| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Rational Function Asymptotes |
| Difficulty | Standard +0.3 This is a standard Further Maths rational function question requiring polynomial division, asymptote identification, and curve sketching. While it involves multiple parts, each step follows routine procedures: finding asymptotes by inspection/division, showing no x-intercepts via discriminant, finding stationary points via differentiation, and sketching. The techniques are well-practiced in FP1 with no novel insights required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical asymptote is \(x = -b\) | B1 | |
| \(x^2 + b = (x+b)(x-b) + b^2 + b\) or long division | M1 | By inspection or long division |
| Oblique asymptote is \(y = x - b\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(y = 0\) then \(x^2 + b = 0\) which has no real root | B1 | Must refer to \(b > 0\) OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{2x(x+b) - (x^2+b)}{(x+b)^2} = 0 \Rightarrow x^2 + 2bx - b = 0\) | M1 | Find \(\frac{dy}{dx}\) and set \(= 0\) |
| Or differentiating \(y = x - b + \frac{b^2+b}{x+b}\) and setting \(\frac{dy}{dx} = 0\) gives \(1 - \frac{b^2+b}{(x+b)^2} = 0\) | ||
| \(b^2 + b > 0\), therefore there are two stationary points on \(C\) | A1 | Use discriminant or \((x+b)^2\) to show two stationary points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph with intersection \((0,1)\) given and asymptotes drawn | B1 | Intersection \((0,1)\) given and asymptotes drawn |
| Each branch correct | B1 B1 | Each branch correct. Penalise at most one mark for poor forms at infinity |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical asymptote is $x = -b$ | B1 | |
| $x^2 + b = (x+b)(x-b) + b^2 + b$ or long division | M1 | By inspection or long division |
| Oblique asymptote is $y = x - b$ | A1 | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $y = 0$ then $x^2 + b = 0$ which has no real root | B1 | Must refer to $b > 0$ OE |
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{2x(x+b) - (x^2+b)}{(x+b)^2} = 0 \Rightarrow x^2 + 2bx - b = 0$ | M1 | Find $\frac{dy}{dx}$ and set $= 0$ |
| Or differentiating $y = x - b + \frac{b^2+b}{x+b}$ and setting $\frac{dy}{dx} = 0$ gives $1 - \frac{b^2+b}{(x+b)^2} = 0$ | | |
| $b^2 + b > 0$, therefore there are two stationary points on $C$ | A1 | Use discriminant or $(x+b)^2$ to show two stationary points |
## Question 6(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph with intersection $(0,1)$ given and asymptotes drawn | B1 | Intersection $(0,1)$ given and asymptotes drawn |
| Each branch correct | B1 B1 | Each branch correct. Penalise at most one mark for poor forms at infinity |6 The curve $C$ has equation
$$y = \frac { x ^ { 2 } + b } { x + b }$$
where $b$ is a positive constant.\\
(i) Find the equations of the asymptotes of $C$.\\
(ii) Show that $C$ does not intersect the $x$-axis.\\
(iii) Justifying your answer, find the number of stationary points on $C$.\\
(iv) Sketch C. Your sketch should indicate the coordinates of any points of intersection with the $y$-axis. You do not need to find the coordinates of any stationary points.
\hfill \mbox{\textit{CAIE FP1 2018 Q6 [9]}}