Pra U STPM 2022 Penggal 2 - Mathematics

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SEMESTER 2 Our Pre-U & STPM Titles: STPM Text Mathematics (T)


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03/02/2022 8:30 AM
CVR_Pre U_STPM_2022 Mathematics T (Semester 2).indd 1-3
CVR_Pre U_STPM_2022 Mathematics T (Semester 2).indd 1-3 03/02/2022 8:30 AM

CONTENTS





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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







iii Content STPM.indd 3 28/01/2022 5:53 PM

CHAPTER
1 LIMITS AND CONTINUITY



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Subtopic Learning Outcome

1.1 Limits (a) Determine the existence and values of the left-hand limit, right-hand limit and
limit of a function.
(b) Use the properties of limits.

1.2 Continuity (a) Determine the continuity of a function at a point and on an interval.
(b) Use the intermediate value theorem.







Bilingual Keywords

continuity – keselanjaran
continuous – selanjar
function – fungsi
interval – selang
left-hand limit – had kiri
limit – had

right-hand limit – had kanan
stationary point – titik pegun


















1






01 STPM Math(T) T2.indd 1 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

1.1 Limits

1 Limit Introduction
to Limit
Limits INFO VIDEO
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Let f be a function defined on an open interval containing a, possibly not defined at a.

The limit (or limiting value) of f(x) as x approaches a, written as lim f(x), is the value that f(x) approaches as
x → a
x approaches a. If the limit of f(x) as x approaches a is l, then we write
lim f(x) = l
x → a
or f(x) → l as x → a



1. If f(x) → l as x → a from the left (i.e. values of x less than a), we write x → a – f(x) = l and this is known
lim
as the left-hand limit.

2. If f(x) → l as x → a from the right (i.e. values of x more than a), we write x → a + f(x) = l and this is
lim
known as the right-hand limit.

3. Take note that x → a – f(x) may not be the same as x → a + f(x).
lim
lim
lim
lim
lim
Hence, x → a f(x) = l if and only if x → a + f(x) = x → a – f(x) = l.
Properties of limits

Let a be any real number and k any constant.

1. lim c = c, c is a constant
x → a
2. lim k f(x) = k lim f(x)
x → a x → a

3. The limit of a sum (or difference) is the sum (or difference) of the limits.

lim [f(x) ± g(x)] = lim f(x) ± lim g(x)
x → a x → a x → a

4. The limit of a product is the product of the limits.
.
.
lim [f(x) g(x)] = lim f(x) lim g(x)
x → a x → a x → a
5. The limit of a quotient is the quotient of the limits

lim
lim f(x) = x → a f(x) , provided lim g(x) ≠ 0
x → a g(x) lim g(x) x → a
x → a






2






01 STPM Math(T) T2.indd 2 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Example 1

Evaluate 1
2
2
(a) lim (x – x – 2) (b) lim x + 2x – 3
2
x → 1 x → 2 x + x + 1
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3
2
(c) lim 36 – x 2 (d) lim x – 2x + 3x
x → 6 6 – x x → 0 x – 2x
2
Solution: (a) lim (x – x – 2) = lim x – lim x – lim 2
2
2
x → 1 x → 1 x → 1 x → 1
= 1 – 1 – 2
2
= –2
Alternative Method
Substitute x = 1,
2
2
lim (x – x – 2) = 1 – 1 – 2
x → 1
= –2
2
2
(b) lim x + 2x – 3 = 2 + 2(2) – 3
2
x → 2 x + x + 1 2 + 2 + 1
2
= 5
7
(c) x → 61 36 – x 2 2
lim
6 – x
The function f(x) = 36 – x 2 is not defined when x = 6
6 – x
If x ≠ 6, then lim 36 – x 2 = lim (6 – x)(6 + x)
x → 6 6 – x x → 6 6 – x
= lim (6 + x)
x → 6
= 6 + 6
= 12
Note : We can take values of x as near as possible to 6, but not equal to 6.
2
2
3
(d) x → 0 1 x – 2x + 3x 2 = x → 0 x(x – 2x + 3)
lim
lim

x – 2x
x (x – 2)
2
2
lim

= x → 0 1 x – 2x + 3 2 , x ≠ 0
x – 2

= 0 – 0 + 3
0 – 2
= – 3
2
Example 2
Determine whether the limit exists in each of the following cases.
2 + e , x , 1

(a) lim f(x) where f(x) = x
x → 1 1 + e – x , x . 1

2
(x – 2) , x < 2
(b) lim f(x) where f(x) =  2
x → 2 1 – , x . 2
x
3



01 STPM Math(T) T2.indd 3 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Solution: (a) lim – f(x) = lim – (2 + e )
x
x → 1 x → 1
= 2 + e
1 lim lim
x → 1 + f(x) = x → 1 _ (1 + e – x)
= 1 + e – 1
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= e
Since lim – f(x) ≠ lim + f(x), lim f(x) does not exist.
x → 1 x → 1 x → 1
(b) lim – f(x) = lim – (x – 2) 2
x → 2 x → 2
= (2 – 2) 2
= 0
+ 1 –
lim
lim + f(x) = x → 2 1 2 2
x → 2 x
= 1 – 2 2
= 0
Since lim – f(x) = lim + f(x) = 0, lim f(x) exist.
x → 2 x → 2 x → 2






Limits at infinity

+
For n  Z , x → ∞ x 1 n = 0.

lim
Example 3

Evaluate
(a) lim 2 + x
x → ∞ 3x + 1
2
(b) lim 3n – n + 1
2
n → ∞ n + 1
Solution: We cannot substitute x = ∞ as ∞ is undefined.
∞
Instead, both the numerator and the denominator are divided by the highest power
of x. 2
1 2
lim

(a) lim 2 + x = x → ∞1 x + 1
x → ∞ 3x + 1 3 + x
= 1
3
1 2
n
2
2
lim


lim
(b) n → ∞1 3n – n + 1 = n → ∞1 3 – 1 + n 1 2
n + 1
2
1 +
= 3 n 2
4





01 STPM Math(T) T2.indd 4 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Exercise 1.1


1. Find the limit of f(x) as x → 2 if f(x) is given by 1
2
2
(a) 3x + 2 (b) 2 + x (c) x – 1 (d) 1 + x – x 3
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2. Find the limit of f(x) as x → 3 if f(x) is given by
2
1 1 x – 5 x – 9
(a) (b) 2 + (c) (d)
x + 1 x x + 1 x – 3
3. Find the limit of f(x) as x → ∞ if f(x) is given by
(a) 1 + 2 – x 3 2 (b) 3x + 1
x
x
3
2
(c) x + 2x – 1 (d) x
2
2
2x + 3x + 1 3x (x – 1)
4. Evaluate
2
(a) x → 0 1 3x + x 2 2 (b) x → 1 1 x + 1 2


lim
lim
x

x + 1
2
2
lim

(c) x → 2 1 x – 4 2 (d) x → ∞ 1 x – 4x + 3 2
lim


x – 2

5x + 2
x 2
x 21
(e) x → 2 1 1 2 + 1 (f) x → ∞ 1 2x + 1 2
1 +

lim
lim
5x + 2

5. Evaluate
3
x – 1
3x + x – 2
2
lim

(a) x → 1 1 2x + 5x + 3 2 (b) x → –1 1 2x – 3x + 1 2

lim
3
3


| x – 3|
, x ≠ 3
6. A function f is defined by f(x) = x – 3
0 , x = 3
Find
(a) lim – f(x), (b) lim + f(x).
x → 3 x → 3
Hence, determine whether lim f(x) exists.
x → 3
7. The function is defined by
3x – 1, x , 0,
f(x) = 0, x = 0,
2x + 5, x . 0.
Evaluate
(a) lim f(x), (b) lim – f(x), (c) lim – f(x),
x → 2 x → 3 x → 0
(d) lim + f(x), (e) lim f(x).
x → 0 x → 0
5
01 STPM Math(T) T2.indd 5 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity
8. A function f is defined by
2x, x < 1,
f(x) = x + 1, 1 , x < 2,
1
–x – 1, x . 2.
Evaluate
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(a) lim – f(x), (b) lim + f(x), (c) lim f(x),
x → 1 x → 1 x → 1
(d) lim – f(x), (e) lim + f(x), (f) lim f(x).
x → 2 x → 2 x → 2




1.2 Continuity



Continuity at a number

A function f is said to be continuous at a number a if the graph of f is unbroken at the point (a, f(a)).
The definition implicitly requires that the following three conditions hold:
(a) f(a) is defined,
(b) lim f(x) exists,
x → a
(c) lim f(x) = f(a)
x → a
By definition, function f is continuous at a
if lim f(x) = f(a)
x → a




y y
y = (x)
f(a)



x x
0 a 0 a

Figure 1.1 Figure 1.2
Function f is continuous at x = a Function f is discontinuous at x = a



Example 4


Sketch the graphs of each of the following functions and state whether the function is a continuous function.
4 + x , x , 0 x – 1, x < 0
2
(a) f(x) = (b) f(x) =
4 – x , x > 0 x , x . 0
2


6






01 STPM Math(T) T2.indd 6 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Solution: (a)
y
8 1
6
4
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2
x
–2 –1 0 1 2 3 4 5 6 7
–2


f is a continuous function since the graph is unbroken in its domain.

(b)
y
4



x
– 3 –1 0 2
–1

– 4


f is not a continuous function.



Example 5

The function f is defined by
x
e + 3 , x , 1,
f(x) = 4 , x = 1,
e – x + 4 , x . 1.

lim
lim
(a) Find x → 1 – f(x) and x → 1 + f(x). Hence, determine whether f is continuous at
x = 1.
(b) Sketch the graph of f.
Solution: (a) x → 1 – f(x) = x → 1 – (e + 3)
lim
lim
x
= e + 3
lim + f(x) = lim + (e – x + 4)
x → 1 x → 1
= e + 3
lim f(x) = e + 3
x → 1
But f(1) = 4
lim
Since x → 1 f(x) ≠ f(1)
\ f is not continuous at x = 1.



7






01 STPM Math(T) T2.indd 7 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

(b) y

1 4

x
f(x) = e + 3 f(x) = e – x + 4
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3


x
0 1 4 + e





Continuity at an endpoint

lim
A function f is continuous from the left at a if x → a – f(x) = f(a);
lim
a function is continuous from the right at a if x → a + f(x) = f(a).


y y





x x
0 0
a a
f is continuous from the left at a f is continuous from the right at a.




Example 6

The function f is defined by
0, x , 0,

f(x) = 1, x > 0
Determine whether f is continuous from the left at 0 and continuous from the right at 0.

lim
lim
Solution: x → 0 – f(x) = x → 0 – (0)
= 0
f(0) = 1
lim
Since x → 0 – f(x) ≠ f(0), f is not continuous from the left at 0.
lim
lim
\ x → 0 + f(x) = x → 0 + (1)
= 1
lim
Since x → 0 + f(x) = f(0) = 1, f is continuous from the right at 0.



8






01 STPM Math(T) T2.indd 8 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity
Continuity on an interval

A function f is continuous on an open interval (a, b) if it is continuous at every number in the interval; a
function f is continuous on a closed interval [a, b] if it is continuous on (a, b) and is also continuous from the 1
right at a and from the left at b.
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Example 7


 2
Show that the function f defined by f(x) = 1 – x is continuous on the closed interval [–1, 1].

Solution: We first show that f is continuous on (–1, 1).
Let –1 , a , 1. Then
lim
lim
x → a f(x) = x → a 1 – x =  x ) = 1 – a = f(a)
 2
2
lim  2
(1 –
x → a
Next, we show that f is continuous from the right at –1 and from the left at x = 1.
lim
lim
+ (1 – x
lim  2
x → –1 + f(x) = x → –1 + 1 – x =  ) = 0 = f(–1)
2
x → –1
lim
lim
2
x → 1 – f(x) = x → 1 – 1 – x =  ) = 0 = f(1)
lim  2
– (1 – x
x → 1
Notice that a polynomial function is continuous on (–∞, ∞). A rational function is continuous on its domain.
x
The functions e , ln x, sin x, cos x, tan x, sin x, cos x and tan x are also continuous on their respective
–1
–1
–1
domains.
Exercise 1.2
1. The function f is defined by
sin x, x < 0,
f(x) =
x, x . 0.
π
Sketch the graph of f for – , x , π , and state whether f is continuous at 0.
2 2
2. The function f is defined by
x(x – 1), 0 < x , 2,
f(x) = 2(3 – x), 2 < x < 3.

Sketch the graph of f, and state whether it is continuous on its domain.
3. Sketch the graph of the function f in each of the following cases, and state whether f is a continuous
function.
(a) y = cos x, 0 < x < 2π (b) y = tan x, 0 < x < π
–x
2
(c) y = e , x  R (d) y = (x + 1) , x  R
(e) y = ln (1 + x), x . –1 (f) y = 2 , x  R, x ≠ 1
x – 1




9






01 STPM Math(T) T2.indd 9 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity
4. The function f is defined by

|5 – x| , x ≠ 5,
1 f(x) = 5 – x
0, x = 5.
lim
lim
(a) Find x → 5 – f(x) and x → 5 + f(x). Hence, determine whether f continuous at 5.
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(b) Sketch the graph of f.
5. The function f(x) is defined as follows:
3x – 1, x , 0,
f(x) = 0, x = 0,
2x + 5, x . 0.

lim
(a) Determine whether x → 0 f(x) exist. Hence, determine if f is continuous at 0.
(b) Sketch the graph of f.

6. The function f is defined by

x(x – 2), 0 < x , 2,
f(x) = 1 – 1 x, 2 < x , 3,
2
1 x – 2, 3 < x , 5.
2
Determine whether f is continuous at 2 and 3. Sketch the graph of f.

7. A function f is defined by
2
2 – x , –3 < x , 0,
2
f(x) = 2 – (x – 2) , 0 < x < 5.

(a) Show that f is discontinuous at 0.
(b) Determine whether f is continuous from the left, from the right, or neither at 0.
(c) Sketch the graph of f.
8. The function f is defined by
2 – x , 0 < x , 1,
f(x) = 3 + x
2
1 + kx , x > 1
where k  R.
lim
Given that x → 1 f(x) exists, find the value of k.
With the value of k, determine whether f is continuous at 1.

9. The function f is defined by
x – 2 , x ≠ 1, x ≠ 2,
2
f(x) = x – 3x + 2
c, x = 2.
If f is continuous at x = 2, find the value of c.






10






01 STPM Math(T) T2.indd 10 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity
10. The function f is defined as follows:
x, 0 < x < 2,
f(x) = C, 2 , x < 4, 1
x – 4x + 2, 4 , x < 5.
2
(a) Determine the value of C which makes f continuous on the interval 0 < x < 5.
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(b) Sketch the graph of f.



Intermediate value theorem
Let f be a function which is continuous on the closed interval [a, b]. Suppose that d is a real number between
f(a) and f(b), then there exists a number c in (a, b) such that f(c) = d.



f(x)
f(b)



d

f(a)
x
0 a c b

Figure 1.3

Corollary: Let f be a function which is continuous on the closed interval [a, b]. If f(a) · f(b) , 0, then there
exists c in (a, b) such that f(c) = 0. In other words f has at least one zero in the interval (a, b).

An application of the intermediate value theorem is to prove the existence of roots of equations.


Example 8


2
3
Show that f(x) = 2x – 5x – 10x + 5 has a zero in the interval (–1, 2).
Solution: We have to show that there is a number c such that –1 , c , 2 and f(c) = 0.

Applying the intermediate value theorem, we need to show that f is continuous and
that d = 0 is between f(–1) and f(2), i.e. f(–1) , 0 , f(2) or f(2) , 0 , f(–1).
f(–1) = 8 and f(2) = –19
we have f(2) , 0 , f(–1)
\ d = 0 is between f(–1) and f(2).
Since f(x) is a polynomial, it is a continuous function.
So by the intermediate value theorem, there must be a number –1 , c , 2 such
that f(c) = 0.
\ The function does have a zero between –1 and 2.




11






01 STPM Math(T) T2.indd 11 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Example 9

1 Use the intermediate value theorem to show that the equation x – x + 2 = 0 has at least one real root.
5

5
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Solution: Let f(x) = x – x + 2 f(x)
f(x) is continuous since it is a polynomial
f(–1) = 2 2
f(–2) = –28
c
Since f(–2) , 0 and f(–1) . 0, the intermediate –2 –1 0 x
value theorem tells us that f(c) = 0 for some c
in the interval (–2, –1).
5
Hence, the equation x – x + 2 = 0 has at least
one real root.
Graph of f in the interval [–2, –1].










–28




Example 10


2
Given f(x) = x – x + x. Show that there is a number c  R such that f(c) = 10.
3
Solution: f(0) = 0 and f(3) = 21.
We have f(0) , 10 , f(3)
Since f is continuous, there must be a number c  R such that f(c) = 10.


Example 11


Show that there is a positive real root of the equation x = x + 1.
7
7
Solution: Let f(x) = x – x – 1
f(0) = –1
f(2) = 125

Since f(0) , 0 and f(2) . 0, the intermediate value theorem tells us that f(c) = 0
for some c in the interval (0, 2), all of whose elements are positive numbers.
7
Hence, there is a positive real root of the equation x = x + 1.



12






01 STPM Math(T) T2.indd 12 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

Exercise 1.3


1. Use the intermediate value theorem to show that there is a root of the given equation in the given interval. 1
3
(a) x – 4x + 1 = 0 ; (0, 1)
5
4
(b) x – 2x – x – 4 = 0 ; (2, 3)
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3
2
(c) x + 3x = x + 1 ; (0, 1)
2
(d) x = x + 2 ; (1, 2)
3
2. Show that the equation x + x – 5 = 0 has at least one root a where 1 , a , 2.
–x
3. Show that the equation e + 2 = x has at least one real root.
x
4
4. Use the intermediate value theorem to show that the equation x = 2 has at least one root.
3
5. Use the intermediate value theorem to show that the equation x + x + 1 = 0 has a root in the interval
(–2, 0).




Summary



lim
lim
lim
1. x → a f(x) exists if and only if x → a – f(x) = x → a + f(x).
2. Properties of limits
Let a be any real number and k any constant.

(a) lim c = c, c is a constant
x → a
(b) lim k f(x) = k lim f(x)
x → a x → a
(c) The limit of a sum (or difference) is the sum (or difference) of the limits.
lim [f(x) ± g(x)] = lim f(x) ± lim g(x)
x → a x → a x → a

(d) The limit of a product is the product of the limits.
.
lim [f(x) g(x)] = lim f(x) . lim g(x)
x → a x → a x → a
(e) The limit of a quotient is the quotient of the limits
lim
lim f(x) = x → a f(x) , provided lim g(x) ≠ 0
x → a g(x) lim g(x) x → a
x → a
lim
3. A function f is continuous at a if x → a f(x) = f(a).
4. A function f is continuous on an open interval (a, b) if it is continuous at every number in the interval;
a function f is continuous on a closed interval [a, b] if it is continuous on (a, b) and is also continuous
from the right at a and from the left at b.




13






01 STPM Math(T) T2.indd 13 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity

STPM PRACTICE 1

1
1. Evaluate 2 1 x + 1 2
x – 1
lim
lim
(a) x → –3 √x + 4 – 1 (b) x → 1 4e x – 1 – 1
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x + 3
2e 1 x + 1 2 – 1
2. (a) State the three conditions in which function f, f(x) is not continuous at point x = c.
(b) A function g is defined by
8 – x , x , 6
7 –
g(x) =  x
2
x – 2x – 6 , x > 6 where k ≠ 0
kx
Determine the set of values of k such that g is not continuous at point x = 6.


3. The function f is defined by
x – a, x < –3,
2
f(x) = bx + 2, –3 , x < 1,
3x + 2, x . 1.
lim
lim
Determine the value of a and the value of b if x → –3 f(x) and x → 1 f(x) exist.
4. The function f is defined by
f(x) = x + 2, –2 < x , 2,
|x| – 2, otherwise.

lim
lim
lim
lim
(a) Find x → –2 – f(x), x → –2 + f(x), x → 2 – f(x) and x → 2 + f(x).
(b) Determine whether f is continuous at –2 and 2.
5. Evaluate
x – 10
lim
lim
(a) x → 3 8 (x – 3) (b) x → 10 √ 7 – √ x – 3
3
x – 27
6. The function f is defined by
x – 4 , x , 2,
2
x – 2
2
f(x) = a – 2, x = 2
|2 – x| + b , x . 2
x – 2
Where a and b are constants.
lim
(a) If x → 2 f(x) exists, find the value of b.
(b) If f(x) is continuous at x = 2, determine the values of a.



14






01 STPM Math(T) T2.indd 14 28/01/2022 5:30 PM

Mathematics Semester 2 STPM Chapter 1 Limits and Continuity
7. The function f is defined by

3 – 2e , x , 0
x
f(x) = 2, x = 0 1
3e – 2, x . 0
x
(a) Determine the existence of limit f(x) as x approaches 0.
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(b) State, with a reason whether f(x) is continuous at x = 0. Hence, determine the intervals on which f
is continuous.

8. The function f is defined by
f(x) = ln x, 0 , x , 1,
2
ax + b, 1 < x , ∞.
Given f(2) = 3, determine the values of a and b for which f is continuous on (–∞, ∞).

9. The function f is defined by
2
x – 9 , x ≠ 3
f(x) = |x – 3|
6, x = 3
Determine whether f is continuous at x = 3.

10. Evaluate
x – 5
4h
lim
lim
(a) h → 0 e – 1 (b) h → –∞  7
2h
2
4x +
e – 1
11. A function f is defined by
f(x) = ln (x + 2), –2 , x , ∞.
(a) Show that f is continuous on its domain.
(b) Sketch the graph of f.
12. Use the intermediate value theorem to show that there exists a solution to the equation cos x = x in the
interval [0, π ].
2
13. Evaluate
2
2
3x + 16
16x +
lim
lim
(a) x → 0  – 4 (b) x → ∞  1
x
2
2x – 1
14. Show that the graph of y = x – 4x + 1 intersects the x-axis in the interval [0, 2]. Can the same be said
2
for the graph of y = 2x – 3 ?
x – 1
15. The function f is defined by
2
f(x) = x + 2x + 8, x , 0
x
2e + c, x > 0
lim
lim
(a) Find x → 0 – f(x) and x → 0 + f(x). Hence, determine the value of c such that function f is continuous
at x = 0.
(b) Describe the continuity of the function f for
(i) x = 0 (ii) x , 0 (iii) x . 0
15






01 STPM Math(T) T2.indd 15 28/01/2022 5:30 PM

CHAPTER DIFFERENTIAL
4 EQUATIONS



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Learning Outcome


(a) Find the general solution of a first order differential equation with separable variables.
(b) Find the general solution of a first order linear differential equation by means of an integrating factor.
(c) Transform, by a given substitution, a first order differential equation into one with separable variables or
one which is linear.
(d) Use a boundary condition to find a particular solution.
(e) Solve problems, related to science and technology, that can be modelled by differential equations.







Bilingual Keywords

boundary condition – syarat sempadan
differential equation – persamaan pembezaan
first order – tertib pertama
general solution – penyelesaian am
integrating factor – faktor pengamir
linear – linear
particular solution – penyelesaian khusus
separable variables – pemboleh ubah terpisahkan
substitution – gantian
transform – jelmakan
variable – pemboleh ubah









126






04 STPM Math(T) T2.indd 126 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

4.1 Differential Equations


Introduction
Differential to Differential
Equation
Differential equations INFO VIDEO Equation
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Consider x as an independent variable and y as a dependent variable. An equation that involves at least one
2
n
dy d y d y
derivative of y with respect to x, e.g. , , …, is known as a differential equation.
dx dx 2 dx n
A few examples of differential equations are as follows.
dy
(a) + xy = sin x
2
dx
d y dy
2
(b) 2 – 3 + 4y = x
dx 2 dx
dy
2
(c) (1 + x ) 1 2 2 + 3y = 0
dx
Order of differential equation

The order of a differential equation is the highest order of the derivative found in the differential equation.
In the above examples, equations (a) and (c) are differential equations of the first order because the equations
dy
consist of only derivatives of the first order, . Equation (b) is a differential equation of the second order
dx
2
d y
since it involves a second order derivative, .
dx 2 4
Degree of differential equation
The degree of a differential equation is determined by the power of the highest order of the derivative in the
equation. Hence, equations (a) and (b) are differential equations of the first degree since no derivatives are
dy
raised to any power except one. Equation (c) is of second degree as the highest derivative is of power 2.
dx
In this chapter, only differential equations of the first order and first degree that can be solved by separating the
variables or that can be transformed into such equations will be dealt with in further details.


Example 1


Determine the order and the degree of the following differential equations.
dy
(a) + y = x 2
dx
dy
2
2
(b) x (1 + y) 1 2 2 – (1 + x)y = 0
dx
2
d y dy — 3
= 1 +
(c) 2 1 2 2
dx dx
127






04 STPM Math(T) T2.indd 127 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Solution: (a) order : 1, degree : 1
(b) order : 1, degree : 2
(c) order : 2
3
d y dy —
2
1
2
From = 1 + 2
dx 2 dx
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3
2
— 2
22 31
1 d y 2 = 1 + dy 2 4
2
dx dx
2
1
22
1 d y 2 = 1 + dy 2 3
dx
dx
Thus, the degree is 2.
Solution of differential equations
The function y = φ(x) is the solution of a differential equation if it satisfies a given differential equation.
Example 2

2
Show that y = 1 x + C is the solution of the differential equation dy = x.
2 dx
Solution: From y = 1 x + C,
2
2
dy = x
dx
2
Thus, y = 1 x + C is the solution of the differential equation dy = x.
4 2 dx


Example 3

dy
2x
Show that y = Ae + (x + 2)e is a solution of the differential equation – y = (x + 3)e .
x
2x
dx
Solution: From y = Ae + (x + 2)e ,
2x
x
dy = Ae + e + 2(x + 2)e 2x
x
2x
dx
= Ae + (2x + 5)e 2x
x
Substitute this into the given equation,
dy

– y = Ae + (2x + 5)e 2x – [ Ae + (x + 2)e ]
x
x
2x
dx = (x + 3)e 2x

x
2x
Thus, y = Ae + (x + 2)e is a solution of the differential equation
dy – y = (x + 3)e .
2x
dx
From these two examples, note that the solution of the differential equation of the first order will contain one
arbitrary constant.
128



04 STPM Math(T) T2.indd 128 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations
A solution of the differential equation that contains an arbitrary constant is known as a general solution.


Example 4

y
2
Show that y = A 1 x – 1 2 is a solution of the differential equation dy = x – 1 . Hence, find the value of A
x + 1
2
dx
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if y = 1 when x = 2.
x – 1
Solution: Given y = A 1 x + 1 2 ...................................
2
dy 2A
Thus, 2y = ......................................
dx (x + 1) 2
From ,
1
A = y 2 x + 1 2
x – 1
Substitute into equation ,
dy 2y 2 x + 1
2y = 2 1 2
dx (x + 1) x – 1
dy y
i.e. =
2
dx x – 1
y
1
Thus, y = A x – 1 2 is a solution of the differential equation dy = x – 1 .
2
2
x + 1
dx
Hence, with the condition y = 1 when x = 2, from equation ,
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1
2
1 = A 2 – 1
2 + 1
A = 3
1
2
Thus, y = 3 x – 1 2 4
x + 1
In Example 4, the value of A can be determined from the given condition.
A general solution that satisfies certain conditions (either initial conditions or boundary conditions) so that
the value of the arbitrary constant in the solution is obtained is
called the particular solution.
Family of curves of the solution y
C= 0
The equation y = 1 x + C represents a family of curves with C = 2 C= 1 C= –1
2
2
similar characteristics. Each value of C will give rise to a
particular curve in the family.
Figure 4.1 shows sketches of four members of the family curves
for the solution corresponding to C = –1, C = 0, C  = 1 and
C = 2.


x
0

Figure 4.1



129






04 STPM Math(T) T2.indd 129 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Exercise 4.1

1. State the order and degree of the following differential equations.
2
2
(a) x 2 d y + x dy + 4y = 0 (b) 2x d y + (sin x) dy – y = cos x
dx 2 dx dx 2 dx
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2
dy
(c) 1 2 2 + x = y (d) d y 2 – x = dy
dx
dx
dx
2
3
dy
32
(e) 1 d y 2 + x d y 2 1 2 5 + y = cos x
4
+
dx
dx
dx
2. State the degree of the following differential equations.
5
—
2
dy d y 2 3
3
(a) = 1 + 1 22 4
dx dx
dy
(b) 1 y – x dy 2 2 = 1 + 1 2 2
dx
dx
d s  2
2
ds
(c) 2  1 2
= 5 +
dt dt
3. Show that y = 3x + 2A is the general solution of the differential equation x dy + y = 6x. Hence, obtain
x
dx
dy
the particular solution of the differential equation x + y = 6x if y = 1 when x = 1.
dx
dy
4. Show that y + 2 = Ax + x is the general solution of the differential equation x – y = 2 + x . Hence,
2
2
dy 2 dx
4 obtain the particular solution of the differential equation x dx – y = 2 + x , given that y = 1 when
x = 1.
4.2 First Order Differential Equations with
Separable Variables
Generally, a first order differential equation can be written as
dy
= f(x, y) .....................................................
dx
In certain cases, function f(x, y) can be written as
.
f(x, y) = u(x) v(y) ..............................................
where u(x) and v(y) are functions of x and y respectively.
By substituting equation  into equation ,
dy
= u(x) v(y)
.
dx
Equation  is a differential equation with separable variables if it can be written as
dy = u(x) dx
v(y)



130






04 STPM Math(T) T2.indd 130 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 5

Determine whether each of the following differential equations is a differential equation with separable variables.
dy
2
(a) = y + 3x y
dx
dy x + y
(b) =
dx x
dy
(c) = e  x + y
dx
dy
2
Solution: (a) From = y + 3x y,
dx
dy = y(1 + 3x )
2
dx Rearranging
dy 2 dy
= (1 + 3x ) dx
= u(x) dx
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v(y)
dy
2
Thus, = y + 3x y is a differential equation with separable variables.
dx
dy x + y
(b) = is not a differential equation with separable variables.
dx x
dy
(c) From = e x + y
dx
dy = e e
x y
dx
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dy = e dx 4
x
e
dy
Thus, = e x + y is a differential equation with separable variables.
dx



dy
To find the solution of a differential equation with separable variables, integrate the equation = u(x) dx
separately. The result of the integration is v(y)
dy
∫ v(y) + C = u(x) dx + C 2
1 ∫
where C and C are integral constants.
2
1
Rearrange it,
dy
∫
∫
v(y) = u(x) dx + C  – C 1
2
∫
= u(x) dx + C C = C – C
2 1
This seems to be the solution of the first order differential equation with separable variables.



131






04 STPM Math(T) T2.indd 131 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 6

dy 1 + y 2
Find the general solution of the differential equation = .
dx 2y
Hence, find the particular solution if y = 0 when x = 0.
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Solution: From dy = 1 + y 2 ,
dx 2y
2y
1 + y 2 dy = dx
2y
∫ 1 + y 2 ∫
= dx
2
ln |1 + y | = x + C where C is an arbitrary constant
1 + y = e x + C
2
x
2
y = Ae – 1 where A = e C
Ae –
Thus, the general solution of the differential equation is y = ± 1 .
x
When x = 0, y = 0, C = 0 and A = 1.
x
e –
Therefore, the particular solution of the differential equation is y = ± 1 .
The particular solution can also be obtained by using definite integration with the initial condition as the lower
limit and x and y as the upper limit.



Example 7

2y
4 The gradient function of a curve at any point (x, y) is given by the equation dy = 1 – x 2 . Find the equation
dx
2
of the curve if the curve passes through the point 1 1 , 1 .
2
Solution: The equation of the gradient of the curve can be written as
dy 2 dx
y = 1 – x 2

2 dx
∫
∫ dy = 1 – x 2
y

ln | y | = ln 1 + x  + C
1 – x
When x = 1 and y = 1, C = – ln 3
2

i.e. ln | y | = ln 1 + x  – ln 3
1 – x
2
Thus, the equation of the curve that passes through the point 1 1 , 1 is
2
1 + x
y = .
3(1 – x)




132






04 STPM Math(T) T2.indd 132 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Exercise 4.2

Find the general solution of the following differential equations.

dy 1 dr 3 + t 2
1. = x + 2. =
2
dx x 2 dt t
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1 dy 1 dx
3. x dx = x – 1 4. x dt = t + 2
2
dy dy
5. = y – 4 6. x = –y 2
2
dx dx
dy 2 + y 2 dy
7. = 8. (x – 2) – y = 0
dx y dx
dy dy 1
9. x = xy + y 10. =
dx dx ye x
y 2 dy dy
11. x dx = ln x 12. x dx = 1 – 2y + y 2
dy dy
2
2
2
13. (y + 1) + 2xy = 2x 14. 2y – xy = x
dx dx
Solve the following differential equations.
dy y dy 1
15. = – 16. =
dx x dx x – 1
dy dy
2
17. = y + 2 18. (x – 2) – 2xy = 0
dx dx
dy dy
19. (x – 1) – y = 0 20. = 2xy + y
dx dx 4
Find the particular solution of the following differential equations with the given initial conditions.

dy dy 1
21. y = x + 2, y = 2 when x = 1 22. = + x, y = 1 when x = 1
dx dx x
dy dy 1 + x
23. = 2y – 3, y = 2 when x = 0 24. = , y = 1 when x = 2
dx dx x – 1
2
dy 2x dy 3y – 1
25. = , y = 0 when x = –1 26. = , y = 1 when x = –1
dx 1 + x 2 dx 6y
–x dy
27. e = 1, y = –1 when x = 0 28. xy dy = ln x, y = 0 when x = 1
dx dx
dy x + y dy
29. + y = 4, y = 0 when x = ln 2 30. e = x, y = 1 when x = 0
2
dx dx
dy dy

2
2
31. y = , y = 0 when x = 0 32. = x 9 , y = 9 when x = 0
x +

9 – 4y
dx dx
dy 1 e y dy x – 1
2
33. + = , y = ln 2 when x = 1 34. xy = , y = 0 when x = 1
dx x x dx y – 1
2 dy
35. (1 + x ) + 2xy = 4x, y = 0 when x = 1
dx
133



04 STPM Math(T) T2.indd 133 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

4.3 First Order Linear Differential Equations


dy
The first order linear differential equation is of the form + f(x)y = g(x) where f(x) and g(x) are functions of
dx
x. It can be solved by multiplying both sides of the equation with e ∫ f(x) dx which is called the integrating factor;
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dy
+ f(x)y = g(x)
dx
4
3 dy + f(x)y e ∫ f(x) dx = g(x)e ∫ f(x) dx
dx
1
3 dy e ∫ f(x) dx + f(x) e ∫ f(x) dx 2 4 ∫ f(x) dx
y = g(x) e
dx
Note that the left hand side is the differentiation of ye ∫ f(x) dx .

2
d d d
i.e. y e ∫ f(x) dx = 1 y e ∫ f(x) dx + y 1 e ∫ f(x) dx 2
dx dx dx
dy
= 1 2 e ∫ f(x) dx + y 1f(x) e ∫ f(x) dx 2
dx
d
4
\ 3 dx 1ye ∫ f(x) dx 2 = g(x)e ∫ f(x) dx
Integrating both sides with respect to x:
∫1 dx ye ∫ f(x) dx 2 dx = g(x)e ∫ f(x) dx dx
d
∫

∫
\ ye ∫ f(x) dx = g(x)e ∫ f(x) dx dx + C
4 ∫ g(x)e ∫ f(x) dx dx
\ y = + Ce –∫ f(x) dx
e ∫ f(x) dx


Example 8

dy
2
Find the general solution to the differential equation x + 2x = y.
dx
dy
2
Solution: x + 2x = y
dx dy
Arrange the differential equation to the form dx + f(x)y = g(x).
dy y
Divide by x: + 2x =
dx x
dy – 1 x 2 y = –2x
1
dx
dy + – 1 y = –2x
1 2
dx x
1
\ f(x) = – and g(x) = –2x
x
1
–—dx
\ The integrating factor is e ∫ x = e –ln x = e ln x –1 = x . (Notes: e ln a = a)
–1

134






04 STPM Math(T) T2.indd 134 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

1
–∫
dx
Multiplying both sides of the differential equation with the integrating factor is e x .
dy e – —dx 2 – 1 e – —dx 2 y = –2x e – —dx 2
1
1
1
1
1
∫ x
∫ x
∫ x
dx x 1
d ye – —dx 2 = –2xe – —dx
1
1
1
∫ x
∫ x
dx
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1
1
1
∫ x
∫ x
= –2xe
\ ye – —dx 2 ∫ – —dx dx
∫
–1
–1
\ y(x ) = – 2x(x ) dx
∫
yx = –2 dx
–1
–1
yx = –2x + A
\ y = –2x + Ax
2
Example 9
dy
Find the general solution to the differential equation = x – 2y.
dx
dy
Solution: = x – 2y
dx
dy
Arrange the differential equation to the form dx + f(x)y = g(x).
dy
+ 2y = x
dx
\ f(x) = 2 and g(x) = x.
4
So the integrating factor is e ∫ 2 dx = e .
2x
Multiplying both sides of the equation with e ;
2x
dy (e ) + 2(e )y = x(e )
2x
2x
2x
dx
\ d (ye ) = xe 2x
2x
dx
∫
2x
\ ye = xe dx + C
2x
Integrating the right side by parts:
du = e and v = x
2x
dx
∫
2x
\ u = e dx = 1 e and dv = 1
2x
2
dx
1
∫
\ ye = 1 xe – e dx + C
2x
2x
2x
2 2
\ ye = 1 xe – 1 e + C
2x
2x
2x
2 4
\ y = 1 x – 1 + Ce –2x
2 4
135
04 STPM Math(T) T2.indd 135 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Exercise 4.3


1. Solve the differential equations
2 dy
(a) (1 + x ) + xy = x + x 3 (b) dy + 3y = x
dx dx
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dy dy
2
(c) x + y = 2x (d) = 2x – y
dx dx
dy
(e) 3 + 15y = 24e 3x
dx
2. Solve the differential equations given the boundary conditions:
dy 2 dy
(a) x – 3y = x(1 + x ); x = 1, y = 1 (b) (1 + x ) + xy = x ; x = 1, y = 0
3
dx dx
dy dy
–x
(c) + 3y = 2e ; x = 0, y = 1 (d) – 2xy = x ; x = 1, y = 1
dx dx




4.4 Transformations Differential Equations


Some first order differential equations which are not of the form where the variables are separable can be reduced
to the separable forms by suitable substitutions.



4
Example 10

dy y
2
Solve the differential equation + x y = – .
dx x
dy y
2
Solution: The variables of the differential equation dx + x y = – are not separable.
x
Let y = v x

Differentiating with respect to x:
dy = x v + v d x –1
–1 d
dx dx dx
dy = x + v(–x )
–1 dv
–2
dx dx
dy 1 dv v
\ = –
dx x dx x 2








136






04 STPM Math(T) T2.indd 136 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

dy y
So + x y = – reduces to
2
dx x
1 dv v 2 v 1 v
x 1 x 2
1 x dx – x 22 + x 1 x 2 = –
1 dv – v + vx = – v

x dx x 2 x 2
1 dv + vx = 0

x dx

\ dv = –vx which is the form where the variables are separable.
2
dx
Separating the variables and integrating with respect to x.
∫
2
\ dv = – x dx
∫ v
\ ln v = – x 3 2Sdn Bhd. All Rights Reserved.
+ A
3
Substituting v = yx
\ ln yx = – x 3 + A is the general solution.
3


Example 11


2
3
dy y – x y
Solve the differential equation = .
dx x + xy 2
3
Penerbitan Pelangi
4
3
dy y – x y
Solution: =
3
dx x + xy 2
Substituting y = vx,
3
v + x dv = v – v dy = v + x dv
dx 1 + v 2 dx dx
3
x dv = v – v – v
dx 1 + v 2
x dv = – 2v
dx 1 + v 2

∫
∫ 1 + v 2 dv = – dx
2
v
x
ln | v | + 1 v = –2 ln | x | + C C is an arbitrary constant.
2
2
2
ln | vx | + 1 v = C
2
2
ln | yx | = – y 2 + C
2x 2
y 2
–—
yx = Ae 2x where A = e C

2
137



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Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 12


dy x + y + 5 dv 2(v + 1)
If v = x + y, show that the equation = x + y – 3 can be reduced to = v – 3 , and hence solve
the equation. dx dx
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dy x + y + 5
Solution: =
dx x + y – 3
Substituting v = x + y or y = v – x,
d v + 5
(v – x) =
dx v – 3
dv – 1 = v + 5
dx v – 3
dv = v + 5 + 1
dx v – 3
= 2(v + 1)
v – 3
∫
∫ v – 3 dv = 2 dx
v + 1
∫
4
∫1 v + 1 2 dv = 2 dx
1 –
v – 4 ln | v + 1 | = 2x + C
x + y – 4 ln | x + y + 1 | = 2x + C
1
ln | x + y + 1 | = 1 (y – x) + D D = – C
4 4
1
x + y + 1 = Ae —(y – x) A = e D
4
1
x + y + 1 – Ae —(y – x) = 0
4
4
Example 13


3
By writing u = 1 , reduce the equation 2 dy – y = y to du + u = –1.
y 2 dx x dx x
Hence solve the differential equation.
Solution: Differentiating u = 1 with respect to x
y 2
du d 2 dy
–2
= y = –
dx dy y 3 dx
dy y
3
Dividing the equation 2 – = y with y .
3
dx x
2 dy – 1 = 1
y 3 dx xy 2
du u
\ – – = 1
dx x
du u
\ + = –1
dx x
1 dx
∫
ln x
The integrating factor is e x = e = x.
138






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Mathematics Semester 2 STPM Chapter 4 Differential Equations

1 dx
∫
Multiplying both sides with the integrating factor e x :
du 1 u 1 1
∫
∫
∫
e x dx + e x dx = –1(e x dx )
dx x
du
x + u = –x
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dx
d
(ux) = –x
dx
∫
\ ux = –x dx + A
\ ux = – x 2 + A
2
Substituting u = 1 ,
y 2
\ x = – x 2 + A
y 2 2

2
\ y = 2x
–x + 2A
2
2x
2

\ y = ± 
2
–x + 2A
Exercise 4.4


1. Solve the differential equations 4
dy x + y dy x – 2y
(a) = (b) =
dx x dx x
2
dy x + 2y 2 dy
2
(c) = (d) xy = x + y 2
dx xy dx
dy dy y – yx 2
3
(e) xy = x – y 2 (f) =
2
dx dx x + y x
3
2
dy xy
2. Show that, by substituting y = vx, the differential equation dx = x + y 2 can be transformed into
2
x dv + v 3 = 0. Hence, find the particular solution if y = 2 when x = 1.
dx 1 + v 2
dy y – x + 1
3. Using the substitution y – x = z, show that the differential equation dx = y – x + 5 can be transformed
into dz = –   4 . Hence, solve the differential equation.
dx z + 5
dy x + y + 2
4. Using the substitution z = x + y, solve the differential equation = .
dx x + y + 1
dy 2x – y – 1
5. Solve the differential equation = by using the substitution v = 2x – y.
dx 2x – y + 3



139






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Mathematics Semester 2 STPM Chapter 4 Differential Equations
dy
2
2
6. Using the substitution y = vx, find the general solution of the differential equation 2xy = y – 4x .
dx
Show that the particular solution which satisfies the condition y = a (a  0) when x = 1 a is
2
y = 4x(a – x).
2
3
3
7. Show that the differential equation xy 2 dy – x – y = 0 can be transformed into the differential equation
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dv 1 dx
x = by using the substitution y = vx. Hence, find the particular solution of the original equation
dx v 2
given that x = 1, y = 1.
dy
2
3
8. Show that the substitution v = x + y transforms the differential equation (x – 1) – 3x – 3y = 0 into
dv 3v dx
the differential equation = . Hence, find the general solution to the original differential equation.
dx (x – 1)
1 dy
9. By substituting z = , transform the differential equation + y = xy into a differential equation
3
y 2 dx
containing z and x. Solve the equation for z, and, hence, show that the solution of the given differential
1
2

equation for which y = 2 when x = 0 is y = 1 2x + 1 2 — 2 .
dy
10. By substituting u = y , transform the differential equation 2xy = y – 4x into a differential equation
2
2
2
dx
containing u and x. Find the particular solution of the given differential equation for which y = a when
u = 1 a.
2
dy
11. By writing u = 1 , reduce the differential equation dx + x y = y to du – u = –1.
2
x
y
dx
Hence, solve the original differential equation when x = 1, y = 1.
4
4.5 Problems Modelled by Differential
Equations
There are many physical situations in which different variables changes at different rates. Differential equations
always arise when we model these physical situations in mathematics. The following situations show a few
physical problems that involve differential equations.
(a) Assuming a particle falls from rest in a medium that causes the velocity to increase at a rate proportional
to its velocity. By denoting velocity as v and time as t, the rate of increase of velocity can be written
as dv . Hence, the velocity of the particle satisfies the differential equation dv = kv.
dx dt
(b) The rate of decay of a radioactive substance is proportional to the amount of remaining substance.
Hence, the amount of remaining substance, x at any time t can be found by solving the differential equation
dx = –kx.
dt
(c) In a chemical process, a certain substance A continuously changes to another substance B. The rate of such
a change is proportional to the mass of A and inversely proportional to the mass of B, at any time t. The
total mass of A and B at any time remains constant and is equal to N. Hence, the mass of B, n, at time t
can be expressed by the differential equation dn = k (N – n).
dt n
2
(d) The rate of increase of a population due to birth is αx and the rate of decrease due to death is βx where
x is the population number at time t. The population number at any time can be found by solving the
2
differential equation dx = αx – βx .
dt
140






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Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 14

The rate at which a substance evaporates is k times the amount of the substance that has not been evaporated.
If the amount of the substance at the beginning is A and the amount that has been evaporated at time t is
x, write down a differential equation involving x, A, k and t.
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(a) Solve this differential equation and sketch a graph of x against t.
(b) Show that the time taken for half the amount to evaporate is 1 ln 2.
k 1
(c) Find the percentage of the amount of substance that has not been evaporated after time ln 2.
2k


Solution: (a) The differential equation is dx = k(A – x).
dt
1
∫
∫ (A – x) dx = k dt
– ln | A – x | = kt + C

When t = 0, x = 0, therefore C = –ln A.
Hence, – ln | A – x | = kt – ln A
x


In A – x = –kt A
A

 A – x = e –kt
A
t
0
–kt
x = A(1 – e )
(b) When half of the amount has evaporated, x = 1 A.
2
Thus, 4
–kt
A = A(1 – e )
2
1 – e –kt = 1
2
e = 1
–kt
2
–kt = ln 1 2
t = 1 ln 2
k
(c) When t = 1 ln 2,
2k
1
–— ln 2
x = A(1 – e )
2
1
–—
= A(1 – 2 )
2
= 0.2929A

Thus, the amount that has not evaporated is 0.7071A.
Hence, 70.71% of the substance has not evaporated.





141






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Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 15


An object moves along a straight line and passes a fixed point O with velocity u in the positive direction of
the x-axis. At time t, the object is at a displacement x from O and the velocity of the object is v. The rate of
change of velocity has magnitude k , where k is a constant, and is directed to towards the fixed point O.
v 2
(a) Write down a differential equation for the motion of the object involving the velocity v and the time t.
Hence, find the velocity v as a function of the time t.

(b) Show that dv = v dv and hence, write down the differential equation for the motion of the object.
dx
dt
Hence, find the velocity v as function of the displacement x.
4
—
3
4
Hence, show that after a time t and the object has moved a distance x, 4kx = u – (u – 3kt) .
3
Solution: (a) The differential equation for the motion of the object involving the velocity v
as a function of the time t is dv = – k .
dt v 2
(the negative sign shows that the rate of change of velocity is directed towards
the fixed point O but the motion of the object is in the opposite direction i.e.
positive direction of the x-axis)
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Separating the variables and integrating both sides:
∫
2
∫ v dv = – kdt
v 3 = –kt + C
3
When t = 0, v = u,

4 u 3 3 = –k(0) + C ⇒ C = u 3 3

Substituting the value of C
v 3 = –kt + u 3
3 v = –3kt + u 3 3
3

3
Penerbitan (b) Using the chain rule of differentiation. dx
–3kt + u
\ v = 
3

dv
dx
dv
1 21 2

=
dx
dt
dt
The rate of change of x with respect to time t,

dv
dv
\

dt = 1 2 v dt = v
dx
dv dv
\ = v
dt dx
142






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Mathematics Semester 2 STPM Chapter 4 Differential Equations

The differential equation for the motion of the object that involves the
change of velocity v with respect to the displacement x and the velocity v is
dv k
v = – .
dx v 2
Separating the variables and integrating both sides:
3Bhd. All Rights Reserved.
∫
3
∫ v dv = – k dx
v 4 = –kx + C
4

When x = 0, v = u
u 4 u 4
= –k(0) + C ⇒ C =
4 4

Substituting the value of C,
v 4 = –kx + u 4
4 4
4
v = –4kx + u 4
4
4
–4kx + u
\ v = 
Equating the velocity found in (a) and that in (b),
4
3
4
–4kx + u
3
–3kt + u
\  = 
Penerbitan Pelangi Sdn
–4kx + u = (u – 3kt) — 4 3
4
4
\ 4kx = u – (u – 3kt) — 4 3
3
4
Example 16
A research is being done on a particular species of bear on an animal reserve territory. Initially there are 50
bears on the reserve. After t years the number of bears, n, satisfies the differential equation
dn = 1 n(k – n) where k is a constant.
dt 20k
(a) Show that k = 200 if it is known that the rate of growth is 0.9 bear per year when n = 20.
(b) What is the maximum rate of growth?
(c) Explain what happens as n approaches 200?
(d) Obtain the solution of the differential equation and sketch the solution curve.
(e) Find
(i) the number of bears after 77 years,
(ii) the time for which the number of bears is 100.








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Mathematics Semester 2 STPM Chapter 4 Differential Equations


Solution: (a) Substitute n = 20 and dn = 0.9
dt
1
0.9 = (20)(k – 20)
20k
0.9k = k – 20 ⇒ k = 200
Hence, the maximum growth rate is 2.5 bears per year.Reserved.
1
(b) \ The differential equation is dn = 4000 n(200 – n).
dt
1
The expression 4000 n(200 – n) is a quadratic expression and has the maximum
1
1
(0 + 200) = 100. The maximum value of
n(200 – n) is
value at n =
2
4000
1
4000 (100)(200 – 100) = 2.5

(c) As n approaches 200, the growth rate of the bears approaches 0, i.e. there will
be no bear born.
(d) Separating the variables and integrating both sides:

dn
1
∫ 1 n(200 – n) 2 ∫ 4000 dt

=
1 1 + 1 2 dn = 1 ∫ dt
∫ 1
200 n (200 – n) 4000
n
(ln – ln (200 – n) = 0.05t + C
n
ln 1 200 – n 2 = 0.05t + C
4 Penerbitan Pelangi Sdn Bhd. All Rights
Substituting t = 0, n = 50

50
ln 1 200 – 50 2 = 0.05(0) + C

⇒ C = ln 1
3
n
\ ln 1 200 – n 2 = 0.05t + ln 1
3
n
1
ln 1 200 – n 2 – ln = 0.05t
3
3n
ln 1 200 – n 2 = 0.05t
3n
1 200 – n 2 = e 0.05t
1 200 – n 2 = –0.05t
e
3n
200 – n = 3ne –0.05t

\ n = 200
1 + 3e –0.05t




144






04 STPM Math(T) T2.indd 144 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations


The solution curve of n = 200 .
1 + 3e –0.05t
When t = 0, n = 50. As t → ∞, e –0.05t → 0, \ n → 200.

n
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200



50
O t

(e) (i) Substitute t = 77
\ n = 200 = 188 bears
1 + 3e –0.05 × 77

3n
(ii) Substitute n = 100 and using the form ln 1 200 – n 2 = 0.05t
300
t = 20 ln 1 200 – 100 2 = 22 years





Example 17

A series of regular interval injections is administered onto laboratory mice to test the efficacy of a new drug.
The rate of destruction of the drug is proportional to the amount of drug present in the mice.
(a) If k is the proportional constant, x is the amount of drug at time t, write a differential equation relating 4
–kt
the amount of drug and the time and hence, show that the general solution is x = Ae where A is an
arbitrary constant.
(b) Initially an amount, D, of the drug is injected onto a mice and after a time t = 1 hour the amount of
3
the drug remaining is 3 D. Show that x = D 1 2 t .
4 4
(c) The drug is injected again onto the mice after t = 1 hour and t = 2 hours. Find the amount of drug
remaining in the body immediately after 2 hours.
(d) If the drug is administered at regular intervals of 1 hour for an indefinite period, find the amount of
drug remaining in the mice.


Solution: (a) The differential equation for amount of drug remaining after a time t is
dx = –kx
dt
(The negative sign shows that the amount, x, is getting less with time, t)
Separating and integrating both sides.
1
∫ dx = – k dt
∫
x
ln = –kt + C
x



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Mathematics Semester 2 STPM Chapter 4 Differential Equations


When t = 0, let x = A
⇒ C = ln A
\ ln x = –kt + ln A

ln x = –kt
A
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x = e –kt
A
\ The solution of the differential equation is x = Ae .
–kt


(b) When t = 0, A = D ⇒ x = De –kt
When t = 1, x = 3 D ⇒ 3 D = De –k
4 4
e = 3
–k
4
–kt
–k t
The solution of the differential equation x = De can be written as x = D(e ) .
3
\ x = D 1 2 t
4
(c) After 2 hours:
The first dose of drug administered at time t = 0 will reduce to the amount of
3
D 1 2 2 .
4
The second dose of drug administered at time t = 1 will reduce to the amount
3
of D 1 2 1 .
4 4
The third dose of drug administered at time t = 2 has amount D.
\ Immediately after 2 hours the total amount of drug remaining is
3
5
3
D + D 1 2 + D 1 2 2 = 2 16 D
4
4
(d) Let the amount of drug remaining after an indefinite period be S.
3
3
3
3
S = D + D 1 2 + D 1 2 2 + D 1 2 3 + D 1 2 4 + …
4
4
4
4
3
3
1
2
S = D 1 + 3 + 1 2 1 2 3 + … 2
+
4
4
4
1
S = D 1 1 – 2
3
4
\ S = 4D
146






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Mathematics Semester 2 STPM Chapter 4 Differential Equations

Exercise 4.5


1. The height, h of a type of tree increases proportionately to the difference between its height at that time,
t and its final height, H. Find the differential equation connecting the variables h and t.
If h = 1 H when t = 0 and h = 1 H at t = T, solve the differential equation.
10 5
If time t is measured in days and T = 3, find the number of days passed before the height is more
than 9 H.
10

2. The manufacturer of a branded shampoo launches an advertisement program which results in the number
of consumers, n at time t to increase at a rate proportional to the square root of n. Write down the
differential equation which describes this relationship between n and t.
If n = N when t = 0 and n = 9 N when t = T, solve the differential equation. Find t when n = 4N.
4
3. Under certain conditions, the rate of cooling of a liquid is proportional to the difference between its
temperature and its surrounding temperature. The liquid is placed in a room of temperature 30°C and the
temperature of the liquid at time t (minutes) is x. Form a differential equation to describe this phenomena.
If the liquid cools from 100°C to 70°C in 8 minutes, find the further time taken for the liquid to reach a
temperature of 31°C. Find also the temperature of the liquid after it has been in the room for 10 minutes.


4. In a community, the number of people, n infected by a certain disease at a certain time increases at a rate
of λn. Form a differential equation connecting n and t.
λt
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Initially, the number of patients is N. Show that at day t, n = Ne . Find the value of λ if n = 2N when
t = 10. Find also
(a) the value of n when t = 20,
(b) the value of t when n = 3N.
4
5. In a certain chemical process, substance X is continuously changed to form substance Y. The total mass
of X and Y at any time is the same and equals to M. The rate of increase of Y at time t is proportional
to the mass of X at that time. If the mass of Y at time t is m, form a differential equation that describes
this chemical reaction. If M = 100 grams and 60 grams of substance X remains after 2 minutes, find the
mass of Y formed in 6 minutes.

6. At time t minutes, the number of micro-organisms in a liquid of a certain colony is x. The rate of increase
of x resulting from natural growth is αx and the rate of decrease due to death is β. Write down the
differential equation that describes the change in the size of the colony.
If the number of organisms is n at the beginning,
0
(a) find the time taken for all the micro-organisms in the colony to extinct if n =  200, α = 2 and
0
β = 500,
(b) find the number of micro-organisms in the colony after half a minute if n = 200, α = 4 and
0
β = 500.
7. At time t, the volume of water in a container is v. Due to leakage, water flows out at a rate of kv where k
is a positive constant. The rate at which water is lost due to evaporation is c. Obtain a differential equation
that describes the decrease in the volume of water. Given that v = V when t = 0, find the volume of water
in the container after t minutes. Find also the time taken before the container is empty.





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Mathematics Semester 2 STPM Chapter 4 Differential Equations
8. The rate of decomposition of a radioactive substance is proportional to the mass of the remaining substance
at that time. If two-thirds of the initial mass remains after 100 days, find the time taken for one-half of
the initial mass to remain. Find the percentage of the mass that remains after 120 days.


9. Heat is supplied to an electric kettle at a constant rate of 2 000 watts whereas heat is lost to the surroundings
–1All Rights Reserved.
at a rate of 20 watts for every Celsius degree difference between the temperature of the kettle and the
surrounding temperature. One watt of heat results in an increase in the temperature of the kettle at a rate
of 1 °C per minute. If the surrounding temperature is 15°C and θ°C is the temperature of the kettle after
50
t minutes, show that dθ = 40 – 2 (θ – 15).
dt 5
How long does it take for the temperature of the kettle to increase from 15°C to 100°C?


10. In a chemical reaction, two substances A and B react together to form another substance C. At time t, the
amount of A and B are a – x and b – x respectively, where a and b are constants and x is a function of
t. The value of x is zero when t = 0. At any time, the rate of decrease of A is proportional to the product
of the amount of A and B at that time. Find a differential equation involving x and t and solve x as a
function of t in each case of (i) a = b, (ii) a  b.

2
11. A water tank is in the shape of a cylinder with a vertical axis. The base area is A m . Initially, the tank
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is empty. Starting at time t = 0, water is poured into the tank at a constant rate of w m  s , and water
–1
3
3
flows out from a small hole at the base at a rate of kx m  s , where k is a constant and x m is the depth
of water in the tank. Form a differential equation and solve it to obtain a function of x in terms of t.
Show that no matter how long this process is repeated, the depth of the water will not exceed w m and
k
if the time taken to reach one-half of the depth of the cylinder is T s, then k = A ln 2.
T
4 12. An object moves along a straight line and passes a fixed point O with velocity u in the positive direction
of the x-axis. At time t the object is at a displacement x from O and the velocity of the object is v. The
rate of change of velocity has magnitude c v , where c is a constant, and is directed to towards the fixed
point O.
(a) Write down a differential equation for the motion of the object involving the velocity v and the time
t. Hence, find the velocity v as a function of the time t.
(b) Show that dv = v dv and hence, write down the differential equation for the motion of the object
dt dx
that involves the change of velocity v with respect to the displacement x and the velocity v.
Hence, find the velocity v as function of the displacement x.
2
—
3
Hence show that after a time t and the object has moved a distance x, 3cx = u – (u – 2ct) and deduce
2
3
2
that 2ct , u .
13. A research has been set up on an island to study a particular species of turtle. Initially there are 25 turtles
1
on the island. After t years the number of turtles x satisfies the differential equation dx = 20k x(k – x)
dt
where k is a constant.
(a) Show that k = 100 if it is known that the rate of growth is 0.45 turtle per year when x = 10.
(b) What is the maximum rate of growth?




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Mathematics Semester 2 STPM Chapter 4 Differential Equations
(c) Explain what happens as x approaches 100?
(d) Obtain the solution of the differential equation and sketch the solution curve.
(e) Find
(i) the number of turtles after 30 years,
(ii) the time for which the number of turtles is 50.
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14. The rate of destruction of a particular drug is proportional to the amount of drug present in the body.
(a) If x is the amount of drug at time, t, write a differential equation relating the amount of drug and
the time and hence, solve the differential equation.
(b) Initially an amount, Q, of the drug is injected to a body and after a time t = 1 hour the amount of
the drug remaining is 2 Q. Obtain an expression for x in terms of Q and t.
3
(c) The drug is injected again to the body after t = 1 hour, t = 2 hours and t = 3 hours. Find that the
amount of drug remaining in the body immediately after 3 hours.
(d) If the drug is administered at regular intervals of 1 hour for an indefinite period, find the amount
of drug remaining in the body.







Summary



1. A differential equation is an equation involving at least one derivative of y with respect to x,

2
n
dy d y d y 4
e.g. , , …, , where x is the independent variable and y the dependent variable.
dx dx 2 dx n
2. The order of a differential equation is the order of the highest derivative found in the differential equation.


3. The solution of a differential equation that contains of an arbitrary constant is a general solution.


4. The solution of a differential equation that satisfies certain conditions (initial or boundary conditions) is
a particular solution.

dy dy
.
5. If the differential equation = f(x, y) can be expressed as = u(x) v(y), where u(x) and v(y) are
dx dx
functions of x and y, it is a differential equation with separable variables.
dy
6. The linear differential equation + f(x) y = g(x), where f(x) and g(x) are functions of x, can be solved
dx
by multiplying both sides of the equation by an integrating factor e ∫ f(x) dx .








149






04 STPM Math(T) T2.indd 149 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations

STPM PRACTICE 4


dy
2
2
1. Solve the differential equation 2xy = x + 2y given that y = 0 when x = 1.
dx
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2 dy
y
2. Find the solution of the differential equation x = 4e that satisfies the condition y = 0 when x = 2.
dx
1 dy y
2
3. Show that the substitution u = transforms the non-linear differential equation – = y into
y
du – u = –1. Solve this linear equation given that y = 2 when x = 1. dx x
dx x
4. Differentiate y e with respect to x. Find the particular solution of the differential equation
2 x
2x
2ye x dy + y e = e if y = 0 when x = 1.
2 x
dx
5. Using the substitution z = sin y, find the general solution of the differential equation
dy + 1 tan y = 1 sec y.
dx x x 2

dy
6. Using the substitution y = vx, solve the differential equation y = 2x + y given that y = 2 when x = 2.
dx

dy 2y
7. Show that the substitution y = vx transforms the differential equation x · = y + x cot 1 2 into
dv dx x
x · dx = cot 2v. Hence, find the particular solution of the given differential equation for which y = 1
4 when x = 0. Express your answer in the form y = f(x).


8. In a study on the effectiveness of a type of insect poison, it was found that the rate of decrease of the
dy 10
insect population, y is given by = – 1 2 , where t is the time taken in hours after the poison is
dt 1 + 5t
administered. Initially, there are 50 insects. Find
(a) the number of insects left 24 hours after the administration of poison,
(b) the time taken to destroy half the insect population.


9. A cultured bacteria of a species multiply at a rate that is directly proportional to the number of cultured
bacteria in the culture. If x is the number of bacteria in the culture at time t seconds, write down the
differential equation that describes the growth of the bacteria.
At the beginning of the experiment, there were 1 000 bacteria of a certain species. It was known that the
cultured bacteria multiply at a rate of 1.5 times per hour. Find the number of bacteria in the culture after
(a) 3 hours,
(b) 5 hours.
Find the time taken for the cultured bacteria to increase to five times the original number.







150






04 STPM Math(T) T2.indd 150 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations
10. At a chemical reaction, there is x kg of chemical reagent X and y kg of chemical reagent Y at time t.
Initially, there were 1 kg of X and 2 kg of Y. The variables of x and y satisfy the following differential
equations.
dx = –x y and dy = –xy .
2
2
dt dt
dy
Find in terms of x and y and express y in terms of x. Hence, write a differential equation relating
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dx
x and t. Express x in terms of t.
11. An object moves in a medium and experiences a resistance so much so that the velocity v decreases at a
rate given by dv = –kv where k is a positive constant.
dt
–1
–1
If the initial velocity of the object is 100 m s and its velocity is 40 m s after 2 seconds, show that
k = 1 ln 5 and find its velocity as a function of t. Hence, find the distance travelled by the object in
2 2
the first 2 seconds.

12. The rate of growth of a certain animal population is directly proportional to the population. If x is the
animal population at time t (measured in years), write down a differential equation relating x and t. Find
the animal population in year 2000 based on the information given below.

Number Year Population
1 1970 3 683 thousands

2 1980 4 453 thousands



13. A firm uses a computer software to control the work of the machines that produce certain electronic
components. It is known that the time T, taken by the computer software increases with the number of 4
machines used, N at a rate given by the equation dT = 1 + ln N.
dN
Given that the computer software takes one second to control 50 machines, find the time taken to control
100 machines.

14. A contagious disease spreads at a rate directly proportional to the product of the number of the population
infected and the remaining population that is not infected. Initially, one-half of the population is infected
and if the rate of infection is kept constant, the whole population will be infected in 24 days. Find the
proportion of the population that will be infected after 12 days.
[Hint: If the proportion of the population infected is x, then the proportion of the population not infected
will be (1 – x).]

15. A tank contains 10 kg of sodium hydroxide in 1 000 litres of water. Water is continuously added to the
–1
tank at a rate of 5 litres min so that the mixture is diluted evenly. At the same time, the solution flows
out at the same rate. Initially, there were 10 grams of sodium hydroxide in every litre of water. How long
will it take for the concentration of the solution to drop to 3.2 grams per litre?

dy
16. Find the solution of the differential equation 3x · – 4y = 1 which satisfies the condition y = 0 when
dx
x = 1. Give your answer in the form y = f(x).




151






04 STPM Math(T) T2.indd 151 28/01/2022 5:44 PM

Mathematics Semester 2 STPM Chapter 4 Differential Equations
17. In a rabbit farm there are 500 rabbbits and two rabbits are infected with Myxomatosis, a devastating viral
infection, in the month of April. The farm owner has decided to cull the rabbits if 20% of the population
is infected. The rate of increase of the number of infected rabbits x at t days is given by the differential
equations dx = kx(500 – x) where k is a constant.
dt
Assuming that no rabbits leave the farm during the outbreak,
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(a) show that x = 1000 ,
498e –500kt + 2
(b) if it is found that, after two days, there are ten infected rabbits, show that k = 1 ln 249 .
1000 49
(c) determine the number of days before culling will be launched.


18. Find the general solution to the differential equation ln x dy = tan y .
dx x

3 dy
4
4
19. Using the substitution y = vx, show that the differential equation xy dx – x – y = 0 may be reduced
3 dv
to v x = 1. Hence, find the particular solution that satisfies y = 1 and x = 1.
dx
3
dy x + x y + y 3
2
20. The variables x and y, x  0 and y  0, satisfy the differential equation x = . Show
2
dx x + y 2
that the substitution y = ux transforms the given differential equation into the differential equation
du 1
x = . Hence, find the solution of the differential equation for which y = 1 and x = 1.
dx 1 + u 2
21. By using the substitution y = vx transform the equation
dy y dv
x = y + x tan 1 2 into x = tan v.
dx x dx
4 Hence, find the solution of the given differential equation satisfying the condition y = π when x = 1.
Give your answer in the form y = f(x). 2

dy
22. Solve the differential equation x · dx = 2x – y with the condition y = 2 when x = 3. Express your answer
in the form of y = f(x).


dy 2x + y – 1
23. The variables x and y are related by the differential equation = . Show that the substitution
dx 2x + y + 5
dV 3V + 9
V = 2x + y transforms the differential equation to = . Hence, find the particular solution of
dx V + 5
the differential equation given that y = 1 and x = 1.

dy
24. Find the general solution of the differential equation cos x · – y sin x = 4 sin x cos x.
dx
dy y – 9
2
25. Find the general solution of the differential equation dx = 6x . Express your answer in the form of
y = f(x).

dy
2
26. Find the particular solution of the differential equation x · dx – y = x (ln x) with the condition y = 3
when x = 1.


152






04 STPM Math(T) T2.indd 152 28/01/2022 5:44 PM

ANSWERS




1 Limits and Continuity (c) y (d) y
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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 0 1 x
4 5 –2
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.
–1
f(x )

(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
x 0 π –1
0 π 2π

continuous not continuous






195






Ans STPM Math T Term2.indd 195 28/01/2022 5:52 PM

Mathematics Semester 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
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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
x
–5
3
f(x ) 3
7. – 1 8. 3 x
x 2 2
1 – —
9. – x 2 3 10. 2 – — 1
x
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
x
1. 2 ln 2 2. 1 2 ln
lim
lim
2. (a) (i) x → c f(x) ≠ x → c f(x) 5 5
–
+
3
lim
lim
(ii) x → c f(x) = x → c f(x) ≠ f(c) 3. 10x + 3 4. 4x – 18x 2
+
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 11. – 2 – 2x 12. – 1
5. (a) 8 (b) –2√ 7 3x 2x
27 13. 10x – 3 + 8 14. 1 + 4
6. (a) b = 3 (b) a = ±√ 6 3 x 2 x 3 x 2 x 3
lim
lim
lim
Penerbitan 3. 8x – 30x + 1 4. 12 – 1 – 2x
–
7. (a) x → 0 f(x) = x → 0 + f(x) = 1, x → 0 f(x) exists. 15. – 2x 3 16. 2 cos x – 3 sin x
lim
18. 13
17. 13
4
f(x) is discontinuous at x = 0
f(x) is continuous in the interval (–∞, 0) < (0, ∞)
1. 2x(2x + 1)
2

8. a = 1, b = –1 Exercise 2.4 2. 24x – 5
3
2
lim
lim
9. x → 3 f(x) = –6 ≠ x → 3 + f(x) = 6 2 x 4 x 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
2
x
9. 1 + ln x
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
196
Ans STPM Math T Term2.indd 196 28/01/2022 5:52 PM

Mathematics Semester 2 STPM Answers
2t
2
(c) –3t (t – 1) 2 (d) 2
2
4
3. 2(2 – x) 4. –3x + 6x + 8x 3t + 1
3
2 2
(x + 2) 3 (x + 2x ) 2. (a) –cot q (b) –tan q
5. 2x cos x – sin x 6. (ln x) – 1 3. 1
2x x (ln x) 2 4. cos t
t
e
x
7. 1 8. e (x – 2) 5. (0, 2), (0, –2)
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1 + cos x (x – 1) 2 9. (2, 2), (2, –2)
2
9. sec x 10. –cosec x
2
11. sec x tan x 12. –cosec x cot x Exercise 2.9
x
13. 2 14. e 1. (a) 2 1 (b) 34
1 – sin 2x (2 + e ) (c) 2 (d) –1
x 2
4
15. 1 1
1
x (1 – x ) 2 2. –2, 2 2
Exercise 2.6 3. (a) – 10 (b) – 3
13
4
1. 3e 3x 2. 2x e x 2 4. (a) (3, –5) (b) (2, –8)
3. 2 4. 6(x + 1) 5 (c) (0, –8)
x – 1
π
2
1
3
5. 36x (2x + 1) 5 6. 8x 5. 2n π + , 1
2
2
x + 1 2
–2
7. 2 sin x cos x 8. – (x + 1) Exercise 2.10
π
2
9. 2x e (x + 3) 10. 3 cos (3x + ) 1. Increase for x . 3,
4 decrease for x , 3
2
11. 4x sin (x + 1) cos (x + 1)
2
12. 2 cot 2x 13. –(e + e )(e – e ) 2. Increase for x . 4,
–x –2
x
x
–x
decrease for x , 4

2
14. –cosec x 15. – 4x sin (2 + 2x )
3. Increase for x , –1,
16. (2x – 3)(43 – 6x) 17. –sin x decrease for x . –1
(3x + 4) 4 1 + cos x 4. Increase for x , –2 and x . 2,
2
18. x 19. 2 sec 2x decrease for –2 , x , 0 and 0 , x , 2
1 + x 2 tan 2x
4
20. 2x + 5 21. tan x Exercise 2.11
2
2 3) 1. – , – 25 2
(x + 5x +
1
1
24. – 1 3 25. –1 1 2 14 4
2 1 1 π – 1 2 — 2 2. 1 3 , – 27 2 , (3, –10)
2
3. (1, 0)
Exercise 2.7 4. (–2, 9)
1. (a) x + y dy = 0 5. (–1, –2), (1, 2)
dx
(b) 8x + 18y dy = 13 6. (2, 108), (0, 0), (5, 0)
dx 1 3
(c) 2x + y + (x + 2y) dy = 0 7. 1 4 ln , 12 2
2
dx
dy π π 5π 5π
(d) x + (y – 3) = 0 8. 1 , + 3 ), ( , – 3 2
2
2
dx 6 6 6 6
2
(e) 3x + y + (2xy – 5x) dy = 5y — 1 2 1
2
1
dx
2 3 9. e , 2e 2
2. (a) 2y – 2xy (b) cot x cot y 1
2 2
3x y – 4xy – 3 10. 1 2 ln 2, 2 2 2
(c) x (d) 2x + 5y + 2
y 2y – 5x + 3 Exercise 2.12
(e) sin 2x – cos 2x – 3y 1. (a) (0, 0) maximum, 1 2 –4 2 minimum
,
3 3x
x e x e 3 27
2 3x
6. 1 (b) (2, 10) point of inflexion
7. 3 (c) (2, 32) maximum, (6, 0) minimum
8. – 1 (d) (–1, –1) minimum, (0, 0) point of inflexion
3 2. (3, 9)
Exercise 2.8 3. (a) (0, 0) (b) (2, –11)
2
1. (a) 1 (b) 3(t + 1) (c) (–1, –25), (3, –173) (d) (2, 0), (1, –1)
t 2t
197
Ans STPM Math T Term2.indd 197 28/01/2022 5:52 PM

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