| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Topic | Curve Sketching |
| Type | Sketch rational with reciprocal terms |
| Difficulty | Standard +0.3 This is a straightforward rational function sketching question requiring standard techniques: finding the vertical asymptote at x=2, horizontal asymptote as x→±∞, stationary points via quotient rule differentiation, and y-intercept. While it involves multiple steps, each is routine A-level calculus with no novel problem-solving required. The reciprocal squared denominator makes it slightly more involved than the simplest rational functions, placing it slightly above average difficulty. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
(i) $y = \frac{12(x+1)}{(x-2)^2} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = 12\left\{\frac{(x-2)^2 - (x+1)\cdot 2(x-2)}{(x-2)^4}\right\}$ **M1 A1**
$= 0$ when $x - 2 = 2x + 2$ i.e. $x = -4$, $y = -1$ **A1 A1** [4]
(ii) VA $x = 2$, HA $y = 0$ Stated or clearly shown on diagram **B1 B1**
Intercepts $(0, 3)$ and $(-1, 0)$ Stated or clearly shown on diagram **B1 B1**
Basic shape correct **B1**
All details correct **B1** [6]
(Allow **ft** on min. point provided it doesn't ruin overall shape or position)5 The curve $C$ has equation $y = \frac { 12 ( x + 1 ) } { ( x - 2 ) ^ { 2 } }$.\\
(i) Determine the coordinates of any stationary points of $C$.\\
(ii) Sketch $C$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q5 [6]}}