Mathematics · May / June 2021 · Paper 21 · Question 7
YMA01
As printed in the paper
Show the answer and mark scheme
7(a)[4]
2
=
log 3
2
3
3log (2
x
1)
log (2
x
1)
3
or
−
=
−
3
3
(
)
log "3 " log (14
2
+
x
−
25)
=
l
og
"3
2
"(1
4
x
−
25
)
3
3
3
(2
x
1)
3
9(14
x
25)
⇒
−
=
−
3
2
⇒
8
x
−
12
x
+
6
x
−=
1
126
x
−
225
⇒
8
x
3
−
12
x
2
−
120
x
+
224
=
0
⇒
2
x
3
−
3
x
2
−
30
x
+
56
=
0*
2( 4)
3
3( 4)
2
30( 4)
56
...
7(b)[2]
±
−
±
−
±
+
=
2( 4)
3
3( 4)
2
30( 4)
56
128
48 120
56
0
−
−
−
−
−
+
= −
−
+
+
=
Hence −4 is a root of the equation.
(
)
Alt (b)
2
x
3
−
3
x
2
−
30
x
+
56
=
(
x
±
4) 2
x
2
+
..
.
±
1
4
(
)
3
2
2
2
x
−
3
x
−
30
x
+
56
=
(
x
+
4) 2
x
−
11
x
+
1
4
so −4 is a root of the equation.
(
)
7(c)[4]
2
x
3
−
3
x
2
−
30
x
+
56
=
0
⇒
(
x
+
4) 2
x
2
+
... 14
±
=
0
(
)
(
x
+
4) 2
x
2
−
11
x
+
14
=
0
(
x
+
4)(2
x
−
7)(
x
−
2
)
=
0
⇒
x
=
..
.
(Equation not defined for x = −4 so) solutions are 2 and 7
2
Notes:
(a)
B1: States or implies
or equivalent. May be gained via
l
og
f x
( )
2
f x
(
)
2
=
log 3
2
=
⇒
=
3
3
M1: Uses a correct log law, either power law on
3log (2
1)
log (2
1)
3
or sum law on
x
−
=
x
−
3
3
(
)
log "3 " log (14
2
+
x
−
25)
=
l
og
"3
2
"(1
4
x
−
25
)
or equivalent work (e.g. difference law if
3
3
3
combining the two log terms on one side first).
dM1: Uses correct log work to remove the logs. Allow slips, but all log work must have been
correct.
A1*: Expands and simplifies to the given cubic. Must see an intermediate step between the
factorised sides and final answer.
(b)
M1: Substitutes ±4 into the cubic and attempts to evaluate it.
A1: Evaluates to 0 with intermediate working shown, and gives conclusion that −4 is a root.
Alt:
M1: Attempts to take a factor of (x ± 4) out of the equation or uses long division. Look for first term
2x and last term 14
2
in the quadratic factor.
±
A1: Correct factorisation with conclusion. If via long division, all work must be correct with zero
remainder found.
(c)
M1: Attempts to take a factor of (
x +
4)
out of the cubic, or attempts long division by x + 4
A1: Correct quadratic factor (either by factorisation or by long division)
Note: the first two marks may be awarded for work seen in part (b).
dM1: Depends on first M. Solves the resulting quadratic (any means).
A1: Correct two solutions only identified. Withhold if −4 is listed as a solution.
Since the question says “hence, using algebra and showing each step of your working”, solutions
derived from a calculator with no working shown score no marks. The intermediate quadratic must
be attempted to gain marks in part (c)
7
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