Mark scheme Pure Mathematics Year 2 (A Level) Unit Test 9: Numerical Methods
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
1a x4 + 2 M1 1.1b 5th
4 2
x2 =
Rearranges x - 8x + 2 =0 to find 8 Understand the
concept of roots
A1 1.1b of equations.
x4 + 2 1 1
x= a b
States 8 and therefore 8 and 4 or states
1 4 1
x= x +
8 4
(2)
1b Attempts to use iterative procedure to find subsequent values. M1 1.1b 6th
Solve equations
Correctly finds: A1 1.1b
approximately
x1 =0.9396 using the method
of iteration.
x2 =0.5894
x3 =0.5149
x4 =0.5087
(2)
1c Demonstrates an understanding that the two values of f(x) to be M1* 2.2a 5th
calculated are for x = –2.7815 and x = –2.7825.
Use a change of
sign to locate
Finds f (- 2.7815) =- 0.0367... and f (- 2.7825) =(+)0.00485... M1 1.1b
roots.
Change of sign and continuous function in the interval A1 2.4
éë- 2.7825, - 2.7815ùû
root
(3)
(7 marks)
Notes
1b
Award M1 if finds at least one correct answer.
1c
Any two numbers that produce a change of sign, where one is greater than –2.782 and one is less than –2.782,
and both numbers round to –2.782 to 3 decimal places, are acceptable. Minimum required is that answer states
there is a sign change in the interval and that this implies a root in the given interval.
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Mark scheme Pure Mathematics Year 2 (A Level) Unit Test 9: Numerical Methods
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
2a æxö
3
æx ö
3 M1 1.1b 5th
1 1
3sin ç ÷ - x - 1 =0 3sin ç ÷ = x +1
è 6 ø 10 è6 ø 10 Understand the
Deduces from that concept of roots
of equations.
æx ö
3
æ1 1 ö M1 1.1b
ç ÷ =arcsin ç + x ÷
States è 6 ø è3 30 ø
Multiplies by 63 and then takes the cube root: A1 1.1b
1 1
x 6 3 arcsin x
3 30
(3)
2b Attempts to use iterative procedure to find subsequent values. M1 1.1b 6th
Solve equations
Correctly finds: A1 1.1b
approximately
x1 =4.716 using the method
of iteration.
x2 =4.802
x3 =4.812
x4 =4.814
(2)
(5 marks)
Notes
2b
Award M1 if finds at least one correct answer.
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Mark scheme Pure Mathematics Year 2 (A Level) Unit Test 9: Numerical Methods
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
3a Finds f (1.9) =- 0.2188... and f (2.0) =(+)0.1606... M1 1.1b 5th
Use a change of
Change of sign and continuous function in the interval A1 2.4 sign to locate
éë1.9, 2.0 ùû roots.
root
(2)
3b Makes an attempt to differentiate f(x) M1 2.2a 6th
Solve equations
2 A1 1.1b
Correctly finds f ¢(x) =- 9sin x cos x + sin x approximately
using the Newton-
Raphson method.
Finds f (1.95) =- 0.0348... and f ¢(1.95) =3.8040... M1 1.1b
M1 1.1b
Attempts to find x1
f (x0 ) - 0.0348...
x1 =x0 - Þ x1 =1.95 -
f ¢(x0 ) 3.8040...
A1 1.1b
Finds x1 =1.959
(5)
(7 marks)
Notes
3a
Minimum required is that answer states there is a sign change in the interval and that this implies a root in the
given interval.
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Mark scheme Pure Mathematics Year 2 (A Level) Unit Test 9: Numerical Methods
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
4a Attempts to sketch both M1 3.1a 5th
2 Understand the
y= x
x - 1 and y =e concept of roots
of equations.
2 A1 2.4
y= x
States that x - 1 meets y =e in just one place, therefore
2
=e x
x- 1 has just one root g(x) =0 has just one root
(2)
4b Makes an attempt to rearrange the equation. For example, M1 1.1b 5th
2 Understand the
- e x =0 Þ xe x - e x =2
x- 1 concept of roots
of equations.
-x A1 1.1b
Shows logical progression to state x =2e +1
2 + ex
x=
For example, e x is seen.
(2)
4c Attempts to use iterative procedure to find subsequent values. M1 1.1b 6th
Solve equations
Correctly finds: A1 1.1b
approximately
x1 1.4463 using the method
x2 1.4709 of iteration.
x3 1.4594
x4 1.4647
(2)
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Mark scheme Pure Mathematics Year 2 (A Level) Unit Test 9: Numerical Methods
4d 2 A1 2.2a 6th
g¢(x) =- - ex
Correctly finds (x - 1)2 Solve equations
approximately
using the
Finds g(1.5) =- 0.4816... and g¢(1.5) =- 12.4816... M1 1.1b
Newton–Raphson
method.
g ( x0 ) 0.4816... M1 1.1b
x1 x0 x1 1.5
Attempts to find x1 : g( x0 ) 12.4816...
A1 1.1b
Finds x1 =1.461
(4)
(10 marks)
Notes
4a
2
g(x) = - ex
Uses their graphing calculator to sketch x- 1 (M1)
States that as g(x) only intersects the x-axis in one place, there is only one solution. (A1)
4c
Award M1 if finds at least one correct answer.
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Mark scheme Pure Mathematics Year 2 (A Level) Unit Test 9: Numerical Methods
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
5a Finds h(19.3) =(+)0.974... and h(19.4) =- 0.393... M1 3.1a 7th
Use numerical
Change of sign and continuous function in the interval A1 2.4 methods to solve
éë19.3, 19.4 ùû problems in
root context.
(2)
5b Makes an attempt to differentiate h(t) M1 2.2a 7th
Use numerical
40 æt ö 1 A1 1.1b
h¢(t) = + 8cos ç ÷- t methods to solve
Correctly finds t +1 è5 ø 2 problems in
context.
Finds h(19.35) =0.2903... and h¢(19.35) =- 13.6792... M1 1.1b
x1 M1 1.1b
Attempts to find
h( x0 ) 0.2903...
x1 x0 x1 19.35
h ( x0 ) 13.6792...
A1 1.1b
Finds x1 =19.371
(5)
5c Demonstrates an understanding that x = 19.3705 and M1 2.2a 7th
x = 19.3715 are the two values to be calculated.
Use numerical
methods to solve
Finds h(19.3705) =(+)0.0100... and h(19.3715) =- 0.00366... M1 1.1b
problems in
context.
Change of sign and continuous function in the interval A1 2.4
é
ë19.3705,19.3715 ù
û root
(3)
(10 marks)
Notes
5a
Minimum required is that answer states there is a sign change in the interval and that this implies a root in the
given interval.
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Mark scheme Pure Mathematics Year 2 (A Level) Unit Test 9: Numerical Methods
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
descriptor
6a States that the local maximum occurs when p¢(t) =0 B1 3.1a 7th
Use numerical
Makes an attempt to differentiate p(t) M1 2.2a methods to solve
problems in
1 1 t 3 1 A1 1.1b context.
p(t ) sin t 2
Correctly finds 10(t 1) 2 2 20
Finds p¢(8.5) =0.000353... and p¢(8.6) =- 0.00777... M1 1.1b
A1 2.4
Change of sign and continuous function in the interval
8.5, 8.6
Therefore the gradient goes from positive to negative and so the
function has reached a maximum.
(5)
6b States that the local minimum occurs when p¢(t) =0 B1 3.1a 7th
Use numerical
Makes an attempt to differentiate p¢(t) M1 2.2a methods to solve
problems in
1 1 t 3 1 A1 1.1b context.
p(t ) 2
cos t 2
Correctly finds 10(t 1) 4 2 40
Finds p¢(9.9) =- 0.00481... and p¢¢(9.9) =0.0818... M1 1.1b
M1 1.1b
Attempts to find t1
p¢(t0 ) - 0.0048...
t1 =t0 - Þ t1 =9.9 -
p¢¢(t0 ) 0.0818...
A1 1.1b
Finds t1 =9.959
(6)
(11 marks)
Notes
6a
Minimum required is that answer states there is a sign change in the interval and that this implies a root in the
given interval.
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