Download Higher Extended Unit 1 Test A

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Higher Still
Higher Mathematics
Extended Unit Tests
(more demanding tests covering all levels)
Pegasys Educational Publishing
HIGHER UNIT 1 EXTENDED TEST A
All questions should be attempted
1.
In the diagram triangle ABC is isosceles with AB  AC . A is the point (-5, 5), B is (-3, -6)
and the equation of the line BC is y  2 x . M is the midpoint of BC.
(a)
Find the equation of the altitude AM.
(b)
Hence establish the coordinates of
the point M.
C
y
(4)
A
(2)
M
(c)
Find the coordinates of the point C.
(2)
o
x
B
2.
A function is defined as:
(a)
(b)
3.
f ( x)  x 3  2 x 2  x  1 .
Find the stationary values of the function and the corresponding values of x.
Determine the nature of each stationary point.
Establish the equation of the tangent to the curve y  f (x) at the point
where x  2 .
(8)
(3)
A patient is given an initial life saving dose of 100mg of a drug. As the effect of the drug
wears off it is topped up hourly according to the recurrence relation:
U n1  aU n  b
After the first top-up there is 90mg of the drug in the patient’s body. After 2 top – ups there
is 84mg.
(a)
Use this information to calculate the values of a and b.
(b)
This treatment is deemed effective if, in the long term, the strength of the drug
in the body doesn’t fall below 46mg, and safe, if the strength is not consistently
above 80mg.
Is this treatment both effective and safe in the long term?
You must show working to justify your answer.
4.
(a)
(4)
Given that f ( x)  2( x 2  1) and g ( x)  2 x  3 find an expression for h(x) where
h( x)  f ( g ( x)) .
(b)
(3)
Find the value(s) of x so that h( x)  0 .
(2)
(2)
5.
A function is defined as f ( x)  x( x  3) 2 .
Part of the graph of y  f (x) is shown
in the diagram.
y
C(1,4)
Make a copy of the diagram.
On your diagram draw the graph of
y  f ( x)  2
o
D(3,0)
x
showing clearly the new positions of C and D.
6.
(3)
Part of the graph of y  a sin nx is shown in the diagram.
y
2
y 1
A
360o
o
x
-2
(a)
Write down the values of a and n.
(1)
The broken line has equation y  1 .
(b)
7.
Establish the coordinates of point A.
y  (2 x  x 3 ) 4 find the value of the gradient of the tangent to the
Given that
curve at the point where x = 1.
[ END OF QUESTION PAPER ]
(4)
(5)
SET A
Marking Scheme - UNIT 1
Give 1 mark for each
1.
(a)
(b)
(c)
Illustration(s) for awarding each mark
ans:
2y  x  5
●1
●2
●3
●4
knowing to use gradient of BC
knowing m1  m2 = -1
subs into ( y  b)  m( x  a)
rearranging to acceptable form
ans:
(1, 2)
●1
●2
knowing to use sim. eqs.
solving to answer
ans:
(5,10)
●1
finding x-coordinate
●1 xM =
finding y-coordinate
●
●
2
(4 marks)
●1
●2
●3
●4
mBC= 2
mAM=  12
( y  5)   12 ( x  5)
2 y  x  5 (or equivalent)
(2 marks)
●1 evidence
●2 (1,2)
(2marks)
2
yM =
3 x
2
6  y
2
 2 x  5
 2  y  10
2.
(a)
Ans:
23
 27
, x  13 , max :
●1
●2
●3
●4
 1, x  1, min
knowing to differentiate
carying out differentiation
making f ' ( x)  0
solving
●5
substituting into f (x)
●6
●7
(8 marks)
……..0
x  13 or 1
●5
( 13 ) 3  2( 13 ) 2  13  1 or
●6
●7
evaluating
setting up table of values
f ' ( x)  .......
3x 2  4 x  1 3
●1
●2
●3
●4
(1) 3  2(1) 2  1  1
23
 27
,1
1
3
f ' ( x)
+
0
1
-
0
+
shape
(b)
3.
(a)
●8
identifies correct nature of each value
ans:
●1
●2
●3
y  5x  9
knows how to find gradient
finds point on tangent
subs into ( y  b)  m( x  a) & arrange
ans:
a = 06; b = 30
●1
●2
●3
establishes values of U0,U1 and U2
knows to set up system of equations
solve to answer
●8
(3marks)
23
max at ( 13 , 27
) ; min at (1, -1)
f ' (2)  3(2) 2  4(2)  1  5  m
f (2)  (2) 3  2(2) 2  2  1  1
●1
●2
●3
( y  1)  5( x  2)
●1
●2
●3
U0 = 100, U1 = 90, U2 = 84
100a  b  90; 90a  b  84
a = 06 and b = 30
(3 marks)
Give 1 mark for each
3.(b)
ans:
safe but not effective
●1
●2
●3
●4
knows that a limit exists
finds limit
realises lower limit exists
conclusion with valid reason
Illustration(s) for awarding each mark
(4 marks)
●1
●2
●3
●4
limit exists since  1  0  6  1
b
l  1a
 75mg
lower limit = 45mg
not effective since 45<46, safe as 75<80
(student may use successive calculations - mark accordingly)
4.
ans:
●1
knows to substitute
●1
f ( g ( x))  2 2 x  3  1
rearranges
●
8( x  3x  2) or equivalent
●
(b)
5.
8( x 2  3 x  2) or equi.
(a)
2
(2 marks)
ans:
●1
●2
x = 2 or 1
makes h( x)  0
solves to answer
(2 marks)
ans:
diagram C(-1,6); D(-3,2)
(3 marks)
●1
●2
●3
reflects in y-axis
moves 2 units up
marks new positions of C and D
2
●1
●2

2

2
8( x 2  3x  2)  0
x = 2 or 1
y
●1
●2
●3
C(-1,6)
D(-3,2)
0
6.
(a)
(b)
ans:
a = 2; n = 3
●1
knows how to find a and n
ans:
(130o, 1)
●1
●2
●3
●4
knows to equate equation to 0
continues poss.soln.beyond 360o
solves for x
states coordinates of A
.
2.
ans:
(1 mark)
●1
a = 2; n = 3.
●1
●2
●3
●4
2 sin 3x  1
3x = 30o, 150o, 390o……….
x = 10o, 50o, 130o………..
A(130o, 1)
(4 marks)
Give 1 mark for each
-4
x
Illustration(s) for awarding each mark
(5 marks)
dy
…..
dx
4(2x  x 3 )......
●1
knows to differentiate
●1
●2
●3
begins to use chain rule
completes use of chain rule
●2
●3
●4
correct substitution into derivative
●4
………..  (2  3x 2 )
dy
 4(2  (1) 3 ).( 2  3(1) 2 )
dx
●5
evaluates to answer
●5
-4
Total: 43marks