| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2019 |
| Session | Specimen |
| Marks | 5 |
| Topic | Curve Sketching |
| Type | Simple rational function analysis |
| Difficulty | Standard +0.8 This question requires rearranging to form a quadratic in x, then applying discriminant conditions (Δ ≥ 0) to find the range of y—a non-standard technique that goes beyond routine curve sketching. Deducing turning points from the range boundary and sketching requires synthesis of multiple ideas, making it moderately challenging but still within typical A-level scope. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives |
**(a)** $y = \frac{x+1}{x^2+3} \Rightarrow y\cdot x^2 - x + (3y-1) = 0$ 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**
**Total: 5**
**(b)** $y = \frac{1}{2}$ substituted back $\Rightarrow \frac{1}{2}(x^2 - 2x + 1) = 0 \Rightarrow x = 1$ $\left[y = \frac{1}{2}\right]$ **M1A1**
$y = -\frac{1}{6}$ substituted back $\Rightarrow -\frac{1}{6}(x^2 + 6x + 9) = 0 \Rightarrow x = -3$ $\left[y = -\frac{1}{6}\right]$ **M1A1**
**Total: 4**6 The curve $C$ has equation $y = \frac { x + 1 } { x ^ { 2 } + 3 }$.
\begin{enumerate}[label=(\alph*)]
\item By considering a suitable quadratic equation in $x$, find the set of possible values of $y$ for points on $C$.
\item Deduce the coordinates of the turning points on $C$.
\item Sketch $C$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2019 Q6 [5]}}