Mathematics · Oct / Nov 2023 · Paper 11 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. The equation 12 15 p x ≠ –p p x 4 ( 2 ) x p where p is a constant, has two distinct real roots.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[3]
12
+
15
p
4(
−
2 )
=
p
x
+
x
p
2
2
(
)
8
x
+
4
px
+
12 15
+
p
−
4
p
=
0
=
8,
=
4 ,
=
12 15
+
−
4
2
e.g.
a
b
p c
p
p
(
4
)
2
4 8 (12 15
4
2
)
(
0
)
p
−××
+
p
−
p
>
3
2
−
10
−>
8
0
p
p
*
6(b)[3]
2
(
)
e.g.
(
−
4) 3
+
2 ( 0)
=
4,
−
p
p
3
2
p <−
or
p >
4
3
Allow part (b) to follow on from (a) without any labelling of specific parts
(a)
M1:
Multiplies by (
+
)
, multiplies out the brackets and collects terms on one side.
x
p
Condone slips and the omission of = 0, but there must be an
...x term.
2
May be implied by further work e.g. their attempt at the discriminant.
dM1: Attempts
2
−
4
for their quadratic in x with their b and their c both being in terms of p.
b
ac
The values/expressions for a, b and c for their quadratic in x must be embedded in the
correct positions in the expression but condone invisible brackets and sign slips miscopying.
It is dependent on the previous method mark.
Condone for this mark if they work in terms of x instead of p provided the coefficients were
actually those in terms of p from their original quadratic. i.e.
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