| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | June |
| Marks | 4 |
| Topic | Curve Sketching |
| Type | Range restriction with excluded interval (linear/mixed denominator) |
| Difficulty | Challenging +1.2 This is a structured range restriction problem requiring algebraic manipulation to find where y=k intersects the curve, then using the discriminant condition (b²-4ac≥0) to find excluded y-values. Part (ii) uses the boundary values as turning points. While it involves multiple steps and the discriminant technique, the question provides clear scaffolding and uses standard A-level methods without requiring novel insight—slightly above average difficulty due to the algebraic manipulation and conceptual link between range restrictions and turning points. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.07n Stationary points: find maxima, minima using derivatives |
**Question 3** [4+4 marks]
**(i)**
$\frac{2x^2-x-1}{2x-3} = k \Rightarrow 2x^2-(2k+1)x+(3k-1)=0$ B1 **(AG)** Shown legitimately
For non-real $x$, $(2k+1)^2 - 8(3k-1) < 0$ M1 (Considering discriminant or equivalent)
$4k^2 - 20k + 9 < 0 \Rightarrow (2k-1)(2k-9) < 0$ M1 (Solving from $\Delta < 0$)
$\Rightarrow$ no curve for $\frac{1}{2} < k = y < \frac{9}{2}$ A1 **(AG)** Must be satisfactorily explained
**[4]**
**(ii)**
TPs at $y = \frac{1}{2}$ and $y = \frac{9}{2}$ M1 (First $y$ ($k$) substituted back)
i.e. $2x^2-2x+\frac{1}{2}=0$ and $2x^2-10x+\frac{25}{2}=0$ M1 (Second $y$ ($k$) substituted back)
$x = \frac{1}{2}$ and $x = \frac{5}{2}$ A1A1
**[4]**
Alternative method: when $\Delta=0$, **M1** $x = "-\frac{b}{2a}" = \frac{2k+1}{4}$
**M1** $\Rightarrow x=\frac{1}{2}\ (y=\frac{1}{2})$ & $x=\frac{5}{2}\ (y=\frac{9}{2})$ **A1 A1**
Note: For finding TPs via $\frac{dy}{dx}=0$, max. **M1 A1** since qn. asks for a "deduce" method3 A curve has equation $y = \frac { 2 x ^ { 2 } - x - 1 } { 2 x - 3 }$.\\
(i) Show that the curve meets the line $y = k$ when $2 x ^ { 2 } - ( 2 k + 1 ) x + ( 3 k - 1 ) = 0$, and hence show that no part of the curve exists in the interval $\frac { 1 } { 2 } < y < \frac { 9 } { 2 }$.\\
(ii) Deduce the coordinates of the turning points of this curve.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q3 [4]}}