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Surds/Indices
Mark Scheme 2
Level
Subject
Exam Board
Module
Topic
Sub Topic
Booklet
A Level
Mathematics (Pure)
AQA
Core 2
Algebra
Surds/indices
Mark Scheme 2
Time Allowed:
49 minutes
Score:
/40
Percentage:
/100
Grade Boundaries:
A*
>85%
A
777.5%
B
C
D
E
U
70%
62.5%
57.5%
45%
<45%
Page 1 of 7
Mark schemes
1
(a)
(i)
{p =} 2
Condone ‘64 = 82’
B1
(ii)
{q =} – 2
Ft on ‘ – p’ if q not correct
B1ft
(iii)
{r =} 0.5
ᶁᶭᶬᶢᶭᶬᶣ͚᷉ᴹᵶ͚ᵻ͚ᵶ0.5’
B1
3
(b)
OE
Using parts (a) & valid index law to stage
8c = 8d (PI)
M1
x – 0.5 = – 2
x = –1.5
Ft on c’s (q + r) if not correct
(Accept correct answer without working)
A1ft
2
ALT: log8x = logk, xlog8 = logk;
x = –1.5
(M1 A1)
[5]
2
(a)
(i)
x2
B1
1
Page 2 of 7
(ii)
Accept either form
B1
1
(iii)
x3
B1
1
(b)
(i)
Index raised by 1
M1
Simplification not yet required
A1
Need simplification and the + c OE
A1
3
(ii)
F(9) – F(1), where F(x) is candidate‘s
answer to (b)(i) [or clear recovery]
M1
= 52
Ft on (b)(i) answer of form kx1.5 i.e. 26k
A1ft
2
[8]
Page 3 of 7
3
(a)
Accept k = 0.5
B1
1
(b)
M1
Accept p = 1.5, q = 0.5
A1
2
(c)
dx=
(+c)
Increases a power of x by 1
M1
ft non-integer p
A1ft
ft non-integer q
A1ft
3
Page 4 of 7
(d)
=
Limits; F(2) – F(1)
M1
…=
Fractional powers to surds
m1
= pr. ans
CSO AG (be convinced)
A1
3
[9]
4
(a)
(1 + x)4 = 1 + 4x + 6x2 + 4x3 +x4
Full method
M1
A1 if four terms correct or just one slip
A2,1
3
(b)
(i)
)4 = 1 + 4
(1 +
+ 6(
)2 + 4(
)3 + (
)4
Substitute. ᴹ5 for x.
M1
=1+4
+ 6(5) + 4(5
) + (25)
Two of 3 terms shown in brackets
A1ft
= 56 + 24
AG Be convinced
A1CSO
3
Page 5 of 7
(ii)
log2 (1 +
)4 = log2 [8(7 + 3
)]
M1
= log2 8 + log2 (7 + 3
)
m1
= 3 + log2 (7 + 3
)
SC B1 Change to base 10 and verify as
per model marked script
A1CSO
3
[9]
5
(a)
......... = x5 – x–3
One power correct
M1
Accept p = 5, q = –3
A1
2
(b)
(i)
f’( x) = 5x
4
ft on pxp – 1
B1ft
+3x–4
ft on –qxq – 1 provided q < 0
B1ft
2
Page 6 of 7
(ii)
M1 Considers sign of f’( x); a statement
“f’( x) > 0 OE” with ‘f increasing’.
M1
f is increasing {function}
ᵿᵯ͚ᶬᶣᶣᶢᶱ͚ᶤ ӷx) of the form ax4 +
,
where a and b both > 0 and no incorrect
statements based on f’( x) at different values of x
A1
2
(c)
At (1, 0), f’(1) = 5 + 3 = 8
Attempts to find f’(1)
M1
grad. of normal = –
Use of m × m ͚ᵻ͚᷅ᵯ͚ᶎᶇ
m1
ft on wrong f’( x)
A1ft
3
[9]
Page 7 of 7