Zeros of Polynomials
To find the zeros of the polynomial P(x), Put P(x) = 0 and solve the corresponding equation
Examples: Find the zeros of the polynomial
- P(x) = x2 – 5x + 6 (b) P2 (x) = x2– 1
Solution
- P1(x) = 0
X2 – 5x + 6 = 0
(x – 3) (x – 2) = 0
Either x = 3 or x = 2
- P2(x) = 0
X2 – 1 = 0
(x + 1) (x -1) = 0
X = -1 or x = 1
The Remainder Theorem
It states that “if a polynomial f(x) is divided by x-a the remainder is f (a)
Proof
The polynomial function f(x) can be written as
f(x) = (x – a) Q(x) + R where x –a is the divisor and Q(x) quotient, R is remainder
Put x = a into (1)
f(a) = (a – a) Q(a)
f (a) = R
Example find the remainder when
- f(x) = (x + 3) (x – 2) (x + 2) is divided x + 1
- f(x) = 3x3 – 4x2 + 2x + 3 is divided by x – 1
Solution
- Let x+1 = 0, x = -1
F(x) = (x + 3) (x + 2) (x + 2)
f(-1) = (-1 + 3) (-1-2) (-1 + 2)
= (2)(-3)(1)
= -6
R = -6
- Let x -1 = 0, x = 1
f(x) = 3x3– 4x2 + 2x + 3
= 3(1) 3 – 4 (1)2 + 2(1) + 3
= 3-4 + 2 + 3
= 4
R = 4
Exercise
- Find the remainder when f(x) = 2x3 + 3x2 – 4x + 1 is divided by 2x – 1
- Show that x + 1 is a factor of f(x) = 2x3 + 3x2 – 5x -6
Assignment: Given that the poly f(x) = 6 –x – x2 is a factor of the poly g(x) = ax3 + 5x2 + bx = 18. Find
- The values of the constants a and b
- The remainder when the polynomial g(x) is divided by x + 2
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WEEK 5
Factorization of Polynomials
Example: if x – 1 is a factor of the polynomial f(x) = 4x3 – 4x2 – x – k where k is a constant.
- Find the value of k
- Factorize f(x) completely and state its zeros
- Find the remainder when g(x) is divided by x + 2
Solution
- Let x-1 = 0, x = 1, f(1) = 0
f(x) = 4x3 – 4x2 – x + k
f(1) = 4(1)3 – 4(1)2 – 1 + k = 0
K = 1
- f(x) = 4x3 – 4x2 – x + 1
4x2 – 1
(-) 4x3 – 4x2
– x + 1
(-) – x + 1
– –
Exercise: When the polynomial f(x) = (p-1) x3 + px2 + qx + r where p, q and r are constants, is divided by x+2 and x – 1 the remainders are -5 and 4 respectively. If x + 1 is a factor of f(x), find the values of p, q and r. Hence factorize f(x) completely
WEEK 6
Roots of Cubic equation
Given the Cubic equation ax3 + bx2 + (x + d = 0 ———- (1)
Where a 0 x3 + + + = 0 ————– (2)
If are the roots of the equation (2)
Then (x ) (x – ) (x – ) = 0
X2 – ( + + ) x2 + (( + + )x – = 0 ————– (3)
Comparing the coefficient of (2) and (3) gives
| = Sum of roots |
| = Product of roots pro |
| + + = Sum of two roots |
Example: Solve the equation x3 – 6x2 + 11x – 6 = 0
Given that 1 is root of the equation
Solution
Let and be two other roots and = 1 then
+ 1 = 6
= 5—————- (1)
(1) = 6
= 6 ——————- (2)
From (2) = ————— (3)
+ = 5
6 + 2 = 5
Exercise: Solve the equation x3 – 3x2 – 4x + 12 = 0 given that two of the roots are equal but the opposite in signs
Assignment: 27 -30 page 38 New further Mathematics Project
WEEK SEVEN
MID TERM BREAK
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WEEK 8
LOGICAL REASONING
An intelligent System: it is a system that senses its environment and learns, for each situation which action permits it to reach its objective.
Basic issues in intelligent system. (Assignment)
Fundamental Definition
- A proposition or statement
It is a sentence that is either true or false. For example “four plus four equals eight” and Obama is president of Nigeria” are either statement or preposition. While the first statement is true the second is false.
Truth Value:
It is the truth or falsify of a statement. A true statement has truth value T while a false statement has the value F.
Exercise: State the truth value of the following statements
- The earth is a planet
- Come out
- What is a great day
- I am a Nigerian
- X + 5 = 9
Connectives
They are words and phrases or systems that are sued to formed compound preposition
| Connective word | Symbol | Compound statement formed | Symbolic form |
| not | Negation | P | |
| And | Conjunction | P q | |
| Or | Disjunction | P q | |
| If — then | Implication | P q | |
| If and only if | biconditional | p q | |
Negation
Given that P is a statement, the negation of p denoted by is a statement that is false when p is true and true when p is false.
Example: P: Abuja is in Nigeria
Abuja is not in Nigeria
Note the following statements are equivalent in meaning
- All human are mortals
- Every human is a mortal
- Each human is a mortal
- Any human is a mortal
The negation of the statement P: All goats are mammals is any of the following
~P: some goats are not mammals
~P: there exists a goat which is not a mammal
~P: at least one goat is not a mammal
Conjunction:
If p and q are two given propositions the conjunction is the compoundproposition denoted p n q. p q is true when both p and q are true
Example: let p: 12 is a multiple of 4
q: 12 is a factor of 24
p n q: 12 is a multiple of 4 and a factor of 24 and p n q is true since both p and q are true.
2. Let a: 3 + 4 = 7
b: 3 – 7 = 4
a b: 3 + 4 = 7 and 3 – 7 =4
a b is false
Exercise: Form the conjunction of p and q
- P: Grace is intelligent q: She is hardworking
- P: The weather is hot q: Rain is falling
- P: 6x – 2 = 40 q: x < 8
Disjunction
The disjunction p q is true if at least one of p or q is true. If is false where both of p and q are false.
Example: Consider the following statement
P: Lagos is in Nigeria
Q: Lagos is the capital of Nigeria
R: Lagos is in Ghana
S: Lagos is the most populated state in Nigeria
Write the following disjunction and state the truth value
- P v q
- P v s
- q v r
- r v s
ii. State the truth value of the following compound statements
- P q
- P s
- Q r
- R s
Conditional Proposition
Given that p and q are propositions. A constitutional proposition denoted P q has the following meaning
- P implies q
- P is sufficient for q
- Q is necessary for p
- P only of q
- If p then q
- Q follows from p
- Q is the consequences of p
Example: P: this month is January
Q: Next month is February
P : of this month is January then next month is February
2. P: Rain falls, q: I will wear a rain coat
P q: I will wear a rain coat if rain falls.
Note: The conditional statement p q is when the hypothesis (p) is true and the conclusion (q) is false
Examples: (1) P: If Lagos is in Nigeria then it is Africa is true
(2) If Lagos is in Nigeria then it is in Europe false.
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WEEK 9
BICONDITIONAL STATEMENTS
Given that p and q are statements the bi conditional statement P q means
- P if and only if q
- P implies q and q implies p
- P is necessary and sufficient for q
Examples: let p and q the statements
P: He is lazy q: He will be a successful business man; write the following statement in symbolic form. “He will be a successful businessman if and only if he is hardworking
Answer: ~p q or q ~P
Exercise: Ex 3, No 4,5 and 20 page 52 and 53
Quantifiers
- Universal quantifier
Examples:
- All man are wise
- Every orange is sweet
- Any Nigerian is either male or female
Let M denoted the set of men and P the predicate has are wise. The preposition “All men are wise” can be written as ( ) P(x)
Exercise: let m be the set of all men and p the predicate “has conscience” write the preposition all men have conscience”
- Existential Quantifier( )
The symbol is called existential quantifier and reads
i. There exist ii. For some iii. For at least
Example: let N be the set of natural numbers and P(x): x + 5 < 9. Determine the true value of x P(x)
Solution
If XEN then {x: x + 5 < 9} = {1, 2, 3}
Hence ( x P (x) is true.
Introduction to proving Theorem
Theorem: A theorem is a statement that has be proved on the basis of previously established statements such as other theorem.
A theorem has two parts namely hypothesis and conclusion.
The proof of a theorem is a logical argument which demonstrates the fact that the conclusion is a necessary consequence of the hypothesis.
The Principal of Mathematic induction
If P(n) is a statement which involves positive integers n = 1, 2, 3…… then P(n) is true for all positive integers n provided
- P(1) is true
- P(k+1) is true when ever P(k) is true.
Exercise: Let P(x) be “x+1>5” depend on the set N of natural members. Determine its truth sets.
Assignment Ex 4 Q 5, 7 and 8 page 61
WEEK 10
Trigonometric Functions
Review of trigonometric ratios
S A
T C
In the first quadrant (0 ≤ ≤ 900)
Sin = cos (90 – )
Cos = sin (90 – )
Tan = cot (90 – )
In the second quadrant (90o≤ ≤ 180)
Sin = sin (180o – )
Cos = -cos(180o – )
Tan = tan (180o– )
In the third quadrant (1800≤ ≤ 270o)
Sin = sin (180o+ )
Cos = -cos(180o + )
Tan = tan (180o+ )
In the fourth quadrant (2700≤ ≤ 3600)
Sin = -sin (360o– )
Cos = cos(360o– )
Tan = -tan (360o– )
Negative angles
Sin(- ) = -sin
Cos (- ) = cos
Tan (- ) = – tan
Special angles
| Angles | 0o | 30o | 45o | 60o | 90o |
| Sin | 0 | 1 | |||
| Cos | 1 | 0 | |||
| Tan | 0 | 1 |
Conversion from degree to Radius
| Rad = o x |
Example: change to following angles from degree to radian 30o, 45o, 60o, 90o
Graphs of trigonometric functions
- Graph of y = sin – 3
y = sin
| 3 |
| 2 |
| – |
| -2 |
| -3 |
- Graph of y = cos – 3
Y = cos
| 0 |
| 4 |
| 3 |
| 2 |
| -2 |
| – |
| -3 |
In general the graph of y = A sin has an amplitude of /A/ and the graph of y = sin is period of e.g. The graph of y = 3sin2 has an amplitude of 3 and period of .
The graph of y=tan – 2 2
In general the graph of y = a cosbx + c has amplitude /a/ and period
Examples
- Draw the graph of y = 3sin2x + 1 in the range 0o at intervals of 30o
- From your graph find the
i. max value of 3sin2x + 1
ii. Corresponding values of x of which 3sin2x + 1 is max
Solution
i. 4 ii. 45o and 225o
Exercise: page 181 Q 17
WEEK 11
GRAPHS OF INVERSE TRIGONOMETRIC FUNCTIONS
y
Y = arc sin x y = arc cos x
Y=arctanx,
Trigonometric identities
Sin2 + cos2 = 1
1 + tan2 = sec2
1 + cot2 = cosec2
Examples: Prove that
- = cosec + cot
- =
Compound angles
Sin (A+B) = sinAcosB + cosA sin B tan A +B =
Sin(A-B) = sin AcosB – cosAsinB
Cos(A+B) = cosAcosB – sinAsinB tan (A-B) =
Cos(A-B) = cosAcosB + sinAsinB
Examples: evaluate in surd form
- Sin 75o (b) cos 75o (c) tan 195o
Solution
Cos 75o = cos (30o + 45o)
Cos 30ocos 45o – sin 30o sin45o
x –
x
( – )
b. cos 75o= ( – )
- Tan 195o = tan (1800 + 15o)
=
= tan 15o
=
Multiple Angles
- Double angles
Sin 2A = 2sinA cosA tan 2A =
= Cos2A – sin2A
Cos 2A 2cos2A – 1
1 – 2sin2A
Half angle
sinA = cosA = , tanA =
where t = tan .
Examples: if cot = where is an acute angle evaluate sin 2
Solution
Cot =
Tan =
Sin =
Cos =
Cos 2 = cos2 – sin2
= – =
=
+
=
= x
=
Exercise: if are acute such that sin = and tan = . Find without tables
- Sin ( )
- Cos ( )
- Tan ( )
- Tan 2
- Sin 2
- Cos 2
2a. show that = tan A
b. cos 2A =
c. = –
d. tan2A =
Assignment: Ex 8 No11, 14 and 15 page 109 and 110
