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Lesson 29

The document contains 4 problems about functions. Problem 1 defines functions f and g and asks to find expressions and state domains. Problem 2 defines function g and asks to find an expression, domain and range. Problem 3 defines functions f and g and asks to find a range, domain, and solve an equation. Problem 4 defines functions f and g and asks to find expressions.

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0% found this document useful (0 votes)
23 views28 pages

Lesson 29

The document contains 4 problems about functions. Problem 1 defines functions f and g and asks to find expressions and state domains. Problem 2 defines function g and asks to find an expression, domain and range. Problem 3 defines functions f and g and asks to find a range, domain, and solve an equation. Problem 4 defines functions f and g and asks to find expressions.

Uploaded by

Chan Yuyan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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1

1 for x > 0 .

The range of f is f > 9.

(a) Find

1 =
(2x +
3)
a= (21 +
3)

2y
.
x = +
)

1-3 y
=

-
2

f (m) / :

- [3]

(b) State the domain of

19
[1]

[Total: 4]

2 The function g is defined by


(4 4) /
g(u) =
-
+

Find an expression for g(x)-1 and state its domain and range.

1
=
(x 4) -
+

domain 1)
4)
(y
+
x = -

g"(n)
=

x
-

1 =

(y
-

4) range

= 4
y
-

x -

1 =

4=
y 1
=
x
-

[4]
4
y
=

x -

1
[Total: 4]
.

g(x) 4
/
=
=
n -
2

3 The functions f and g are defined by

(a) State the range of g.


(x) =
4(0)" 9 -

9 -

=
9

9
g(n)a
-

[1]

(b) Find the domain of gf.

177

[1]
3

(c) Showing all your working, find the exact solution of .

g(5x 2) 45
-

4(5x -
2) 9 45 =

4(5x -

2) =

54
54
(5x 2)2 -
:

54 2
=
-

5x 2
c
= +

x =

j2 +

[3]

[Total: 5]

4 The functions f and g are defined for real values of x by

for ,

.
4

Find an expression for

(a) ,

+
y
=

eY -
2

=
[2]

(b) ,

g(i
+ 1) =

m 1)
+ + 2

[2]

(c) .

f(x) 2)
+ =

y
+
2
+

[2]
5

(d) Show that and solve

+ (i
: x + 2

1)
+ 3
+ =

- +
1 2 +
x
x))

2
3x +
2 = 1)(2 x) +

= T

2 +
1 3x + 2 = 2(n + (2
Ch

x + 2x -
x
-
2 =

21) 2 + 3)
- +

2 + 1) 2 + 1) x 1 -
2 =
0

2x) 2 + 1)
2)(x 1) 0
+
- (x - + =

2 + 1)
1= 2 Ch = -

34 + 2

(reg)
=

2 +
1)

[4]

[Total: 10]

5 The function is divisible by and leaves a remainder


of 20 when divided by .

(a) Show that b = 6 and find the value of a.

3 ) a(b)
+ +
+
b = ( (1) + + al -

1) b + = 2

2 +
9 +
3a + b =
-

+
1 -

a +
b =
20
[4]
3a + b =
-

3) -
1
-

a + b =
20 - 2

1 -
2 3a -

( a)- = -

36 -

20

4a = -

5) -

( 14) b
- +
= 20

a =
-
1 b : 20 -

14

b =
6
6

(b) Using your value of a and taking b as 6, find the non-integer roots of the equation
in the form ,where p and q are integers.

f(x) 1 =
+ 1 -
141+ 6

( + 4x) -
2
x + 4x -
2 =
0
x -

)(o
( +
y -

14x +
6
43 - 3) -
45 4" -

4(11) 2) -

7) = .

2()
*
44 -
141
4x2 -
121
47/16
-

= +
8
-
21 +
6
2
-

2x + 6
41 24
-

-
.

= - 41 216
2

= -

27 6
.

[5]

[Total: 9]

6 The polynomial has a factor of and a remainder of 5 when divided


by .

a (2)" 15(2) b(2) +


-

2 :
a(l)) 15(1)" b(1)
-
+ -
2 =

15 b2 5
i 3
a

8
+ :

-
+ -
2 :

a + b =
22 2
xj
a -
30 +
4616 :

a+ 46 46 = I

1 -
24b -
b 4) : -
22 a
+
8 =
2)

3b : 2 a = k

b =
0
7

(a) Show that b = 8 and find the value of a.

[4]

(b) Using the values of a and b from part (a), express in the form , where is a
quadratic factor to be found.

f(x) =
4x3 -

151 + 0x -
2
L

x -
4x7 + 2

29 -

141" -
15a" + Bi-2 (2x) -

1)(7x -

4x + 2)
14123 7x2
-

8x +
01
-

8x + 41)

4x
-
2

4x -
2

[2]

(c) Show that the equation has only one real root.

(21-19779" 41 2) -
+
=

2x- 1 :
0 79 -
412 + 2 =

Cl =

! b 4a =
( 4) 4(7)(2)
=
1656 [2]
= -

40
[Total: 8]
b" 4al
-

7 It is given that x – 2 is a factor of .


8

(a) Find the value of the integer k.

2 k(2)"
+ -

0(2) 0 - =

0 +
41 16 - 8 =

4k =
16
k=

[2]

(b) Using your value of k, find the non-integer roots of the equation f(x) = 0 in the
form , where a and b are integers.

1 44 8n -0
+
=

" +
6114 x 6x
+ +
4 =

x -

2/a t
4, 0x8
"21"
-
6= 6 -
4(1)(4)
=

61x2 -

81 2(1)

6x" -
121)
=
-
61 20
.

4x -

8 2
41 -

8
-
=
-
6= 2/5

=
-

3 = [5

[5]

[Total: 7]
9

8 It is given that has a factor of x + 2 and leaves a remainder of 27


when divided by x – 1.

(a) Show that b = 40 and find the value of a.

[4]

(b) Show that , where p, q and r are integers to be found.

[2]

(c) Hence solve f(x) = 0.

[2]

[Total: 8]
10

9
The line intersects the curve at the points A and B. The x-coordinate of
A is less than the x-coordinate of B. Given that the point C lies on the line AB such that
AC : CB = 1 : 2, find the coordinates of C.

4 8
16 41
y
= -
=

3 16-412
= 4

4 2
=

4
x)" x
-

((4 -

4(4-1) 21 =
K(4-1
16 -

41217 413-4 :

12 101 +
16 =

I 81(1-2) :

x =
8 1 =
2

1 =
16 4(8)
-

y
:
16 -

4(2)
=
16 32 =
16-8
2)
-

(1
16
,

j
= -
=

& 2
A(2 8) ,
B(8 - 16) ,

2(2) +
1(8) 2(8) 1( 16) +
-

x
y
= =

1 +
2
1 + 2

4 +8
=

=
16-16
3 J
12
I
8
I

=
3 3

14, 0)

[8]
11

[Total: 8]

10 Find the coordinates of the points of intersection of the curve and the line .

5 3
4 + +
-

(5x + 18 3

+j

↓ 3
4 + + =

311+ 2 1
x(3x +
2)
1x(3x 2)1()13(3x 2)
+ + =

124084 +
11941J =

124 1 1813 :
6 0 =

2x + 3x + 1 =

(2x +
1((x +
1) :

a=
21 =

y 1572) 15) 1) 18
=
1
y
+ =
-
+

2 =
=
-
5

(2 2) ( 1 5
-
-

,
,

[6]

[Total: 6]
12

11 Find the coordinates of the points of intersection of the curve and the line .

8 10 x= 9 -
1

=
-

q
Y
-
-

y(9 y)8
-

1019 1) (9 1) 9 ( 15) 6
=

9
-
-

.
x = - -

x =
-

90 9-1 24 S
By
=

101
+ =
=

(24 15) (3 6)
By 9y 10y
90
-

,
1 + -
+ - ,

y 19 90
= :
.

153)y6) 6
4 + :

151
y
: - =

[6]

[Total: 6]
13

12 The line cuts the curve at the points A and B. Find the
length of the line AB.

[7]

[Total: 7]
14

13 (a) On the axes below, sketch the graph of showing the coordinates of the points
where the graph meets the coordinate axes.

1
y
= =

310) 14(0) E 31" -

141 -

5 :

y
-

= -

(3x +
1((x -

5) =

=
j
1: -

'y 1 :
5

I
-

5
-

3 + 5
a
=

2 3

: 373) 14(5) -
-
5 :
64

[4]
15

13 (b) Find the exact value of k such that has 3 solutions only.

64
k =

[3]

[Total: 7]

14 (a) Sketch the graph of on the axes below, showing the coordinates of the points where the
graph meets the coordinate axes.
( =

y
=

i =
5(0) -

] [1 3
= -

=
3 51 -

3 =

3
=
j 3

[3]
16

14 (b) Solve the equation .

593 = 2() 52 =
(2(
5x2 + 1 = 2+ ) 51 -
3 = 2 + 1)

61 = 5 x 3
(2 - = -
2 +

5
x =

4)) =
6
a= 1
[3]

[Total: 6]

15 (a) Sketch the graph of on the axes below, showing the coordinates of the points where the
graph meets the axes.

1) = 0
y
=

y
4x 2 :

4(0)
-

2
y
= -

43 = 2
=
2 2
17 :
4
1

=

O I x
2

[3]
17

15 (b) Solve the equation .

41 -
2 : 1) 4x - 2 =
1)

3 2

2
=

2
1=
3

[3]

[Total: 6]
18

16 Solve .

oga (5y-10) logaly-1)" :

loge
I
10)
(y
-

=
2
< 1
Y
-

by 2)
2y
10 = - +
1

2.
3y
- 10 : -
4. + 2
"+
4y
-
12 :

(16)(y -

2) =

Yi

[5]

[Total: 5]
19

17 Solve the equation .

1) 2
(g(x 192
2)
(g(3x
+ + + =
-

2((x + ) 2
g2(3x
-
=

2(3x 2((x - +
1) :

(3x 2((k 1)
-
+
=
5

3 1312x 2 -
-

50 =

3x + x) -

52 :

(3x +
1)((x -

4) =

= -

1 =
<

[5]
Crej)
[Total: 5]
20

18 Solve the following equation.

2
(294-15) J
oga
E

&
loga
(29-15)
2og
" 3
ogc
=

logull"
(29-15 3
g
· -
=
.

194-15
:2
ogs ,
291) -
15
= 2
12

29x -

15 =
84
89 29x +
15 0=

(8x -

j((x -
3) =

a =

% () 3
=

[5]

[Total: 5]
21

19
Q

5 cm

0.8 rad P
O R
x cm

The diagram shows a sector OPQ of a circle with centre O and radius x cm. Angle POQ is
0.8 radians. The point S lies on OQ such that OS = 5 cm. The point R lies on OP such that
angle ORS is a right angle. Given that the area of triangle ORS is one-fifth of the area of
sector OPQ, find

(a) the area of sector OPQ in terms of x and hence show that the value of x is 8.837 correct to
4 significant figures,

24 10.0) = 0 .4 12(5((3 48)(120 : .


8 =
6 . 247

1030 .
8 =
OR
:
247
50 44)
J 6 .
.

OR 5 COS Dij
=
-

Or =
3 .
48 x
= 6 .
247(5)
0 .
4
6 247(5)
a = .

-0 .
4

1= 8 .
83

[5]
22

(b) the perimeter of PQSR,

[3]

(c) the area of PQSR.

[2]

[Total: 10]

20 The diagram shows a sector, AOB, of a circle centre O, radius 12 cm. Angle AOB = 0.9 radians. The point C
lies on OA such that OC = CB.
23

0.9 rad
O 12 cm B

(a) Show that OC = 9.65 cm correct to 3 significant figures.

[2]

(b) Find the perimeter of the shaded region.

[3]
24

(c) Find the area of the shaded region.

[3]

[Total: 8]

21 The number of combinations of n items taken 3 at a time is 6 times the number of combinations of n items
taken 2 at a time. Find the value of the constant n.

[4]

[Total: 4]
25

22 The number of combinations of n items taken 3 at a time is 92n. Find the value of the constant n.

[4]

[Total: 4]

23 Jess wants to arrange 9 different books on a shelf. There are 4 mathematics books, 3 physics books and
2 chemistry books. Find the number of different possible arrangements of the books if

(a) there are no restrictions,

[1]

(b) a chemistry book is at each end of the shelf,

[2]
26

(c) all the mathematics books are kept together and all the physics books are kept together.

[3]

[Total: 6]

24 Jack has won 7 trophies for sport and wants to arrange them on a shelf. He has 2 trophies for cricket, 4 trophies
for football and 1 trophy for swimming. Find the number of different arrangements if

(a) there are no restrictions,

[1]
27

(b) the football trophies are to be kept together,

[3]

(c) the football trophies are to be kept together and the cricket trophies are to be kept together.

[3]

[Total: 7]
28

25 The diagram shows the graph of radians.

Find the value of each of the constants a, b and c.

[4]

[Total: 4]

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