I.G.C.S.E. Functions & Vectors Index: Please click on the question number you want

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 You can access the solutions from the end of each question

Question 1 The following functions Æ , g and h are defined as follows:

: 4 1x xâ âÆ , 2 2: 3

xg x ââ , 4:h x x

â .

Find a. ( )7Æ , ( )7âÆ , ( )17Æ . b. g(4), g(-3), ( )12g . c. h(10), h(-4), ( )18h â Click here to read the solution to this question Click here to return to the index

Solution to question 1 a. : 4 1x xâ âÆ ( ) ( )4 1 28 1 27 77 = â = â =Æ ( ) ( ) 27 74 2 1 91 8= â = â ââ = ââÆ ( ) ( )1 17 7 37474 1 1= â = = ââÆ

b. 2 2: 3

xg x ââ

( ) 2 2 16 2 14( 24

33 4)

4 3 3

g â

â

= = = =

( )2 2 9 2 7( ) 3 3 3

3 1 3

3 2g â

â

= = = =

â

â

( )12g . ( ) 2 1 7

41 4 1 2

2

2 2( ) 3 3

7 3 12

g â

â â

= â= = =

c. 4:h x x

â

( ) 4 210)

5 (

10 h = =

( ) 4( 14) 4

h â

= = ââ

( )18h â ( ) 1 8 1

8

4( ) 32h â â

= = â

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Question 2 For the following functions : 2 3x xâ âÆ and 2: 2g x x xâ â , find a. the value of x such that ( ) 3x = âÆ b. the values of x such that ( ) 3g x = Click here to read the solution to this question Click here to return to the index

Solution to question 2 a. : 2 3x xâ âÆ

( ) 3x = âÆ â2 3 3 3

5 5

3

x x

x

â = â

â = â

=

b. 2: 2g x x xâ â ( ) 3g x = â 2

2

2 3 2 3 0

x x x x

â =

â â =

product = -3 sum = -2 factors -3, 1

( ) ( ) ( )( )

2 3 0 3 3

3 0

3 1 0

x x x x

x

x x

x

â =

â + â =

â +

â

=

+

â 3x = or 1x = â Click here to read the question again Click here to return to the index

Question 3 Given that the function Æ is defined as : x ax bâ +Æ , where a and b are constants. If ( )2 3=Æ and ( )4 9â =Æ . Click here to read the solution to this question Click here to return to the index

Solution to question 3

: x ax bâ +Æ ( ) 32 =Æ â 12 .3a b+ = â¦

( )4 9â =Æ â 24 .9a b+ =â â¦

Solving equations 1 and 2 simultaneously we have 1 2â 1.

2. 2 3 4 9 6 6

a b a b a

+ =

â + =

= â

â¦ â¦

â 1a = â Substitute 1a = â into equation 1 we have

( )2 3 2

5

1 3

b b b

+ =

â + =

=

â

Check in equation 2 we have ( ) ( )4 5 91 45â + = + =â Hence 1a = â and 5b = . Click here to read the question again Click here to return to the index

Question 4

Given the functions ( ) 3 2l x x= â , ( ) 22 4m x x= + and ( ) 4n x x

= find

a. ( )lm x b. ( )ml x c. ( )lmn x d. ( )2lmn â . Click here to read the solution to this question Click here to return to the index

Solution to question 4

( ) 3 2l x x= â , ( ) 22 4m x x= + and ( ) 4n x x

=

a. ( ) ( ) ( ) ( )2 2 22 22 4 2 4 2 33 2 6 1 2 1 52 6 0lm x l xx x x x= = â = + â = + =+ ++ b. ( ) ( ) ( ) ( )

( ) 2

2

2

2 2

2 4 2 9 12 4 4

18

3 2

6 3 424 8

3 2

24 18 24 12 x

xml x m x x

x xx x x

x= = + = â + +

= â + â+ = â + =

â

+

â

c. ( ) ( )

2 32 324 44 4 962 4 3 2 12 2

96 96 101 2

0 44 5

x lmn x lm

x

l l

x x x x

xx x x ï£® ï£¹ï£« ï£¶ ï£« ï£¶ ï£« ï£¶ ï£« ï£¶

= = + = = â = + âï£¯ ï£ºï£¬ ï£· ï£¬ ï£· ï£¬ ï£· ï£¬ ï£· ï£ ï£¸ ï£ ï£¸ ï£ ï£¸

+ + ï£ ï£¸ï£¯ ï£ºï£° ï£»

+ +

+ = = =

d. ( ) ( )( ) 2 24

2 2

3 5

4 4

lmn âï£® ï£¹+ ï£»

â

ï£° = â=â

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Question 5

Given the functions : 7 3x xâ âÆ and 3 5: 4

xg x ââ find the inverse

functions ( )-1 xÆ and ( )1g xâ . Click here to read the solution to this question Click here to return to the index

Solution to question 5

: 7 3x xâ âÆ Let 7 3

3 7 3

7

y x y x y x

= â

+ =

+ =

Interchange the x and y to give ( )-1 3 7

x x= +Æ

3 5:

4 xg x ââ

Let 3 5 4

4 3 5 4 5 3 4 5

3

xy

y x y x y x

â

=

= â

+ =

+ =

Interchange the x and y to give ( )1 4 5 3

g x xâ = +

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Question 6 OABC is a parallelogram where OA a=

"""# # and OC c=

"""# # .

Find in terms of a

# and c

# only

a. AB

"""# b. BC

"""# c. CA

"""# d. BO

"""# .

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a #

c #

C B

AO

Solution to question 6 OABC is a parallelogram where OA a=

"""# # and OC c=

"""# # .

a. AB OC c= =

"""# """# #

b. BC OA aâ= â =

"""# """# #

c. CA CO OA c a a c= + â= â + =

"""# """# " # # #"" ##

d. BO BC CO a c= + â= â

"""# """# "" #" ##

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a #

c #

C B

AO

Question 7 OPQR is a trapezium where 3RQ OP= and M is the point on PQ such that

2PM MQ= .

Given that OP p= """# #

and OR r= """# #

find in terms of p #

and r #

only

a. PR """#

b. RQ """#

c. PQ """#

d. PM """"#

e. MR """"#

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M

p #

r #

R Q

PO

Solution to question 7 OPQR is a trapezium where 3RQ OP= and M is the point on PQ such that

2PM MQ= .

OP p= """# #

and OR r= """# #

a. PR PO OR p r r p= + â= â + =

"""# """# " # # #"" ##

b. 3 3RQ OP p= =

"""# """ ##

c. 23PQ PO OR RQ p p rr p= + + = â + + = +

"""# """# """# """# # # # ##

d. ( )1 1 2 3 3

2 1 3 3

pP PQ r rM p= = + = + """"# """# # # ##

e. ( )2 2 4 22 3 23 3 3 3 3

5 3 3

MR MQ QR PQ QR p r p rp p pr= + = + = + â + â= â = """"# """"# """# """# """# # # # # # # ##

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M

p #

r #

R Q

PO

Question 8

Given the following vectors 2 3

a âï£« ï£¶

= ï£¬ ï£· ï£ ï£¸

# and

4 5

b ï£« ï£¶= ï£¬ ï£· âï£ ï£¸

# . Find the following giving

your answers exactly, wherever appropriate. a. a b+

## b. a bâ

## c. 3 2a bâ

## d. b

# e. 3 2a bâ

##

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Solution to question 8

2 3

a âï£« ï£¶

= ï£¬ ï£· ï£ ï£¸

# and

4 5

b ï£« ï£¶= ï£¬ ï£· âï£ ï£¸

# .

a. 24

5 2 2 3

a b âï£« ï£¶ ï£« ï£¶

+ = + =ï£¬ ï£· ï£¬ ï£· â

ï£« ï£¶ ï£¬ ï£· â ï£¸ï£ ï£ ï£ï£¸ ï£¸

##

b. 64

5 8 2 3

a b âï£« ï£¶ ï£« ï£¶

â = â =ï£¬ ï£· ï£¬ ï£· âï£ ï£¸ ï£

â

ï£· ï£¸ï£¸

ï£« ï£¶ ï£¬ ï£

##

c. 2 4 6 8

3 2 3 2 3 5 9 10

14 19

a b â âï£« ï£¶ ï£« ï£¶ ï£« ï£¶ ï£« ï£¶

â = â = â =ï£¬ ï£· ï£¬ ï£· ï£¬ ï£· ï£¬ ï£· â âï£ ï£¸ ï£ ï£¸ ï£ ï£¸ ï£

âï£« ï£¶ ï£¬ ï£· ï£ ï£¸ï£¸

##

d. ( )224 5 4116 25b = + â = + =# e. ( )2 23 2 14 19 196 3 576 51a bâ = â + = + =## Click here to read the question again Click here to return to the index