CAIE FP1 2018 November — Question 6 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeRational functions with parameters: analysis depending on parameter sign/range
DifficultyStandard +0.8 This is a Further Maths FP1 question requiring multiple techniques: polynomial division for oblique asymptote, discriminant analysis, differentiation using quotient rule, and synthesis into a sketch. While each component is standard, the combination and the need to handle the parameter 'a' correctly across all parts elevates this above typical A-level questions. The stationary points analysis requires careful algebraic manipulation and sign reasoning with the constraint a>1.
Spec1.02n Sketch curves: simple equations including polynomials1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives

The curve \(C\) has equation $$y = \frac{x^2 + ax - 1}{x + 1},$$ where \(a\) is constant and \(a > 1\).
  1. Find the equations of the asymptotes of \(C\). [3]
  2. Show that \(C\) intersects the \(x\)-axis twice. [1]
  3. Justifying your answer, find the number of stationary points on \(C\). [2]
  4. Sketch \(C\), stating the coordinates of its point of intersection with the \(y\)-axis. [3]

Question 6:

AnswerMarks Guidance
6(i)Vertical asymptote is x=−1. B1
x2 +ax−1=( x+1 )( x+a−1 ) −a
x+a−1
AnswerMarks Guidance
or x+1 x2 +ax−1M1 By inspection or long division.
Thus the oblique asymptote is y=x+a−1A1
3

AnswerMarks Guidance
6(ii)a2 +4>0 B1
1

AnswerMarks
6(iii)( )
( x+1 )( 2x+a ) − x2 +ax−1
=0⇒ x2 +2x+a+1=0
( x+1 )2
AnswerMarks
Discriminant =4−4 ( a+1 )<0M1
Therefore there are no stationary points on C.A1
2
AnswerMarks Guidance
Quuestion Answee
Question 6:
--- 6(i) ---
6(i) | Vertical asymptote is x=−1. | B1
x2 +ax−1=( x+1 )( x+a−1 ) −a
x+a−1
or x+1 x2 +ax−1 | M1 | By inspection or long division.
Thus the oblique asymptote is y=x+a−1 | A1
3
--- 6(ii) ---
6(ii) | a2 +4>0 | B1
1
--- 6(iii) ---
6(iii) | ( )
( x+1 )( 2x+a ) − x2 +ax−1
=0⇒ x2 +2x+a+1=0
( x+1 )2
Discriminant =4−4 ( a+1 )<0 | M1
Therefore there are no stationary points on C. | A1
2
Quu | estion | Answee | r | Marks | Guidance
The curve $C$ has equation
$$y = \frac{x^2 + ax - 1}{x + 1},$$
where $a$ is constant and $a > 1$.

\begin{enumerate}[label=(\roman*)]
\item Find the equations of the asymptotes of $C$. [3]

\item Show that $C$ intersects the $x$-axis twice. [1]

\item Justifying your answer, find the number of stationary points on $C$. [2]

\item Sketch $C$, stating the coordinates of its point of intersection with the $y$-axis. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q6 [9]}}