Unit 1 Extended Test 2

Higher Still - 2007/ 2008
MATHEMATICS
Higher Grade Extended Unit Test - UNIT 1
Time allowed - 50 minutes
Read Carefully
1.
2.
3.
4.
Full credit will be given only where the solution contains appropriate working.
Calculators may be used.
Answers obtained by readings from scale drawings will not receive any credit.
This Unit Test contains questions graded at all levels.
© Pegasys 2007
Section A
In this section the correct answer to each question is given by one of the alternatives A, B, C or D.
Indicate the correct answer by writing A, B, C or D opposite the number of the question.
Rough working may be done on the paper provided. 2 marks will be given for each correct answer.
1.
2.
If A is the point (-5, -2) and B is the
point (-2, 4) then the gradient of AB is
A
! 72
B
1
2
3.
4.
Which graph is most likely to be that of
the function f ( x) = x 2 ( x + 3) ?
y
(-2, 4)
C
0
D
2
The derivative of
A
x
O
-3
1
is
2x 3
1
6x 2
A
5.
B
(2, 4)
y
3
2x 4
B
!
C
! 6x 2
D
!
3
2x 2
The limit of the sequence defined by the
recurrence relation U n +1 = 0 ! 25U n + 12
is
A
-16
B
9!6
C
16
D
48
3 x
O
y
C
-3
O
x
(-2, -4)
D
(-2, 4)
y
The rate of change of the function
f ( x) = 3 x 2 when x = 3 is
A
3
B
18
C
27
D
54
© Pegasys 2007
O
-3
-1
x
Section B
ALL QUESTIONS SHOULD BE ATTEMPTED
In this section credit will be given for all correct working.
6.
In the diagram A, B and C are the points (-5, 2), (-3, -2) and (4, -1) respectively.
y
A(-5, 2)
x
O
C(4, -1)
B(-3, -2)
(a)
Find the equation of the line through C parallel to the line AB.
3
(b)
Find the equation of the line perpendicular to BC which passes through the
point A.
3
Find the coordinates of T, the point of intersection of these two lines.
4
(c)
7.
Two functions are defined on suitable domains and are given as
f ( x ) = 3 ! x and g ( x ) = x ! 3 .
(a)
Find an expression, in its simplest form, for g ( f ( x ))
2
(b)
Show that g ( f ( x )) ! f ( g ( x )) = !6
2
© Pegasys 2007
8.
A recurrence relation is defined as U n +1 = a U n + b , where a and b are constants.
Given that U 2 = 14 , U 3 = 9 ! 2 and U 4 = 5 ! 36 , find the values of the
constants a and b.
3
(b)
Hence explain why this recurrence relation has a limit.
1
(c)
Establish the value of U 1 .
2
(a)
9.
Find the equation of the tangent to the curve y = x 3 ! 3 x at the point where x = 2 .
5
10.
Express the function f ( x) = 3 x 2 ! 6 x + 11 in the form p ( x ! q ) 2 + r .
3
11.
(a)
The point (125, k) lies on the graph of y = log 5 x . Find the value of k.
(b)
The diagram show part of the graph of y = a x . State the value of a.
1
y
P(2, 9)
1
o
END OF QUESTION PAPER
© Pegasys 2007
x
1
OPTIONAL MAXIMA/MINIMA QUESTION
12.
The diagram shows an open top box with a square base of x cm and height h cm.
h cm
x cm
x cm
The box has to be made from 1350 cm2 of card.
(a)
Show that, in terms of x, the height, hcm, of the box can be expressed as
1350 ! x 2
4x
(b)
(c)
Show clearly that the volume of the box, in terms of x, can be expressed as:
V ( x) = 14 x(1350 ! x 2 )
3
Hence, or otherwise, find the value of x, so that the volume is a maximum, leaving
your answer as a surd in its simplest form. Justify your answer.
5
END OF QUESTION PAPER
© Pegasys 2007
2
Higher Grade Unit Tests 2007/2008
Marking Scheme - UNIT 1
Give 1 mark for each
Illustration(s) for awarding each mark
1
D
2
B
3
C
Award 2 marks for each
correct answer
4
B
10 marks
5
A
6(a)
ans:
(b)
(c)
knows to find gradient of AB
_1
_2
finds gradient
_2
_3
substitutes values in equation
_3
y 2 ! y1
x 2 ! x1
!2!2
mAB =
= -2
!3+5
y + 1 = !2( x ! 4)
mBC =
y + 7 x = -33
ans:
mAB =
(3 marks)
_1
finds gradient of BC
_1
_2
_3
takes perpendicular gradient
substitutes values in equation
_2
_3
!1+ 2 1
=
4+3
7
mPERP = -7
y ! 2 = !7( x + 5)
_1
_2
_3
_4
evidence
x = -8
y = 23
(-8, 23)
_1
_2
g (3 ! x)
3 ! x ! 3 = !x
_1
_2
3 ! ( x ! 3) = 3 ! x ! 3 = 6 ! x
! x ! (6 ! x) = !6
ans: (-8, 23)
-x
(2 marks)
substitutes
simplifies
ans: proof
_1
_2
(4 marks)
knows to use simultaneous equations
finds value for x
finds value for y
states coordinates
ans:
_1
_2
(b)
(3 marks)
_1
_1
_2
_3
_4
7(a)
y + 2x = 7
finds expression for f ( g ( x))
simplifies to answer
© Pegasys 2007
(2 marks)
Give 1 mark for each
8(a)
ans:
_1
_2
_3
(b)
9
10
ans:
-1 < 0!8 < 1
_1
_2
U2 = 0!8 U1 -2; 14 = 0!8 U1 -2
0!8U1 = 16; U1 = 20
(5 marks)
knows to differentiate
_1
_2
_3
_4
_5
finds derivative
substitutes x = 2 in derivative
finds point on the line
substitutes in equation
_2
_3
_4
_5
dy
=
dx
3x 2 ! 3
3(2)2 – 3 = 9
y = (2)3 -3(2) = 8 – 6 = 2; (2, 2)
y ! 2 = 9( x ! 2)
_1
_2
_3
3(x2 – 2x) + 11
3[(x – 1)2 – 1] + 11
3(x – 1)2 – 3 + 11 = 3(x – 1)2 + 8
_1
k = log5125; k = 3
_1
9 = a2; a = 3
ans: 3(x – 1)2 + 8
ans:
_1
(b)
_1
_1
_1
_2
_3
11(a)
9!2 = 14a + b; 5!36 = 9!2a + b
a = 0!8
b = -2
(2 marks)
substitutes for U2
solves for U1
y = 9x - 16
_1
_2
_3
(1 mark)
states condition for limit
ans: 20
_1
_2
(3marks)
forms a system of equations
finds value for a
finds value for b
ans: -1 < 0!8 < 1
_1
(c)
a = 0!8 ; b = -2
Illustration(s) for awarding each mark
ans:
_1
(3marks)
takes common factor
completes square in bracket
simplifies
k=3
(1 mark)
substitutes and solves for k
a=3
substitutes and solves for a
(1 mark)
Total: 40 marks
© Pegasys 2007
Give 1 mark for each
Illustration(s) for awarding each mark
Optional Question 12
(a)
ans: proof
1
_
2
_
(b)
(c)
(2 marks)
_1
finds expression for S.A.
2
rearranges to answer
ans: proof
_
x2 + 4xh = 1350
1350 ! x 2
4xh = 1350 – x ; h =
4x
2
(3 marks)
& 1350 ' x 2
V = x ( x ( $$
4x
%
& 1350 ' x 2 #
!!
V = x ( $$
4
%
"
_1
finds expression for volume
_1
_2
cancels
_2
_3
completes rearranging
_3
V =
dV
=0
dx
1350 3 2
! x
4
4
x = 450
ans: 15"2
1
x 1350 ! x 2
4
(
#
!!
"
)
(5 marks)
_1
knows to differentiate and equal 0
_1
_2
differentiates
_2
_3
_4
_5
solves for x
expresses as a surd
justifies answer
_3
_4
_5
15 2
or other acceptable method
Total: 50 marks
© Pegasys 2007

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