AQA Further Paper 2 2020 June — Question 5 5 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeRational inequality algebraically
DifficultyStandard +0.3 This is a rational inequality requiring rearrangement to a single fraction, finding critical points, and sign analysis. While it involves multiple steps (combining terms, factoring, testing intervals), it's a standard Further Maths technique with no novel insight required. The 5 marks reflect the working needed, but the method is routine for Further Maths students, making it slightly easier than average.
Spec1.02g Inequalities: linear and quadratic in single variable

Solve the inequality $$\frac{2x + 3}{x - 1} \leq x + 5$$ [5 marks]

Question 5:
AnswerMarks
5Selects a correct
approach which would
lead to solving the
inequality
eg
Multiplies the inequality by
2
or
(π‘₯π‘₯βˆ’1)
Rearranges to an
inequality with 0 on LHS
or RHS
or
Replaces β€œ ” with β€œ=” and
multiplies by
AnswerMarks Guidance
≀1.1a M1
(π‘₯π‘₯βˆ’1) (2π‘₯π‘₯+3) 2
≀ (π‘₯π‘₯βˆ’1) (π‘₯π‘₯+5)
π‘₯π‘₯βˆ’1
2
(π‘₯π‘₯βˆ’1)(2π‘₯π‘₯+3) ≀ (π‘₯π‘₯βˆ’1) (π‘₯π‘₯+5)
(π‘₯π‘₯βˆ’1){(π‘₯π‘₯βˆ’1)(π‘₯π‘₯+5)βˆ’(2π‘₯π‘₯+3 )} β‰₯ 0
2
(π‘₯π‘₯βˆ’1){π‘₯π‘₯ +2π‘₯π‘₯βˆ’8} β‰₯ 0
Consideri(nπ‘₯π‘₯gβˆ’ cu1b)(icπ‘₯π‘₯ c+ur4v)e(:π‘₯π‘₯ βˆ’2) β‰₯ 0
or between -4 and 1
π‘₯π‘₯Buβ‰₯t 2
S o π‘₯π‘₯ β‰  1 or
π‘₯π‘₯ β‰₯ 2 βˆ’4 ≀ π‘₯π‘₯ < 1
Manipulates their
(π‘₯π‘₯βˆ’1)
equation/inequality to
allow the critical values to
AnswerMarks Guidance
be found1.1a M1
Obtains critical values of
AnswerMarks Guidance
-4, 1 and 21.1a M1
Gives one correct region
from
Condone
π‘₯π‘₯ β‰₯ 2,βˆ’4 ≀ π‘₯π‘₯ < 1
Must have three critical
βˆ’4 ≀ π‘₯π‘₯ ≀ 1
AnswerMarks Guidance
values.1.1b A1
Obtains correct solution1.1b A1
Total5
QMarking Instructions AO
Question 5:
5 | Selects a correct
approach which would
lead to solving the
inequality
eg
Multiplies the inequality by
2
or
(π‘₯π‘₯βˆ’1)
Rearranges to an
inequality with 0 on LHS
or RHS
or
Replaces β€œ ” with β€œ=” and
multiplies by
≀ | 1.1a | M1 | 2
(π‘₯π‘₯βˆ’1) (2π‘₯π‘₯+3) 2
≀ (π‘₯π‘₯βˆ’1) (π‘₯π‘₯+5)
π‘₯π‘₯βˆ’1
2
(π‘₯π‘₯βˆ’1)(2π‘₯π‘₯+3) ≀ (π‘₯π‘₯βˆ’1) (π‘₯π‘₯+5)
(π‘₯π‘₯βˆ’1){(π‘₯π‘₯βˆ’1)(π‘₯π‘₯+5)βˆ’(2π‘₯π‘₯+3 )} β‰₯ 0
2
(π‘₯π‘₯βˆ’1){π‘₯π‘₯ +2π‘₯π‘₯βˆ’8} β‰₯ 0
Consideri(nπ‘₯π‘₯gβˆ’ cu1b)(icπ‘₯π‘₯ c+ur4v)e(:π‘₯π‘₯ βˆ’2) β‰₯ 0
or between -4 and 1
π‘₯π‘₯Buβ‰₯t 2
S o π‘₯π‘₯ β‰  1 or
π‘₯π‘₯ β‰₯ 2 βˆ’4 ≀ π‘₯π‘₯ < 1
Manipulates their
(π‘₯π‘₯βˆ’1)
equation/inequality to
allow the critical values to
be found | 1.1a | M1
Obtains critical values of
-4, 1 and 2 | 1.1a | M1
Gives one correct region
from
Condone
π‘₯π‘₯ β‰₯ 2,βˆ’4 ≀ π‘₯π‘₯ < 1
Must have three critical
βˆ’4 ≀ π‘₯π‘₯ ≀ 1
values. | 1.1b | A1
Obtains correct solution | 1.1b | A1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
Solve the inequality
$$\frac{2x + 3}{x - 1} \leq x + 5$$
[5 marks]

\hfill \mbox{\textit{AQA Further Paper 2 2020 Q5 [5]}}