| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Range restriction with discriminant (quadratic denominator) |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths question requiring multiple advanced techniques: finding the range of a rational function via discriminant analysis to locate stationary points (non-standard approach), understanding the relationship between a function and its reciprocal, and synthesizing information across parts. The connection between the range condition and stationary points is particularly insightful and not a routine textbook exercise. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = \dfrac{5x^2 - 12x + 12}{x^2 + 4x - 4}\) | M1 | Eliminates \(y\) |
| \(k(x^2 + 4x - 4) = 5x^2 - 12x + 12\) leading to \((k-5)x^2 + 4(k+3)x - 4(k+3) = 0\) | M1 | Obtains quadratic in form \(Ax^2 + Bx + C = 0\); PI by later work |
| \(b^2 - 4ac =\) (in terms of \(k\)) | A1F | \(y = k\) intersects \(C_1\) so roots of (A) are real; FT 'their' quadratic provided first M1 awarded |
| \([4(k+3)]^2 - 4(k-5)(-4(k+3)) \geq 0\) leading to \(16(k+3)^2 + 16(k-5)(k+3) \geq 0\) | M1 | Obtains inequality including \(\geq 0\), where \(k\) is only unknown; FT 'their' discriminant provided both M1 marks awarded |
| \(16(k+3)(k+3+k-5) \geq 0 \Rightarrow (k+3)(2k-2) \geq 0 \Rightarrow (k+3)(k-1) \geq 0\) | R1 | Completes rigorous argument to show \((k+3)(k-1) \geq 0\); only available if all previous marks awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Critical values are \(-3\) and \(1\) | B1 | Obtains critical values |
| \(k \leq -3\) (or \(k \geq 1\)) satisfy inequality | R1 | Deduces \(k = -3\) for maximum |
| Sub \(k = -3\) in (A) gives \(-8x^2 = 0\) | M1 | Substitutes \(k\) into 'their' quadratic from (a)(i); FT only if first M1 awarded in (a)(i) |
| Max pt of \(C_1\) is \((0, -3)\) | A1 CAO | States coordinates of max pt; NMS 0/4; must be using (a)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((-12)^2 - 4(5)(12) < 0\) | E1 | Uses discriminant to determine solution |
| \(k \neq 0\); denominator \(5x^2 - 12x + 12\) of \(\dfrac{1}{f(x)}\) is never \(0\) so \(C_2\) has no vertical asymptotes | R1 | Deduces no vertical asymptotes with clear reasoning with reference to denominator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 1\) | B1 | Obtains \(y = 1\) |
## Question 12(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = \dfrac{5x^2 - 12x + 12}{x^2 + 4x - 4}$ | M1 | Eliminates $y$ |
| $k(x^2 + 4x - 4) = 5x^2 - 12x + 12$ leading to $(k-5)x^2 + 4(k+3)x - 4(k+3) = 0$ | M1 | Obtains quadratic in form $Ax^2 + Bx + C = 0$; PI by later work |
| $b^2 - 4ac =$ (in terms of $k$) | A1F | $y = k$ intersects $C_1$ so roots of (A) are real; FT 'their' quadratic provided first M1 awarded |
| $[4(k+3)]^2 - 4(k-5)(-4(k+3)) \geq 0$ leading to $16(k+3)^2 + 16(k-5)(k+3) \geq 0$ | M1 | Obtains inequality including $\geq 0$, where $k$ is only unknown; FT 'their' discriminant provided both M1 marks awarded |
| $16(k+3)(k+3+k-5) \geq 0 \Rightarrow (k+3)(2k-2) \geq 0 \Rightarrow (k+3)(k-1) \geq 0$ | R1 | Completes rigorous argument to show $(k+3)(k-1) \geq 0$; only available if all previous marks awarded |
---
## Question 12(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Critical values are $-3$ and $1$ | B1 | Obtains critical values |
| $k \leq -3$ (or $k \geq 1$) satisfy inequality | R1 | Deduces $k = -3$ for maximum |
| Sub $k = -3$ in (A) gives $-8x^2 = 0$ | M1 | Substitutes $k$ into 'their' quadratic from (a)(i); FT only if first M1 awarded in (a)(i) |
| Max pt of $C_1$ is $(0, -3)$ | A1 CAO | States coordinates of max pt; NMS 0/4; must be using (a)(i) |
---
## Question 12(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(-12)^2 - 4(5)(12) < 0$ | E1 | Uses discriminant to determine solution |
| $k \neq 0$; denominator $5x^2 - 12x + 12$ of $\dfrac{1}{f(x)}$ is never $0$ so $C_2$ has no vertical asymptotes | R1 | Deduces no vertical asymptotes with clear reasoning with reference to denominator |
---
## Question 12(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 1$ | B1 | Obtains $y = 1$ |12 A curve, $C _ { 1 }$ has equation $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 5 x ^ { 2 } - 12 x + 12 } { x ^ { 2 } + 4 x - 4 }$\\
The line $y = k$ intersects the curve, $C _ { 1 }$
12
\begin{enumerate}[label=(\alph*)]
\item (i) Show that $( k + 3 ) ( k - 1 ) \geq 0$\\[0pt]
[5 marks]\\
12 (a) (ii) Hence find the coordinates of the stationary point of $C _ { 1 }$ that is a maximum point.\\[0pt]
[4 marks]
12
\item Show that the curve $C _ { 2 }$ whose equation is $y = \frac { 1 } { \mathrm { f } ( x ) }$, has no vertical asymptotes.\\[0pt]
[2 marks]\\
12
\item State the equation of the line that is a tangent to both $C _ { 1 }$ and $C _ { 2 }$.\\[0pt]
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 Q12 [12]}}