CAIE Further Paper 1 2023 November — Question 7 16 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionNovember
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeMultiple transformations including squared
DifficultyChallenging +1.2 This is a comprehensive curve sketching question requiring asymptote finding, differentiation for stationary points, and transformation analysis. While it involves multiple parts and the reciprocal transformation in parts (d)-(e), each component uses standard A-level techniques (quotient rule, asymptote identification, inequality solving). The rational function is straightforward, and the reciprocal transformation is a familiar concept. More demanding than average due to length and the final inequality, but no novel insights required.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.07n Stationary points: find maxima, minima using derivatives

7 The curve \(C\) has equation \(y = f ( x )\), where \(f ( x ) = \frac { x ^ { 2 } } { x + 1 }\).
  1. Find the equations of the asymptotes of \(C\).
  2. Find the coordinates of any stationary points on \(C\).
  3. Sketch \(C\).
  4. Find the coordinates of any stationary points on the curve with equation \(\mathrm { y } = \frac { 1 } { \mathrm { f } ( \mathrm { x } ) }\).
  5. Sketch the curve with equation \(y = \frac { 1 } { f ( x ) }\) and find, in exact form, the set of values for which $$\frac { 1 } { \mathrm { f } ( x ) } > \mathrm { f } ( x ) .$$ If you use the following page to complete the answer to any question, the question number must be clearly shown.

Question 7(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(x = -1\)B1 Vertical asymptote.
\(y = \frac{(x+1)(x-1)+1}{x+1}\)M1 Oblique asymptote.
\(y = x - 1\)A1
3
Question 7(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2} = 0\)M1 Sets \(\frac{dy}{dx} = 0\).
\((0, 0),\ (-2, -4)\)A1
2
Question 7(c):
AnswerMarks Guidance
AnswerMarks Guidance
Axes and asymptotes shown correctlyB1 Axes and asymptotes.
Left branch correctB1 Left branch correct.
Right branch correctB1 Right branch correct.
3
Question 7(d):
AnswerMarks Guidance
AnswerMarks Guidance
\(\left(-2, -\frac{1}{4}\right)\)B1 B1 B1 for each correct coordinate. SC B1 for \(\left(-2, -\frac{1}{4}\right)\) and \((0,0)\).
2
Question 7(e):
AnswerMarks Guidance
AnswerMarks Guidance
Left branch correctB1 Left branch correct.
Right branch correctB1 Right branch correct.
\(\frac{x^2}{x+1} = 1\) or \(\frac{x^2}{x+1} = -1\), leading to \(x^2 - x - 1 = 0\)M2 Finds critical points, award M1 for each case.
\(x = \frac{1}{2} - \frac{1}{2}\sqrt{5}\) or \(x = \frac{1}{2} + \frac{1}{2}\sqrt{5}\)A1
\(x < -1,\ \frac{1}{2} - \frac{1}{2}\sqrt{5} < x < \frac{1}{2} + \frac{1}{2}\sqrt{5},\ x \neq 0\)B1 Condone missing \(x \neq 0\).
6
## Question 7(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = -1$ | B1 | Vertical asymptote. |
| $y = \frac{(x+1)(x-1)+1}{x+1}$ | M1 | Oblique asymptote. |
| $y = x - 1$ | A1 | |
| | **3** | |

## Question 7(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2} = 0$ | M1 | Sets $\frac{dy}{dx} = 0$. |
| $(0, 0),\ (-2, -4)$ | A1 | |
| | **2** | |

## Question 7(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Axes and asymptotes shown correctly | B1 | Axes and asymptotes. |
| Left branch correct | B1 | Left branch correct. |
| Right branch correct | B1 | Right branch correct. |
| | **3** | |

## Question 7(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(-2, -\frac{1}{4}\right)$ | B1 B1 | B1 for each correct coordinate. SC B1 for $\left(-2, -\frac{1}{4}\right)$ and $(0,0)$. |
| | **2** | |

## Question 7(e):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Left branch correct | B1 | Left branch correct. |
| Right branch correct | B1 | Right branch correct. |
| $\frac{x^2}{x+1} = 1$ or $\frac{x^2}{x+1} = -1$, leading to $x^2 - x - 1 = 0$ | M2 | Finds critical points, award M1 for each case. |
| $x = \frac{1}{2} - \frac{1}{2}\sqrt{5}$ or $x = \frac{1}{2} + \frac{1}{2}\sqrt{5}$ | A1 | |
| $x < -1,\ \frac{1}{2} - \frac{1}{2}\sqrt{5} < x < \frac{1}{2} + \frac{1}{2}\sqrt{5},\ x \neq 0$ | B1 | Condone missing $x \neq 0$. |
| | **6** | |
7 The curve $C$ has equation $y = f ( x )$, where $f ( x ) = \frac { x ^ { 2 } } { x + 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of $C$.
\item Find the coordinates of any stationary points on $C$.
\item Sketch $C$.
\item Find the coordinates of any stationary points on the curve with equation $\mathrm { y } = \frac { 1 } { \mathrm { f } ( \mathrm { x } ) }$.
\item Sketch the curve with equation $y = \frac { 1 } { f ( x ) }$ and find, in exact form, the set of values for which

$$\frac { 1 } { \mathrm { f } ( x ) } > \mathrm { f } ( x ) .$$

If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q7 [16]}}