Why is energy never lost or destroyed?

Essay by Ed995High School, 11th gradeA, October 2014

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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 ƒ , g and h are defined as follows:

: 4 1x x→ −ƒ , 2 2: 3

xg x −→ , 4:h x x

→ .

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

Solution to question 1 a. : 4 1x x→ −ƒ ( ) ( )4 1 28 1 27 77 = − = − =ƒ ( ) ( ) 27 74 2 1 91 8= − = − −− = −−ƒ ( ) ( )1 17 7 37474 1 1= − = = −−ƒ

b. 2 2: 3

xg x −→

( ) 2 2 16 2 14( 24

33 4)

4 3 3

g −

−

= = = =

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

3 1 3

3 2g −

−

= = = =

−

−

( )12g . ( ) 2 1 7

41 4 1 2

2

2 2( ) 3 3

7 3 12

g −

− −

= −= = =

c. 4:h x x

→

( ) 4 210)

5 (

10 h = =

( ) 4( 14) 4

h −

= = −−

( )18h − ( ) 1 8 1

8

4( ) 32h − −

= = −

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Question 2 For the following functions : 2 3x x→ −ƒ and 2: 2g x x x→ − , find a. the value of x such that ( ) 3x = −ƒ 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 x→ −ƒ

( ) 3x = −ƒ ⇒2 3 3 3

5 5

3

x x

x

− = −

− = −

=

b. 2: 2g x x x→ − ( ) 3g x = ⇒ 2

2

2 3 2 3 0

x x x x

− =

− − =

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

− =

− + − =

− +

−

=

+

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

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

Solution to question 3

: x ax b→ +ƒ ( ) 32 =ƒ ⇒ 12 .3a b+ = …

( )4 9− =ƒ ⇒ 24 .9a b+ =− …

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

2. 2 3 4 9 6 6

a b a b a

+ =

− + =

= −

… …

⇒ 1a = − Substitute 1a = − into equation 1 we have

( )2 3 2

5

1 3

b b b

+ =

− + =

=

−

Check in equation 2 we have ( ) ( )4 5 91 45− + = + =− Hence 1a = − 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= − , ( ) 22 4m x x= + and ( ) 4n x x

= find

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

Solution to question 4

( ) 3 2l x x= − , ( ) 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= = − = + − = + =+ ++ 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= = + = − + +

= − + −+ = − + =

−

+

−

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         

= = + = = − = + −              

+ +    

+ +

+ = = =

d. ( ) ( )( ) 2 24

2 2

3 5

4 4

lmn − + 

−

 = −=−

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Question 5

Given the functions : 7 3x x→ −ƒ and 3 5: 4

xg x −→ find the inverse

functions ( )-1 xƒ and ( )1g x− . Click here to read the solution to this question Click here to return to the index

Solution to question 5

: 7 3x x→ −ƒ Let 7 3

3 7 3

7

y x y x y x

= −

+ =

+ =

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

x x= +ƒ

3 5:

4 xg x −→

Let 3 5 4

4 3 5 4 5 3 4 5

3

xy

y x y x y x

−

=

= −

+ =

+ =

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

g x x− = +

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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

"""# .

Click here to read the solution to this question Click here to return to the index

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 a−= − =

"""# """# #

c. CA CO OA c a a c= + −= − + =

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

d. BO BC CO a c= + −= −

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

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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 """"#

Click here to read the solution to this question Click here to return to the index

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= + −= − + =

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

b. 3 3RQ OP p= =

"""# """ ##

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

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

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= + = + = + − + −= − = """"# """"# """# """# """# # # # # # # ##

Click here to read the question again Click here to return to the index

M

p #

r #

R Q

PO

Question 8

Given the following vectors 2 3

a − 

=    

# and

4 5

b  =   − 

# . Find the following giving

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

## b. a b−

## c. 3 2a b−

## d. b

# e. 3 2a b−

##

Click here to read the solution to this question Click here to return to the index

Solution to question 8

2 3

a − 

=    

# and

4 5

b  =   − 

# .

a. 24

5 2 2 3

a b −   

+ = + =    −

    −    

##

b. 64

5 8 2 3

a b −   

− = − =    −  

−

 

   

##

c. 2 4 6 8

3 2 3 2 3 5 9 10

14 19

a b − −       

− = − = − =        − −      

−     

##

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