Mathematics · May / June 2023 · Paper 11 · Question 10
YMA01
As printed in the paper
The question as text
10. y l2 l1 P C O x Figure 5 Figure 5 shows a sketch of the quadratic curve C with equation y = – 1 4 (x + 2)(x – b) where b is a positive constant The line l1 also shown in Figure 5, has gradient 1 • 2 • intersects C on the negative x-axis and at the point P
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10(a)(i)[1]
1
(
)
Equation is
=
+
2
y
x
2
10(a)(ii)[3]
1
(
)(
)
1
(
)
1l intersect parabola
−
+
2
−
=
+
2
=
...
x
x
b
x
x
4
2
x
=
b
−
2
1
=
y
b
2
10(b)[2]
2l passes through (b, 0) and has gradient −2
y
=
...
(
)
−
0
= −
2
−
y
x
b
10(c)[2]
1
(
)
So equation is
y
−
"
b
"
= −
" 2"
x
−
"
b
−
2"
2
1
5
y
−
b
= −
2
x
+
2
b
−
4
y
=
−
2
x
+
b
−
4
*
2
2
10(d)[2]
5
5
y
= −
2
x
+
2
b
=
−
2
x
+
b
−
4
2
b
=
b
−
4
b
=
.
.
.
2
2
or
5
5
2
4,
,
0
0
2
4
...
y
= −
x
+
b
−
x
=
b y
=
= −
b
+
b
−
b
=
2
2
or
1
(
)(
)
(
)
( )
−
+
2
−
=
−
2
−
=
... 6
−
2
=
6
=
...
x
x
b
x
b
x
b
b
4
b = 8
Notes:
(a)(i)
1
1
1
(
)
(
)
B1: Correct equation in any form e.g.
. Do not allow e.g.
y
−
0
=
x
+
2
,
y
=
x
+
1,
y
=
x
−−
2
2
2
2
1
1
1
(
)
+ . Apply isw if necessary, so allow e.g.
if followed by
l
=
x
1
y
=
x
+
2
y
=
x
+
2
1
2
2
2
(a)(ii) Allow the marks for (a)(ii) to appear anywhere in their solution provided they are
solving their 1l with the given parabola.
Do not allow methods in (a)(ii) that assume that the y coordinate of P is the same as the y
intercept of C unless this is justified by further work.
1
(
)(
)
1
(
)
M1: Sets
−
+
2
−
=
their
+
2
and makes some attempt to solve for x (or y) however
x
x
b
x
4
2
poor – e.g. allow if there is a slip in the use of the quadratic formula. Condone copying errors
provided the intention is clear. Note that some candidates do not spot the cancelling of (x + 2) and
end up solving a quadratic equation.
A1: Correct simplified x
A1: Correct simplified y
(b) Allow the work for (b) to appear anywhere in their solution
M1: Uses a gradient of −2 and the point (b, 0) in an attempt to find the equation of
2l .
If using y = mx + c they must proceed as far as finding c in terms of b. The coordinates (b, 0) must
be correctly placed.
(
)
(
)
A1: Any correct equation. E.g.
−
0
= −
2
−
,
= −
2
−
,
= −
2
+
2
. Apply isw if
y
x
b
y
x
b
y
x
b
necessary.
(c) Allow the work for (c) to appear anywhere in their solution including in part (b) provided
they have attempted part (a)(ii) and have coordinates for P.
M1: Uses a gradient of −2 and their coordinates of P in terms of b from (a) in an attempt to find the
equation of
2l . If using y = mx + c they must proceed as far as finding c in terms of b. The
coordinates of P must be correctly placed.
A1*: Correct equation reached from fully correct working with full marks in (a)(ii).
(d)
M1: Equates intercepts in the two equations for the line
2l and solves for b. They must use the given
equation in (c) and their attempt at the equation using (b, 0) (however labelled) or uses the equation
(
)
2
from (c) with x = b when y = 0 to find b or equates the parabola to the line
y
= −
x
−
b
, solves for
x, equates to their x coordinate for P and then solves for b.
A1: b = 8
Correct answer only in (d) scores both marks.
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