Week 1 – SS2 Second Term Further Mathematics Notes

SECOND TERM E-LEARNING NOTE

 SUBJECT: FURTHER MATHEMATICS                    CLASS: SS2
SCHEME OF WORK

 WEEK     TOPIC

1. Differentiation: Limits of Function and First Principle, Differentiation of Polynomial
2. Differentiation (Continued): Rules of Differentiation
3. Differentiation of Transcendents: Derivative of Trigonometric Functions and Exponential Functions.
4. Application of Differentiation: Rate of Change, Equation of Motion, Maximum and Minimum Points and Values of Functions.
5. Conic Sections: Equation of Circles, General Equation of Circles, Finding Centre and Radius, Equation and Length of Tangents to a Circle.
6. Conic Sections: The Parabola, Hyperbola and Ellipse
7. Review of First Half Term
8. Statistics Probability: Sample Space, Event Space, Combination of Events, Independents and Dependent Events.
9. Permutation and Combination
10. Dynamics: Newton’s Laws of Motion
11. Work, Energy, Power, Impulse and Momentum
12. Revision and Examination.

 REFERENCES
Further Mathematics Project 2 and 3.

 WEEK ONE
TOPIC : LIMITS OF FUNCTIONS AND DIFFERENTIATION FROM THE FIRST PRINCIPLE
The followings are the properties of limits:

1. lim k = k i e

x²a
The limit of a constant is the constant itself

1. lim [f(x) + f ₂ (x) + f_{3 (x)} + … f_n(x)]

   = lim f₁(x) + lim f₂ (x) + lim f₃ (x) +limf_n(x)

xa xa xaxa
i.e
The limit of the sum of a finite number of functions is equal to the sum of their respective limits
lim [f₁(x) – f₂(x)] = limf₁(x) – limf₂(x)
xa xa xa xa
i.e
The limit of the difference of two functions is equal to the difference of their limits.

1. lim [f₁(x) f₂ (x) f₃ (x) + ….. f_n(x)]

xa
= lim f₁(x) lim f₂ (x) lim f₃ (x) lim f (x)
xa xa xa    xa
i.e
The limit of the product of infinite number of functions is equal to the product of their respective limits.

1. x a[]= lim f₁(x)

   lim f₂(x)
   Provided lim f₂ (x) 0 i.e
   The limit of the quotient function is equal to the quotient of their limits provided the limit of the divisor is not equal to zero

2. lim k f (x) = k lim f (x)

xax a
i.e
Limit of the product of a constant and a function is equal to the product of the constant and the limit of the function

 Example 1
Evaluate lim( 7 – 2x + 5x² – 4x³)
Solution
lim {7 – 2x + 5x² – 4x³)
x a
= lim 7 2 lim x + 5 lim x² 4 lim x
x ax ax a x a
= 7 0 + 0 = 7

 Example 2
Limx² + 5x + 9
x0 2x² – 3x + 15
Solution
lim x² + 5x + 9 = limx² + 5x + 9
x2x² – 3x + 15lim2x² – 3x+15
lim x² +5lim x+ lim 9
x0 x0 x0
2 lim x² – 3lim x + lim 15
x0 x0 x0
=
^=
=

 Example
Evaluate limx²– 25
xx– 5
Solution
Lim x²– 25 = lim =
xx – 5
= lim (x + 5)
x – 5
= lim x + lim 5
x – 5x – 5
= 5 + 5
= 10

 Example
Evaluate lim 3x³+2x²+x+1
x – 5 x³ + 2x+ 5
Solution
We know that lim = 0
x – 0
lim3x³ + 2x² + x + 1
x – 0x³ + 2x + 5
x³(3 + + + )
limx x= 0
x0 x³ ( + )
^{x x}
lim 3 + 2 lim + lim + lim
=x – 0x – 0x – 0x – 0
lim1 + 2 lim+ lim + 5 lim
x – 0x – 0x – 0x – 0
= 3+ 0 + 0 + 0
1 + 0 + 0
=
= 3

 EVALUATION
Evaluate lim -> 4 x³ +4x 6
Evaluastelim x -> -2 x+6/ 2x +4

 Differentiation From first Principle
The technique adopted in unit 11.3 in finding the derivative of a function from the consideration of the limiting value is called differentiation from first principle.

 Example
Find the derivative of f(x) = x² from first principle.
Solution
f (x) = x²
f(x + x) = ( x + x)²
= x² + 2xx + (x)²
f(x + x) – f (x) = (x + x)² – x²
= x²+ 2xx + (x)² – x²
= 2xx + (x)²
= 2x +
lim= 2x
0
f¹(x) = 2x

 Example
Find the derivative of y = x³ from first principle
Solution
y = x³
        y + y = (x + x)³
            = x³ + 3x³x + 3x (x)²+ (x)³
        y = (x +x)³ – x³
            = 3x² + 3x(x)²+ (x)³
= 3x² + 3xx + (x)²
lim = 3x²
x0
Hence = 3x²

 Example
Find the derivative of y = from first principle.
Solution
y =
     y + y = –
y =
=
=
= =
=
lim = –
=
Hence = –

 Example
Find the derivative of y = c, where c is a constant, from first principle.
Solution
y = c
    y+ = c ( since c is a constant)
    = 0
     = 0
lim = 0
= = 0

 Hence the derivative of a constant is zero.
EVALUATION
Find the derivative of the following using first principle.
1. y=3x² + 4 (2) y= x³ -2x² + 2x -5

 GENERAL EVALUATION
1) Evaluate (i) lim x-> 0 x⁴ + 5x / x² + 3 (ii) lim x-> 2 3x + 7
2) Differentiate from the first principle y= 2x² +3x + 5
3) Find the gradient function of y = x² +3x +1 (4) Differentiate y =5x⁴ +7x³ + 6x² – 9x +4

 READING ASSIGNMENT:New further Maths Project 2 page 113- 120

 WEEKEND ASSIGNMENT
1) Evaluate lim_{x-> 1} 4x² + 3x a) 4 b) 3 c) 7 d) 0
2) Evaluate lim_{x-> 0} x² + 9 a) 3 b) 9 c) 6 d) 1
3) Evaluate lim_{x-> 0} ( x+3) ( 3x-3) a) 27 b) 6 c) 9 d) -9
4) Differentiate 8x² + 10 a) 8x b) 16x c) 10 d) 18x
5) Find the derivative of y = b where b is a constant a) 0 b) bx c) x d) 1

 THEORY
1) Evaluate lim_{x-> -2} 3x³ +4 / x² +4 (2) Differentiate from the first principle y = 7x³ + 5x² – 6x +5