| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Rational functions with parameters: analysis depending on parameter sign/range |
| Difficulty | Standard +0.8 This is a comprehensive Further Maths curve sketching question requiring multiple techniques: finding asymptotes (vertical and oblique via polynomial division), analyzing intersections with axes, finding and classifying stationary points using the quotient rule and discriminant analysis, and producing an accurate sketch. While each individual step is standard FP1 material, the multi-part nature, the need to handle a parameter (a > 1), and the requirement to justify the number of stationary points (likely involving sign analysis of the derivative) elevate this above average difficulty. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical asymptote is \(x = -1\) | B1 | |
| \(x^2 + ax - 1 = (x+1)(x+a-1) - a\) | M1 | By inspection or long division |
| Thus the oblique asymptote is \(y = x + a - 1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a^2 + 4 > 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(x+1)(2x+a) - (x^2+ax-1)}{(x+1)^2} = 0 \Rightarrow x^2 + 2x + a + 1 = 0\); Discriminant \(= 4 - 4(a+1) < 0\) | M1 | |
| Therefore there are no stationary points on \(C\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph with correct \(y\)-intercept and asymptotes drawn | B1 | Correct \(y\)-intercept and asymptotes drawn |
| Each branch correct | B1 B1 | Each branch correct |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical asymptote is $x = -1$ | B1 | |
| $x^2 + ax - 1 = (x+1)(x+a-1) - a$ | M1 | By inspection or long division |
| Thus the oblique asymptote is $y = x + a - 1$ | A1 | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a^2 + 4 > 0$ | B1 | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(x+1)(2x+a) - (x^2+ax-1)}{(x+1)^2} = 0 \Rightarrow x^2 + 2x + a + 1 = 0$; Discriminant $= 4 - 4(a+1) < 0$ | M1 | |
| Therefore there are no stationary points on $C$ | A1 | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph with correct $y$-intercept and asymptotes drawn | B1 | Correct $y$-intercept and asymptotes drawn |
| Each branch correct | B1 B1 | Each branch correct |6 The curve $C$ has equation
$$y = \frac { x ^ { 2 } + a x - 1 } { x + 1 }$$
where $a$ is constant and $a > 1$.\\
(i) Find the equations of the asymptotes of $C$.\\
(ii) Show that $C$ intersects the $x$-axis twice.\\
(iii) Justifying your answer, find the number of stationary points on $C$.\\
(iv) Sketch $C$, stating the coordinates of its point of intersection with the $y$-axis.
\hfill \mbox{\textit{CAIE FP1 2018 Q6}}