| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Topic | Curve Sketching |
| Type | Range restriction with excluded interval (linear/mixed denominator) |
| Difficulty | Standard +0.8 This question requires rearranging to form a quadratic in x, then applying the discriminant condition (Δ ≥ 0) to find the range of y—a technique that's non-standard and requires insight beyond routine calculus. Deducing turning points from the range adds another conceptual layer, making this moderately challenging for A-level. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.07n Stationary points: find maxima, minima using derivatives |
**(i)** $y = \frac{x+1}{x^2+3} \Rightarrow y\cdot x^2 - x + (3y-1) = 0$ Creating a quadratic in $x$ **M1**
For real $x$, $1 - 4y(3y-1) \geq 0$ Considering the discriminant **M1**
$12y^2 - 4y - 1 \leq 0$ Creating a quadratic **inequality** **M1**
For real $x$, $(6y+1)(2y-1) \leq 0$ Factorising/solving a 3-term quadratic **M1**
$-\frac{1}{6} \leq y \leq \frac{1}{2}$ **cso** **A1**
NB lack of inequality earlier with unjustified correct answer loses only the 3rd **M** mark
**[5]**
**(ii)** $y = \frac{1}{2}$ substd. back $\Rightarrow \frac{1}{2}(x^2 - 2x + 1) = 0 \Rightarrow x = 1$ $[y = \frac{1}{2}]$ **M1 A1**
$y = -\frac{1}{6}$ substd. back $\Rightarrow -\frac{1}{6}(x^2 + 6x + 9) = 0 \Rightarrow x = -3$ $[y = -\frac{1}{6}]$ **M1 A1**
Allow alternative approach via calculus:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x^2-2x+3}{(x^2+3)^2}$ **M1** Solving quadratic to find 2 values of $x$ **M1**
Then **A1 A1** each pair of correct $(x,y)$ coordinates
**[4]**4 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 $C$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2012 Q4 [9]}}