Mathematics
Higher
Block 3 Assessment, Revision B
Read carefully
1. Calculators may be used.
2. Full credit will be given only where the solution contains appropriate working.
3. Answers obtained from reading from scale drawings will not receive any credit.
Block 3 Practice Assessment 2
Questions
Assessment
Standard
1
2
RC1.3
RC1.4
3
RC1.4
Subskill details
Differentiating k sin x, k cos x
Integrating functions of the form f ( x)  ( x  q) n , n  1
Integrating functions of the form f ( x)  p cos x and
f ( x)  p sin x
4
RC1.4
5
EF1.4
6
7
EF1.4
EF1.4
8
EF1.4
9
EF1.1
10
EF1.1
11
EF1.2
12
RC1.2
Calculating definite integrals of polynomial functions with
integer limits
Determining the resultant of vector pathways in three
dimensions
Working with collinearity
Determining the coordinates of an internal division point of a
line
Evaluate a scalar product given suitable information and
determine the angle between two vectors
Simplifying a numerical expression, using the laws of
logarithms and exponents
Solving logarithmic and exponential equations, using the laws
of logarithms and exponents
Converting a cos x  b sin x to k cos( x   ) or k sin( x   ) , α
in 1st quad k  0
Solve trigonometric equations in degrees, including those
involving trigonometric formulae or identities, in a given interval
1
3
dy
Given y   cos x, find
.
4
dx
(1)
RC1.3
2
Integrate ( x  8)4 , where x  8, with respect to x .
3
3
Find   sin xdx. .
4
(2)
RC1.4
(1)
RC1.4
4
Evaluate

4
1
( x  3)4 dx.
(4)
RC1.4
5
ABCD is a tetrahedron as shown below. M is the midpoint of BC.
 3
 2
  and
 
AB   3 
AC   5 
 4
 3 
 
 
Find AM .
6
(3)
EF1.4
The points P, Q and R have coordinates  5, 2, 2 ,  7, 2, 4 , 10, t,13 respectively.
(a) Write down the components of PQ .
(b) P, Q, and R are collinear, compare PQ and QR to find the value of
(1)
t.
(3)
EF1.4
7
The points P, Q and R lie in a straight line, as shown. Q divides PR in the ratio 1:5.
Find the coordinates of Q.
R(17, -13, 28)
Q
P(5, -7, -2)
(3)
EF1.4
8
A(5, 2, 3), B(−1, 0, 2), C(-3, -3, 0) and D(3, -1, 1) are the four vertices of a
quadrilateral ABCD, where AB  CD  41 and BC  AD  17 .
9
(a)
Express BA and BC in component form.
(2)
(b)
Hence show that ABCD is not a rectangle.
(c)
Calculate the size of angle ABC.
(2)
EF1.4
(a)
Simplify log 4 30 p  log 4 5q .
(1)
EF1.1
(b)
Express log a x8  log a x5 in the form k log a x.
(2)
EF1.1
(#2.1, 1, #2.2)
10
Solve e x  8.5
11
Express 6cos x  8sin x in the form k cos  x  a   , where k > 0 and 0  a  360 .
(2)
EF1.1
(5)
EF1.2
12
Solve cos x cos 70  sin x sin 70 
1
for 0° < 𝑥 < 360°
7
(#R2.1, 3)
RC1.2
Question
Points of expected response
1
1 differentiates correctly
2
1 starts integration correctly
2 completes integration correctly
Illustrative scheme
1
1
3
sin x
4
 x  8
3
 x  8
3
 c **
3
Note: If a candidate differentiates (x + 8)4 then •1 can be awarded for (x + 8)5.

2
** To achieve block 2 question 10 4 the constant of integration must appear at least once
associated with a correct integral in either question 10 (block2), 2 (block 3) or 3 (block 3).
3
3
1 integrates correctly
1
cos x  c **
4
5
4
1 starts integration
1  x  3 
2
.. 
3 substitutes limits
3
1
1
5
5
 4  3  1  3
5
5
4
33
3
or 6 or 6.6
5
5
4 evaluate definite integral
Notes:
1
5
2 completes integration
3 and 4 are only available as a result of an attempt at integration.
Simply substituting these into the integrand gains no marks.
Candidates who differentiate the original expression cannot gain 3 and 4.
5
1 Finds BC

2 Finds a correct pathway
1
2
ie
 complete calculation for AM
3
3
Note:
Do not award •3 if the answer is given as coordinates.
 2
1
  1 
Alternatively, for • , AM   5    8 
 3  2  7 
 
 
1
1
ie AM  AC  CB or AM  AC  BC
2
2
2
 1 
 
BC   8 
 7 
 
 3
 1 
  1 
AM   3    8 
 4  2  7 
 
 
1
AM  AB  BC
2
 52 
 
AM   1 
 1 
 2
6 a)
•1 Finds components of PQ
 2
 PQ   4 
 6 
 
1
b)
 3 


 QR   t  2 
 9 


2
•2 Finds second vector
3 Compare two vectors
3 PQ 
4 Process value of p
4 -4 
2
QR
3
2
(t+2) so t  8 .
3
Note:
Candidates may be awarded all marks for comparing any two appropriate vectors with a
correct communication statement.
An answer of t = –8 without any working receives no credit.
7
1 Interprets ratio
1 eg PQ 
1
PR
6
 Starts to evaluate components
 5
 12 
1


2 eg q =  7    6 
 2  6  30 
 
 
3 Finds coordinates
3 Q(7, -8 )
2
Notes:
1
•1 is awarded for any correct interpretation of the given ratio, eg PQ  QR .
4
•2 is awarded for any correct path which makes use of the given ratio.
Candidates who make use of the section formula and correctly find Q receive full credit.
 6
 
Do not accept  0  for •3.
 2 
 
8 a)
1 Finds BA
 5   1  6 
     
1 BA   2    0    2 
 3   2   1
     
2 Finds BC
 3   1  2 
     
2 BC   3    0    3 
 0   2   2 
     
8 b)
3 Evaluates BA.BC
3
BA.BC  6  (2)  2  (3)   1  (2)  16
#2.1 Valid strategy
#2.1 Knows to use scalar product.
#2.2 Explains solution
#2.2 Since BA.BC  0 , ABCD is not a
rectangle.
•3 may be awarded for evaluating any suitable
scalar product, eg AB. AD .
For #2.2 do not accept statements such as
‘ BA.BC  2 so ABCD is not a
rectangle’.
For candidates who elect to use the converse
of Pythagoras’ theorem:
 8 
 
•3 eg AC   5  and
 3 
 
AC  (8) 2  (5) 2  (3) 2  98
#2.1
Knows to use converse of Pythagoras
– LHS and RHS independently
 41   17 
2
evaluated eg
and
#2.2
Notes:


2
 58
 98 .
Since 58  98 , ABCD is not a
rectangle by the converse of
Pythagoras.
 41   17    98  .
2
Do not award #2.1 for a statement such as
98
2
2
2
#2.2 is only available if ‘converse of Pythagoras’ appears.
8 c)
•1 Uses scalar product
•1
•2 Finds angle
BA.BC
BA BC
•2 127.3 or 2.22 radians (accept 127 or 2.2
rads)
Notes:
Candidates who use the cosine rule can be awarded full credit:
cos  ABC  
9(a)
( 17) 2 

 
2
41 
98

2
2  17  41
1 use log a x  log a y  log a
x
y
1 log 4 (30 p  5q)  log 4 6 p
q
(b)
2 log a x m  m log a x
2 8 log 𝑎 𝑥 +5 log 𝑎 𝑥 𝑜𝑟 log 𝑎 𝑥13
OR log a x m  log a x n  log a x m  n
 simplify to k log a x
3
Note:
10
3 13log a x
For 1 the final answer must be simplified to log 4 6 p .
q
 Convert from exponential to
logarithmic form
1
 Solve equation
2
 x  loge 8.5 or x  ln 8.5
1
2 x  2.14 (correct to 2 d.p.)
11
1 expand k cos  x  a  
1 k cos x cos a  k sin x sin a stated
explicitly
2 compare coefficients
2 k cos a  6 and k sin a  8 stated
explicitly
3 process k
3 k  10
4 process a
4 a  53.1
5 state final form
5 10 cos  x  53.1
Note:
k  cos x cos a  sin x sin a is an acceptable response for •1.
At •1 treat k cos x cos a  sin x sin a as bad form unless k omitted at •2.
Do not accept an unsimplified k  100 at •3.
The value of a must be consistent with •2.
10cos x cos a 10sin x sin a is acceptable for •1 as is k substituted at •2.
Any form of the wave function is permissible provided the final answer is stated in the
form kcos(x - a).
12
#2.1 Express in standard form
cos( x  a)
#2.1 cos( x  70) 
1 Solve equation for one value of
x  70
1 81.8º
2 Process second solution
2 278.2º
3 Process solutions for x
3 151.8º and 348.2º
1
7