PRE-U STPM MATHEMATICS (T) TERM 2

CONTENTS






Chapter
1 LIMITS AND CONTINUITY 1
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
1.1 Limits 2
1.2 Continuity 6

Chapter
2 DIFFERENTIATION 16
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
2.1 Derivatives 17
2.2 Applications of Differentiation 39


Chapter
3 INTEGRATION 80
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
3.1 Indefinite Integrals 81
3.2 Definite Integrals 104

Chapter
4 DIFFERENTIAL EQUATIONS 126
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
4.1 Differential Equations 127
4.2 First Order Differential Equations with Separable Variables 130
4.3 First Order Linear Differential Equations 134
4.4 Transformations of Differential Equations 136
4.5 Problems Modelled by Differential Equations 140

Chapter
5 MACLAURIN SERIES 153
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
5.1 Maclaurin Series 154
5.2 Applications of Maclaurin Series 167

Chapter
6 NUMERICAL METHODS 172
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
6.1 Numerical Solution of Equations 173
6.2 Numerical Integration 186


STPM Model Paper (954/2) 193
Answers 195


iii






x-Content STPM.indd 3 02/11/2018 12:42 PM

CHAPTER
2 DIFFERENTIATION

















Subtopic Learning Outcome

2.1 Derivatives (a) Identify the derivative of a function as a limit.
(b) Find the derivatives of x (n  Q), e , In x, sin x, cos x, tan x, sin x, cos x,
–1
n
–1
x
tan –1 x, with constant multiples, sums, differences, products, quotients and
composite.
(c) Perform implicit differentiation.
(d) Find the first derivatives of functions defined parametrically.
2.2 Applications of (a) Determine where a function is increasing, decreasing, concave upward and
differentiation concave downward.
(b) Determine the stationary points, extremum points and points of inflexion.
(c) Sketch the graphs of functions, including asymptotes parallel to the coordinate
axes.
(d) Find the equations of tangents and normals to curves, including parametric
curves.
(e) Solve problems concerning rates of change, including related rates.
(f) Solve optimisation problems.




Bilingual Keywords

asymptote – asimptot
concave downward – cekung ke bawah
curve – lengkung
decreasing – menyusut
derivative – terbitan
differentiation – pembezaan
extremum point – titik ekstremum
implicit function – fungsi tersirat
increasing – menokok
limit – had
point of inflexion – titik lengkok balas
stationary point – titik pegun












02 STPM Math(T) T2.indd 16 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation

2.1 Derivatives


Feynman's
Differentiation Differentiation
Derivative of a function INFO VIDEO

Let A(x, y) be a fixed point on the curve y = f(x) and B be a y y = f(x)
neighbouring point with coordinates (x + x, y + y),
where x (read as ‘delta-ex’) denotes a small increase in x and B(x + δx, y + δy)
y denotes a corresponding small increase in y. T
In Figure 2.1, AC = x, BC = y. 2
BC y
Gradient of chord AB is = .
AC x A(x, y) C
As the point B is moved along the curve towards the fixed
point A, x → 0 and the direction of the chord AB approaches
closer and closer to the direction of AT, the tangent to the
curve at A. 0 x
Figure 2.1
Thus,
y
gradient of curve at A = lim (read as ‘the limit as x tends to 0’)
x → 0 x
dy
= (read as ‘dee’ y by ‘dee’ x)
dx
dy f(x + x) – f(x)
If y = f(x), then = f(x) defined by f(x) = lim
dx δx → 0 x
This formal definition of a derivative can be used to differentiate any function.

This process is known as differentiation with respect to x (abbreviated to w.r.t.) from first principles.
dy or f(x) is the ‘derivative of f(x)’, the ‘differential coefficient’ or ‘derived function’ of f(x) w.r.t. x.
dx
dy
Note: dy and dx do not have any meaning in themselves; in particular we cannot think of as dy ÷ dx.
dx

Example 1

Find the derivatives of the following functions with respect to x, from the first principles.
2
(a) f(x) = x (b) f(x) = 1
x
d f(x + x) – f(x)
Solution: (a) Using the definition f(x) = lim  4
dx x → 0 x
d 2 lim  (x + x) – x 2 4
2
dx (x ) = x → 0 x
2
2
= lim  x + 2x x + (x) – x 2 4
x → 0 x



17





02 STPM Math(T) T2.indd 17 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation


= lim  2xx + (x) 2 4
x → 0 x
= lim (2x + x)
x → 0
= 2x
1 _ 1
d 1 x + x x
(b) 1 2 = lim  4
dx x x → 0 x
= lim  x – (x + x) 4
x → 0 (x + x)x (x)
2
= lim  –1 4
x → 0 x(x + x)
= – 1
x 2



Example 2

dy
2
If y = 3x + x + 1, find from the first principles.
dx
2
2
dy 3(x + x) + (x + x) + 1 – (3x + x + 1)
Solution: = lim  4
dx x → 0 x
2
2
2
= lim  3x + 6xx + 3(x) + x + x + 1 – 3x – x – 1 4
x → 0 x
2
= lim  6x x + 3(x) + x 4
x → 0 x
= lim [6x + 3 x + 1]
x → 0
= 6x + 1




Exercise 2.1


Find the derivatives of the following functions from the first principles.
3
4
1. y = x 2. y = x 3. y = 5x 2
2
2
4. y = 1 5. y = x + 5x 6. y = x – x + 3
x 2
2
3
7. y = 4x 8. y = 1 9. y = 2x + 3
x 3
10. y = 2x – 3x + 1
2



18





02 STPM Math(T) T2.indd 18 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation
Differentiation of standard functions

Derivative of a constant
Consider the function y = c, where c is a constant. y
dy f(x + x) – f(x)
From the derived definition = lim
dx x → 0 x
As f(x + x) = c = f(x), c y = c
∴ f(x + x) – f(x) = c – c
x x
= 0 0 x
dy 2
Hence, = lim 0 Figure 2.2
dx x → 0
= 0

Geometrically, y = c represents a straight line parallel to the x-axis.

Therefore, its gradient is zero.
d (c) = 0, where c is a constant
dx



Derivative of af(x) where a is a constant
From the derived definition,
d a f(x + x) – a f(x)
[a f(x)] = lim
dx x → 0 x

= lim  a[f(x + x) – f(x)] 4
x → 0 x
= lim a  lim f(x + x) – f(x)] 4
x → 0 x → 0 x
d
= a [f(x)]
dx

d d
Hence, [a f(x)] = a [f(x)]
dx dx


Derivative of x n

Consider the following differentiation from first principles.
d (x + x) – x 2
2
x = lim
2
dx x → 0 x
= lim (2x + x)
x → 0
= 2x
= 2x 2 – 1




19





02 STPM Math(T) T2.indd 19 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation
d (x + x) – x 3
3
x = lim
3
dx x → 0 x
2
2
= lim [3x + 3x x + (x) ]
x → 0
= 3x 2
3 – 1
= 3x
d (x + x) – x –1
–1
–1
x = lim
dx x → 0 x
= lim – 1
x → 0 x(x + x)
1
2 = – x 2
= –x –2
= –1(x –1 – 1 )
Hence,
n
d x = nx n – 1 , n  R.
dx
Example 3

Differentiate with respect to x.
(a) x –3 (b)   x
(c) –5x 4 (d) 3
x 6
Solution: (a) Let y = x –3
dy 3
–4
= –3x –3 – 1 = –3x = –
dx x 4
(b) Let y =   x = x — 1 2
dy = 1 x — – 1 = 1 x = 1
1
1
2
–—
2
dx 2 2 2  x
(c) Let y = –5x 4
dy = –5(4x 4 – 1 ) = –20x 3
dx
(d) Let y = 3 6 = 3x – 6
x
dy = 3[–6x –6 – 1 ] = –18x = – 18
–7

dx x 7
Exercise 2.2

Find the derivatives of the following functions with respect to x.


5
1. 5 2. 3x 3. –4x 4. x 5. x –4
1
—
x
3
6. x 7. x 8.  9. 1 10. 3 x 2
–1
3
  x
3
10
–3
11. 5x 12. 7x 13. 5 14. 7 x 15. –2
4x 4 8 x 8
20
02 STPM Math(T) T2.indd 20 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation
Derivative of e x

x
Consider the exponential function f(x) = a in Figure 2.3 where a is a positive constant. For any value of a,
0
a = 1. Hence, the curve passes through the point A(0, 1).
The gradient at point A is given by

f(0) = lim f(0 + x) – f(0)
x → 0 x
= lim f(x) – f(0) y y = a x
x → 0 x
x
= lim a – a 0
x → 0 x 2
x
= lim a – 1 A(0, 1)
x → 0 x
x
0
For f(x) = a x
d a x + x – a x Figure 2.3
x
(a ) = lim
dx x → 0 x
x
= lim a x 1 a – 1 2
x → 0 x
x
a – 1
= a x lim
x → 0 x
x
= a f(0) f(0) = x → 0 a – 1
x
lim
x
d
x
Hence, (a ) = a f(0)
x
dx
Now there must be a value of a for which f(0) = 1, i.e. the gradient of the graph at (0, 1) is 1. We call this
value e.
x
Thus d (e ) = e x
dx
e is an irrational number. e ≈ 2.718.
e is known as the exponential function.
x


Derivative of ln x
Let y = f(x), where f(x) is any function of x,
dy y
lim
hence = x → 01 2
dx x
= lim 1
x → 0 1 2
x
y






21





02 STPM Math(T) T2.indd 21 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation
But y → 0 when x → 0,
dy 1
=
dx lim 1 2
x
x → 0 y
dy 1
=
dx dx
dy

When ln x = y, From the definition of logarithm i.e. if log y = x, then y = a .
x
x = e y a
Differentiate both sides w.r.t. y,
2 dx d
y
y
= (e ) = e = x
dy dy
dy 1 1
= =
dx dx x
dy
d 1
Hence, (ln x) =
dx x

Derivative of a x

Let y = a x
Taking log of both sides to base e,
ln y = ln a x
= x ln a

Differentiate both sides w.r.t. y,
d dx
(ln y) = ln a
dy dy
1 dx
y = dy ln a

dx 1
=
dy y ln a
dy
dx = y ln a
= a ln a
x
Hence, d (a ) = a ln a
x
x
dx
Example 4

Differentiate each of the following with respect to x.
x
(a) 3 (b) 10 x
Solution: (a) Let y = 3 x (b) y = 10 x
dy dy
x
Hence, = 3 ln 3 Hence, = 10 ln 10
x
dx dx

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02 STPM Math(T) T2.indd 22 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation
Differentiation of trigonometric functions


Derivative of sin x
Let y = sin x
dy sin (x + δx) – sin x
= lim
dx x → 0 δx
2 cos (x + δx) + x sin (x + δx) – x Using the formula

lim  2 2 4 sin A – sin B
=
A – B
A + B
x → 0 δx = 2 cos ––––– sin –––––
2
2
2 cos  x + δx 2 sin δx
1

lim 2 2 2
=
x → 0 δx
δx
sin

lim cos x + δx 2
1
2
=
x → 0 2 δx


2
sin δx
In the limit, as δx → 0, cos x + δx 2 → cos x and δx 2 → 1
1
2
2
d
Hence dx (sin x) = cos x, where x is in radians.
Derivative of cos x
Let y = cos x
dy cos (x + δx) – cos x
= lim
dx x → 0 dx
Using the formula
–2 sin (x + δx) + x sin (x + δx) – x cos A – cos B

lim  2 2 4 = –2 sin ––––– sin –––––
A + B
A – B
=
x → 0 δx 2 2
–2 sin  1 x + δx 2 sin δx
= lim 2 2
x → 0 δx
sin δx
1
= lim –sin x + δx 2 2
x → 0 2 δx

2
sin δx
In the limit, as δx → 0, sin 1 x + δx 2 → sin x and 2 → 1
2
δx
2
d
Hence dx (cos x) = –sin x, where x is in radians.





23





02 STPM Math(T) T2.indd 23 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation

Derivative of tan x
Let y = tan x
dy tan (x + δx) – tan x
= lim  4
dx x → 0 δx

= lim  sin (x + δx) – sin x 4 1
x → 0 cos (x + δx) cos x δx

= lim  sin (x + δx) cos x – cos (x + δx) sin x 4
x → 0 cos x cos (x + δx) · δx

sin [(x + δx) – x]

Using the formula
2 = x → 0 cos x cos (x + δx) · δx sin A cos B – cos A sin B = sin (A – B)
lim
sin δx
= x → 0 cos x cos (x + δx) · δx

lim
1 sin δx
= lim ·
x → 0 cos x cos (x + δx) δx
In the limit, as δx → 0, cos (x + δx) → cos x and sin δx → 1.
δx

d 1
2
Hence, dx (tan x) = cos x = sec x
2


Derivative of arcsin x (sin x)
–1
–1
Let y = sin x
x = sin y

Differentiating w.r.t. y, dx = cos y
dy
dy 1
∴ =
dx cos y
2
2
2
but cos y = 1 – sin y = 1 – x
dy 1
Hence, =
dx (1 – x )
2
d 1
–1
i.e. (sin x) =
dx (1 – x )
2
π
Note: y = sin x ⇒ – < y < π , since y is a principal value angle.
–1
2 2
For this range of values of y, cos x > 0
2
2
∴ cos x = (1 – x ) [not – (1 – x )]



24





02 STPM Math(T) T2.indd 24 02/11/2018 12:43 PM

Mathematics Term 2 STPM Chapter 2 Differentiation
–1
Derivative of arc cos x (cos x)
–1
Let y = cos x
x = cos y
dx
= –sin y
dy
dy 1
= –
dx sin y
2
2
but sin y = 1 – cos y
= 1 – x 2
dy 1
∴ = –
dx (1 – x )
2
2
d 1
–1
Hence, (cos x) = –
dx (1 – x )
2
–1
Derivative of arc tan x (tan x)
Let y = tan x
–1
x = tan y
2
dx = sec y
dy
dy 1
=
2
dx sec y
= 1
2
1 + tan y
= 1
1 + x 2

–1
Hence, d (tan x) = 1
dx 1 + x 2

Rules of differentiation

Differentiation of sums and differences of functions
Consider two functions of x, p(x) and q(x), and let f(x) = p(x) + q(x).
From the derived definition,
d [p(x + x) + q(x + x)] – [p(x) + q(x)]
[f(x)] = lim
dx x → 0 x
[p(x + x) – p(x)] + [q(x + x) – q(x)]
= lim
x → 0 x
p(x + x) – p(x) q(x + x) – q(x)
= lim + lim
x → 0 x x → 0 x
d d
= [p(x)] + [q(x)]
dx dx

d d d
Hence, [f(x)] = [p(x)] + [q(x)]
dx dx dx


25





02 STPM Math(T) T2.indd 25 02/11/2018 12:43 PM

ANSWERS




1 Limits and Continuity (c) (d) y
y
Exercise 1.1
1. (a) 6 (b) 4
(c) 3 (d) –3 1
2. (a) 1 (b) 7 0 x x
4 3 –1 0
1
(c) – (c) 6 continuous continuous
2
3. (a) 1 (b) 3 (e) y (f) y
(c) 1 (d) 1
2 3
4. (a) 3 (b) 1 (c) 4
(d) ∞ (e) 15 (f) 2 –1 0 x –2 0 1 x
5
4
5. (a) 1 (b) –1
5
6. (a) –1 (b) 1 (c) does not exist continuous not continuous
4. (a) 1, –1; Not continuous
7. (a) 9 (b) –10 (c) –1 (b)
(d) 5 (e) does not exist y
8. (a) 2 (b) 2 (c) 2
(d) 2 (e) –3 (f) does not exist 1
Exercise 1.2 5 x
1.
f(x ) –1

(c) does not exist
x 5. (a) does not exist; f is not continuous
_ π 0 π _
– (b)
2 2 y
5
continuous
2. f(x)
2 x
0
–1
1

0
1 x
– – 1 2 3
4 6. Yes
continuous y
3. (a) (b) y 1
2
y x
1 0 1 2 3 4 5
–
x 2
0 π
x –1
0 π 2π

continuous not continuous




195





Ans STPM Math T Term2.indd 195 02/11/2018 12:47 PM

Mathematics Term 2 STPM Answers

7. (b) right 2 Differentiation
f(x )
Exercise 2.1
2 1. 3x 2 2. 4x 3
x 3. 10x 4. – 2
–3 0 5 x 3
–2
5. 2x + 5 6. 2x – 1
7. 12x 2 8. – 3
x
4
not continuous 9. 4x 10. 4x – 3
8. k = – 3 ; Continuous
4
9. c = 1 Exercise 2.2
1. 0 2. 3
10. (a) 2 3. –4 4. 5x 4
(b) 5. –4x 6. 1 – — 2
–5
x
3
f(x ) 3
7. – 1 8. 3 x
x 2 2
1 – —
x
9. – x 2 3 10. 2 – — 1
3
2 3
x 11. –15x –4 12. 70x 9
0 2 4 5
13. – 5 14. 21 2
x
continuous x 5 8
15. 16
STPM Practice 1 x 9
1. (a) 1 (b) 3 Exercise 2.3
2 1 x 1
1. 2 ln 2 2. 1 2 ln
x
lim
lim
2. (a) (i) x → c f(x) ≠ x → c f(x) 5 3 5 2
+
–
lim
lim
(ii) x → c f(x) = x → c f(x) ≠ f(c) 3. 10x + 3 4. 4x – 18x
4
–
+
lim
lim
(iii) x → c f(x) = x → c f(x) but f(c) is not defined 5. 6x – 5 1 6. 2x – x 2
–
+
2
x
3
(b) {k: –∞ < k < 0 or 0 < k < 3 or < k < ∞} 7. – x 3 – x 2 8. e + cos x
2 2
3. a = –32, b = 3 9. 2 10. 8x + x 2 2 + x 9 4
x
4. (a) 1, 0, 2, 1 (b) Yes, No 2 1
11. – – 2x 12. –
5. (a) 8 (b) –2√ 7 3x 2x
27 13. 10x – 3 + 8 14. 1 + 4
6. (a) b = 3 (b) a = ±√ 6 x 2 x 3 x 2 x 3
lim lim lim 15. – 3 16. 2 cos x – 3 sin x
7. (a) x → 0 f(x) = x → 0 + f(x) = 1, x → 0 f(x) exists. 2x
–
3
lim
(b) x → 0 f(x) = 1 ≠ f(0) = 2 17. 13 18. 13
4
f(x) is discontinuous at x = 0
f(x) is continuous in the interval (–∞, 0) < (0, ∞) Exercise 2.4
1. 2x(2x + 1) 2. 24x – 5

2
8. a = 1, b = –1 12 1
2
3. 8x – 30x + 1 4. – – 2x
3
lim lim x 4 x 2
–
9. x → 3 f(x) = –6 ≠ x → 3 + f(x) = 6 2
f(x) is discontinuous at x = 3 5. 3x – 4x – 1 6. 5(cos x – x sin x)
sin x
8. (cos x) ln x +
10. (a) 2 (b) – 1 7. 2x(2 sin x + x cos x) 10. e (x + 2x – 1) x
2
9. 1 + ln x
x
2
13. (a) 3 (b) 2 11. e (tan x – cos x + sec x + sin x)
2
x
8
2
14. No, not continuous at 1 12. sin x (1 + sec x)
13. x(2 cos x – x sin x)
lim lim
–
15. (a) x → 0 f(x) = 8, x → 0 + f(x) = 2 + c, c = 6 14. 2x (sin x + cos x) + x (cos x – sin x)
2
(b) (i) f(x) continuous at x = 0 only when c = 6
(ii) f(x) continuous at x , 0. (quadratic) 15. cos 2x
(iii) f(x) continuous at x . 0. (exponential function) Exercise 2.5
1. –6x 2. 2
(2x – 3) 2 (x + 1) 2
2
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Ans STPM Math T Term2.indd 196 02/11/2018 12:47 PM