Mathematics · Oct / Nov 2025 · Paper 11 · Question 6

Shape in quadrant 1

Fully correct shape and

position

x = 2k

−2

Cuts the yaxis at −2

Vertical asymptote at x = 2k

(a) Notes

Marks cannot be scored without a sketch.

M1: Correct shape in quadrant one for the right hand branch. It must not cross either axis but be

tolerant of functions that don't go down as far as the xaxis and do extend as far left as the y

axis.

Condone curves that touch the x or yaxis as long as they do not cross the axis.

Ignore any asymptotes, correct or incorrect, for this mark.

A1: Correct shape and position (for whole curve). Be tolerant of slips of the pen but the intention

should be clear. Look for the curve in quadrants 1, 3 and 4 with a vertical asymptote in

quadrant 1 and 4. The left hand branch must extend into quadrant 3 and should not

intentionally dip away from the xaxis to e.g. create an intentional maximum.

There should be no horizontal asymptotes other than the xaxis.

Do not allow this mark if the branches clearly overlap vertically. E.g.

B1: Cuts or touches the negative yaxis once at −2

y = − and condone (

)

Must be in the correct position but allow as −2 or

−

2, 0

as long as it is

2

−

in the correct place. Condone missing brackets e.g. 0,

2

Allow to appear away from the sketch but it must be fully correct e.g. (

)

−

0,

2

If there is any ambiguity, the sketch takes precedence.

=

B1: For one vertical asymptote to the right of the yaxis marked

x

2

k

It must be labelled as an equation and not just a value on the yaxis and it must correspond with

the sketch.

The asymptote does not need to be drawn as a dotted line but the curve must be asymptotic to

the correct line for either branch.

=

Allow

x

2

k

to appear away from the sketch but it must be correct and correspond with the

sketch. If there is any ambiguity, the sketch takes precedence.

4

k

2

k =

−

=

−

−

+

6

2

x

4

k

6

x

12

k

2

x

4

kx

o.e.

x

−

2

(

)

2

2

(

)

2

−

+

+

=

b

−

4

ac

=

4

k

+

6

−××

4 2 16

k

2

x

4

k

6

x

16

k

0

1 9

2

4

k

−

20

k

+

9

=

0

(2

k

−

1)(2

k

−

9)

=

0

k

=

,

2 2

1

9

so

0

<

k

≤

,

k

≥

2

2

(b) Notes

M1: Equates curve with given line and attempts to form a quadratic equation.

The terms do not need to be collected.

4

k

Look for

k =

6

−

2

x

cross multiplies (condoning slips) and proceeds to a quadratic

−

x

2

equation.

A1:

Correct quadratic equation with brackets expanded which may be unsimplified e.g.

4

k

2

k =

−

=

−

−

+

6

2

x

4

k

6

x

12

k

2

x

4

kx

x

−

2

2

2

dM1: Correct attempt to find the discriminant

b

−

4

ac

for their

ax

+

bx

+

c

=

0

with both b and c

expressions in terms of k. May be seen embedded in an attempt at the quadratic formula.

It is dependent upon the previous Method mark.

2

−

ddM1: Solves their 3 term quadratic equation/inequality in k resulting from

b

4

ac

...0

the outside region for their critical values.

Allow calculator solutions for solving their 3TQ and condone factors that are inconsistent with

their equation so just look for the correct or correct ft roots with an attempt at the outside

1

9

1

9

2

regions e.g. condone

4

k

−

20

k

+

9

=

0

k

−

k

−

=

0

k

=

,

k

<

2

2

2

2

1

9

2

−

+

=

≤

≥

The roots may be embedded so allow e.g.

4

k

20

k

9

0

k

,

k

to imply this mark.

2

2

Condone the boundaries not being included in the inequality and condone the omission of k > 0

Condone the use of x rather than k.

It is dependent upon both previous Method marks.

1

9

A1:

CSO

0

<

k

≤

,

k

≥

o.e. Must be in terms of k.

2

2

The following are acceptable:

1

9

1

9

1

9

1

0

<

k

≤

,

k

≥

0

<

k

≤

and

k ≥

0

<

k

≤

or

k ≥

0

<

k

≤

∪≥

2

2

2

2

2

2

2

✓

1

9

1

9

0,

,

,

∞

k

:0

<

k

≤

,

k

≥

2

2

2

2

1

9

<

≤

∩≥

But do not allow

0

k

k

.

2

2

Do not isw and mark their final answer.