Download Relationships and Calculus Assessment Standard 1

Questions Assessment
Standard
Description
1
RC1.1
Factorise cubic a≠1
2
RC1.1
Given nature use discriminant to find
coefficient
3
EF1.2
Trig identities
4
EF1.2
Application of addition and double angle
formulae
5,6
RC1.2
Solve equations in degrees or radians using trig
formulae/identities
7
A1.2
Determine and use the equation of a circle
8
A1.2
Use properties of tangency
9
RC1.4
Integrate algebraic functions expressed in
powers of x
10
A1.4
Find the area between curve and x axis
11
A1.4
Find the area between line and curve or two
curves
Relationships and Calculus Assessment Standard 1.1
1
(x – 2) is a factor of the polynomial f (x) = x3 + 2x2 – kx + 42.
(a)
(b)
2
Show that k = 29.
Fully factorise f (x).
(3)
(3)
The x-axis is a tangent to the graph of y = px2 – 2px + 3.
Determine, algebraically,
the value of p.
#2.1
(3)
Expressions and Functions Assessment Standard 1.2
3cos x  sin x  cos x  3sin x   3  5sin 2x .
3
Show that
4
The diagram shows two right-angled triangles.
#2.1
(3)
(a) Write down the values of sin x and cos y.
(1)
16
(b) Show that the exact value of cos( x  y ) is
.
65
(3)
Relationships and Calculus Assessment Standard 1.2
5
6
1
for 0  x  180 º.
2
Show that 2  cos 2 x  3sin 2 x  1  sin 2 x.
Solve the equation cos 2 x 
(a)
(b)
1
Hence solve 2  cos 2 x  3sin 2 x  , 0  x  360.
4
(2)
(2)
(5)
Applications Assessment Standard 1.2
7 (a) A circle has radius 4 units and centre (-3,-2).
Write down the equation of the circle.
2
2
(b) A circle has equation x  y  6 x  8 y  24  0 .
Write down the coordinates of its centre and the length of
its radius.
(2)
(2)
(#2.2
(c) Determine whether the circles in a) and b) intersect. Justify your
)
answer.
8
Determine whether the line y = x is a tangent to the circle
( x – 2)2 + ( y + 1)2 = 1.
Justify your answer
(3)
#2.2
Relationships and Calculus Assessment Standard 1.4
9
Find
 5
  3  x7  dx, x  0
(3)
Applications Assessment Standard 1.4
10
The curve with equation
(4)
Y = 2x3 -5x2 + x + 2 is shown in the diagram
with Q at (1,0). Calculate the shaded area
11
The line with equation y  3  x and the curve
with equation y  3  x 2 are shown in the diagram.
The line and curve meet at the points
where x  0 and x  1 .
Calculate the shaded area shown in the diagram.
(4)
1 (a)
•1 Strategy to start
process of evaluating
polynomial, eg direct
substitution or synthetic
division
•1 f (2) = … and begins substitution
•2 Evaluates polynomial
and equates to zero
•2 f (2) = 58  2k = 0
or
or
2 1 2
k
42
2
8
16  2k
1 4 k  8 2k  58
and –2k + 58 = 0
•3 Completes proof
•3 k = 29
Notes: Candidates who substitute k with 29 at •1 and then evaluate to
obtain zero can be awarded full credit — for showing that the converse is
true — provided a statement such as ‘so (x  2) is a factor when k  29 ’
appears at the end.
(b)
2
•1 First linear factor
•1 (x – 2)(……….)
or
(x – 3)(……….)
or
(x + 7)(……….)
•2 Extracts quadratic
factor
•2 (x2 + 4x – 21) or other consistent
with •1
•3 Completes
factorisation
•3 (x – 2)(x – 3)(x + 7)
#2.1 Correct strategy
#2.1 Evidence of using b2 – 4ac = 0
•1 Sets up correct
equation
•1 4p2 – 12p = 0
•2 Factorises
•2 4p(p – 3) = 0
•3 Solves equation
•3 p not equal to zero so p = 3
Notes: An answer of p = 3 with no working receives no credit.
… = 0 must appear somewhere before •3 otherwise #2.1 and •1 are
unavailable.
•2 is only available for attempting to solve the equation by factorising or by
using the quadratic formula.
For •2 accept an ‘incomplete’ factorisation, eg 2p(2p – 6) or p(4p – 12) but
not 4(p2 – 3p).
Candidates must clearly reject the solution p = 0 in order to receive •3.
Also, for equation 2 p 2  12 p  0 lose •1
•2 available for 2 p( p  6)  0 or p(2 p  12)  0 but not 2( p 2  6 p)  0
•3 available for p  0 so p  6
3
#2.1 Correct strategy
#2.1 Uses a trigonometric identity
•1 Expands LHS
•1 3cos 2 x  10sin x cos x  3sin 2 x
•2 Starts process
•2 3  cos 2 x  sin 2 x   ... leading to 3  ...
•3 Completes process
•3 10sin x cos x  5  2sin x cos x  5sin 2x
Notes: •1 may be awarded for 3cos 2 x  9sin x cos x  sin x cos x  3sin 2 x .
•2 can only be awarded where there is clear evidence of using
cos 2 x  sin 2 x  1
•3 can only be awarded for clearly demonstrating the equivalence of
10sin x cos x and 5sin 2x .
Candidates who use ‘=’ instead of ‘  ’ should not be penalised.
4 (a)
4
12
sin x  and cos y 
5
13
•1 Interpret diagram for
sin x
•1
•2 Interpret diagram for
•2
cos y
4(b)
sin x 
4
5
cos y 
12
13
16
65
•3 Use compound angle
formula
•3
cos(x  y) = cos x cos y  sin x sin y
•4 Substitute and complete
•4
3 12 4 5 16
   
5 13 5 13 65
Notes:
In (a) The evidence for •1 and •2 may not be evident until (b).
In (b) •3 may be stated or implied by •4.
Simply stating the formula for cos(A + B) with no further working gains no marks.
Calculating approximate angles using arcsin(sin 1 ) and arccos(cos1 ) gains no credit.
5
x  30, 150
•1 Solve equation for 2x
1
2

2 x  60 and 300
1
•2 Process solutions for x
2
x  30
and 150
Notes: Answers shown in the marking scheme (•1 and •2) can be marked
horizontally or vertically.
π 5π
Accept ,
or decimal equivalent 0·523598…and 2·617993…
6 6
6(a) •1 Substitutes for
cos2x
•2 Completes process
(b) •3 Rearranges equation
•4 Takes square root of
both sides
•5 Solves equation
•6 Knows to consider
the negative square
root
•7 Completes solution
•1 2  1  2 sin 2 x   3sin 2 x
•2 2  1  2sin 2 x  3sin 2 x
•3 sin 2 x  3
4
3
•4 sin x  
2
5
• x  60, 120
•6 sin x  
3
stated explicitly
2
•7 x  240, 300
Notes: Do not award •1 if brackets are omitted, ie 2  1  2 sin 2 x...
If candidates obtain all correct answers to (b) purely from sin x 
3
,
2
only 3 points of process may be awarded.
Do not penalise solutions outwith 0  x  360 .
Do not accept, eg 180  60 for 240 .
7(a)
( x  3)2  ( y  2) 2  16
7(b)
7(c)
•1 Interpret centre
•1
( x  (3))2  ( y  (2))2 ....
•2 Interpret radius and
complete equation
•2
...  16
Centre  3, 4  radius 1 unit
•3 State centre of circle
•3  3, 4 
4
• Know how to and find
radius of circle
•4 1
The circles do not intersect, with valid explanation.
#2.2 Finds distance
#2.2 Distance between centres is 6, sum
between centres, finds sum
of radii is 5, 6 > 5 so the circles do
of radii and communicates
not intersect
Note: In (a) •2 is not awarded for 42, this must be simplified to 16.
8
•1 set up imtersection
•1 (x-2)2 + (x +1)2 = 1
equation
9
•2 simplifies
•2
•3 calculates the
discriminant
#2.2 interprets solution
5
5
 3  x7 dx  3x  6 x6  c
•3 b2 – 4ac = 12 -4x1x2 = -7
#2.2 -7<0 so no points of contact so line
is not a tangent to the circle
•1 Express in integrable form
•1
5x 7
•2 Integrate the constant
•2
3x
•3 Integrate term with a
negative power
•3
5 6
x
6
2x2- 2x + 4 = 0
Note: •1 is the only mark available to candidates who differentiate.
10
•1 know to integrate with
correct limits
•1
•2 integrates
•2 ½ x4 - 5/3 x3 + ½x2 +2x
•3 substitutes in the
limits
•3 ½x14 - 5/3 x13 + ½x12 +2x1-(0)
•4
•4 4/3
1
 (...)dx
0
Note:
•3 and 4 are only available to those who integrate
11
1
or 0.25 square units
4
•1 Know to integrate and state
limits
•2 Use ‘upper – lower’
•1
1
 ...dx
0
•2
 'upper  lower' stated or
implied by •3
•3 Interpret upper – lower
•3
 x3  x
•4Integrate
•4
•5 Substitute limits
•5
 14 12    0 4  0 2 





 4
  4
2
2 

 
•6 Process limits
•6
1
or equivalent
4

x4 x2

4 2
Notes:
Do not penalise candidates who work with 3  x3  3  x throughout in an
unsimplified form.
Candidates who use 3  x3  3  x leading to  x 3  x lose •3 but all other marks
are still available.
Differentiation loses •4, •5 and •6.
•2 is lost for subtracting the wrong way round and subsequently •6 may be lost for
statements such as:
1
1
1 1
1
or  sq units ’ or ‘  =
’ or ‘  so ignore negative’
4
4
4 4
4
9
9
•6 may be gained for statements such as ‘  so area is ’
2
2
‘
0
 'lower  upper ' is correct and so all six marks are still available
1