| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | Specimen |
| Marks | 9 |
| Topic | Curve Sketching |
| Type | Simple rational function analysis |
| Difficulty | Standard +0.8 This question requires rearranging to form a quadratic in x, then using discriminant analysis to find the range of y-values—a non-routine technique for A-level. Deducing turning points from the range boundary and sketching requires connecting multiple concepts. More sophisticated than standard curve sketching but accessible with problem-solving insight. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives |
**(i)** $y=\frac{x+1}{x^2+3} \Rightarrow yx^2 - x + (3y-1) = 0$ — Creating a quadratic in $x$ **M1**
For real $x$, $1-4y(3y-1)\geqslant 0$ — Considering the discriminant **M1**
$12y^2-4y-1\leqslant 0$ — Creating a quadratic **inequality** **M1**
For real $x$, $(6y+1)(2y-1)\leqslant 0$ — Factorising/solving a 3-term quadratic **M1**
$-\frac{1}{6}\leqslant y\leqslant\frac{1}{2}$ CAO **A1**
**(ii)** $y=\frac{1}{2}$ substituted back $\Rightarrow \frac{1}{2}(x^2-2x+1)=0 \Rightarrow x=1$ $\left[y=\frac{1}{2}\right]$ **M1 A1**
$y=-\frac{1}{6}$ substituted back $\Rightarrow -\frac{1}{6}(x^2+6x+9)=0 \Rightarrow x=-3$ $\left[y=-\frac{1}{6}\right]$ **M1 A1**
**Total: 9 marks**6 The curve $C$ has equation $y = \frac { x + 1 } { x ^ { 2 } + 3 }$.\\
(i) By considering a suitable quadratic equation in $x$, find the set of possible values of $y$ for points on $C$.\\
(ii) Deduce the coordinates of the turning points on $C$.\\
(iii) Sketch $C$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q6 [9]}}