Question 3:
3 | w + 1
w = 4 x − 1  x =
4 | B1 | 3.1a
 w + 1  3  w + 1  2  w + 1 
a + b − 1 9 − b ( = 0 ) or
4 4 4
( 4 x − 1 ) 3 − 9 ( 4 x − 1 ) 2 − 9 7 ( 4 x − 1 ) + c ( = 0 ) | M1 | 3.1a
a w 3 + ( 3 a + 4 b ) w 2 + ( 3 a + 8 b − 3 0 4 ) w + ( a − 6 0 b − 3 0 4 ) = 0
or 6 4 x 3 - 1 9 2 x 2 - 3 0 4 x + 8 7 + c = 0 | M1 | 1.1b
Divides by a and equates the coefficients of w2 and w
3a+4b 3a+8b−304
=−9 =−97
a a
and solves simultaneously to find a value for a or a value for b
Note: 12a+4b=0 and 1 0 0 a + 8 b = 3 0 4
or
Divides through by ‘16’ leading to values of a and b
8 7 + c
4 x 3 - 1 2 x 2 - 1 9 x + = 0
1 9 | M1 | 3.1a
a − 6 0 b − 3 0 4 8 7 + c
c = = . . . or = 1 2 Þ c = . . .
a 1 9 | M1 | 1.1b
a = 4 b = − 1 2 c = 1 0 5 | A1 | 1.1b
(6)
(6 marks)
Notes:
w+1
B1: Selects the method of making a connection between x and w by writing w = 4 x − 1 or x =
4
w + 1
M1: Applies the process of substituting their x = into ax3+bx2 −19x−b=0 or w = 4 x − 1
4
into w3−9w2 −97w+c=0. Must be substitution of the correct variable into the opposing equation
but may be scored if the initial linear equation is incorrect (e.g. x=4w−1 into the first equation).
Note that the “ = 0 “ can be missing for this mark.
M1: Expands the brackets and collects terms in their equation (in x or w). Note that the “ = 0 “ can
be missing for this mark.
M1: A complete method for finding a value for a or b. See scheme, it involves dividing through by
an appropriate factor for their equation to balance the w3 or −19x terms, then equating other
coefficients and solving equations if necessary.
M1: A complete method for finding a value for c. They must have divided through by an appropriate
factor as per the previous M before attempting to compare the constant coefficient (and use their a
and b if appropriate).
A1: a = 4 b = −12 c =105
Alternative
b 1 9 b
At least two of        + + = − + + = − =
a a a | B1 | 3.1a
 b 
New sum =4 (++)−3=9  4 − − 3 = 9  b = − 3 a
a | M1 | 3.1a
New pair sum= 1 6 ( ) 8 ( ) 3 9 7       + + − + + + = −
 1 9   b 
 1 6 − − 8 − + 3 = − 9 7
a a | M1 | 1.1b
 1 9 
 1 6 − − 8 ( 3 ) + 3 = − 9 7  a = . . .
a | M1 | 3.1a
New product 6 4 ( ) 1 6 ( ) 4 ( ) 1 c        − + + + + + − = −
 b   1 9 
 6 4 − 1 6 − + 4 ( 3 ) − 1 = − c  c = ...
a a | M1 | 1.1b
a = 4 b = − 1 2 c = 1 0 5 | A1 | 1.1b
(6)
Alternative Notes
B1: Selects the method of giving at least two correct equations containing , a n d  
M1: Applies the process of finding the new sum to generate an equation in a and b. Must be
substituting in the correct places.
M1: Attempts the new pair sum to generate another equation connecting a and b. Must be
substituting in the correct places.
M1: Solves their equations to find a value for a or b.
M1: Uses the new product with their values to find values for a, b and c
A1: a = 4 b = −12 c =105
Question | Scheme | Marks | AOs
4(i) (a)
(b) | It is possible as the number of columns of matrix A matches the
number of rows of matrix B. | B1 | 2.4
It is not possible as matrix A and matrix B have different dimensions
o.e. different number of columns | B1 | 2.4
(2)
(ii) (a) | 5  = | B1 | 2.2a
a=1, b=2 | B1 | 2.2a
(b) |  0 5 0 
1
Inverse matrix = 2 1 2 − 1
5
− 1 − 1 1 3 | B1 ft | 3.1a
(3)
(iii) | A complete method to find the determinant of the matrix and set
equal to zero. | M1 | 1.1b
Determinant = 1 ( s i n s i n 2 c o s c o s 2 ) 1 ( 0 ) 1 ( 0 ) 0     − − + = | A1 | 1.1b
Uses compound angle formula to achieve c o s 3 0  = leading to ...  =
or
use of s i n 2 q = 2 s i n q c o s q and c o s 2 q = 1 - 2 s i n 2 q (e.g. to achieve
c o s q ( 4 s i n 2 q - 1 ) = 0 ) leading to ...  =
or
use of s i n 2 q = 2 s i n q c o s q and c o s 2 q = 2 c o s 2 q - 1 (e.g. to achieve
4cos3q- 3cosq= 0) leading to ...  = | M1 | 3.1a
5   
, ,  =
6 2 6 | A1 | 1.1b
(4)
(9 marks)
Notes:
(i)(a)
B1: Comments that the number of columns of matrix A (2) equals the number of rows of matrix B
(2) therefore it is possible. Accept other terminology that is clear in intent e.g. “length of A” and
“height of B”
(b)
B1: Comments that matrix A and matrix B have different dimensions therefore it is not possible.
(ii)(a)
B1: Deduces the correct value for =5
B1: Deduces the correct values for a and b
(b)
B1ft: Identifies and applies a correct method find the inverse matrix. May multiply from the given
equation, in which case follow through on their value of lambda. Alternatively, award for a correct
matrix found by calculator or long hand having found a and b and using these values in the matrix.
(iii)
M1: A complete method to find the determinant of the matrix and sets it equal to 0
A1: Correct equation
M1: Uses appropriate correct trig identities to solve the equation and finds a value for q
PMT
  5
A1: All three correct values = , , and no others in the range.
6 2 6
Question | Scheme | Marks | AOs