| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.8 This is a Further Maths question requiring understanding of rational function asymptotes (finding a and b from asymptote conditions), then solving a rational inequality which requires careful sign analysis and consideration of the discontinuity. The multi-step nature, algebraic manipulation, and rigorous justification of inequality solutions place it moderately above average difficulty. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| \(2 \times \frac{9}{2} + b = 0 \Rightarrow b = -9\) | B1 | Deduces correct value of \(b\) |
| \(\frac{a}{2} = 3 \Rightarrow a = 6\) | B1 | Deduces correct value of \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((x+2)(2x-9)^2 \geq (6x-5)(2x-9)\), multiply by square of denominator | M1 | Selects suitable method; multiplies by square of denominator OR sketches graphs of \(y=f(x)\) and \(y=x+2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((2x-9)(x+1)(2x-13) \geq 0\) | M1 | Simplifies inequality/equation OR indicates points of intersection of two graphs |
| Critical values: \(x=-1,\ x=\frac{9}{2},\ x=\frac{13}{2}\) | A1F | Obtains at least two critical values |
| Exclude \(x=\frac{9}{2}\) (not in domain) | A1 | PI by final answer |
| \[-1 \leq x < \frac{9}{2} \quad \text{or} \quad x \geq \frac{13}{2}\] | A1F | Deduces correct solution set; condone inclusion of \(x=\frac{9}{2}\) |
| Completely correct solution with each step clearly shown | R1 | OE |
## Question 7(a):
$2 \times \frac{9}{2} + b = 0 \Rightarrow b = -9$ | B1 | Deduces correct value of $b$
$\frac{a}{2} = 3 \Rightarrow a = 6$ | B1 | Deduces correct value of $a$
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## Question 7(b):
$(x+2)(2x-9)^2 \geq (6x-5)(2x-9)$, multiply by square of denominator | M1 | Selects suitable method; multiplies by square of denominator OR sketches graphs of $y=f(x)$ and $y=x+2$
$(2x-9)\{(x+2)(2x-9)-(6x-5)\} \geq 0$
$(2x-9)\{2x^2-5x-18-6x+5\} \geq 0$
$(2x-9)\{2x^2-11x-13\} \geq 0$
$(2x-9)(x+1)(2x-13) \geq 0$ | M1 | Simplifies inequality/equation OR indicates points of intersection of two graphs
Critical values: $x=-1,\ x=\frac{9}{2},\ x=\frac{13}{2}$ | A1F | Obtains at least two critical values
Exclude $x=\frac{9}{2}$ (not in domain) | A1 | PI by final answer
$$-1 \leq x < \frac{9}{2} \quad \text{or} \quad x \geq \frac{13}{2}$$ | A1F | Deduces correct solution set; condone inclusion of $x=\frac{9}{2}$
Completely correct solution with each step clearly shown | R1 | OE
---7 The function f is defined by
$$\mathrm { f } ( x ) = \frac { a x - 5 } { 2 x + b } \quad x \in \mathbb { R } , x \neq \frac { 9 } { 2 }$$
where $a$ and $b$ are integers.\\
The graph of $y = \mathrm { f } ( x )$ has asymptotes $x = \frac { 9 } { 2 }$ and $y = 3$\\
7
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$\\
7
\item Solve the inequality
$$\mathrm { f } ( x ) \leq x + 2$$
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q7 [8]}}