| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 6 |
| Topic | Curve Sketching |
| Type | Range restriction with excluded interval (linear/mixed denominator) |
| Difficulty | Standard +0.8 This question requires finding range restrictions by analyzing when a rational function intersects a horizontal line (leading to a discriminant condition), then using this to deduce the turning point—a non-standard approach requiring insight beyond routine calculus. The sketch requires synthesizing multiple features (asymptotes, intercepts, turning point). More conceptually demanding than typical curve sketching but accessible with solid understanding. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02p Interpret algebraic solutions: graphically1.07n Stationary points: find maxima, minima using derivatives |
**(i)** $k(x^2+6x+9) = 4x+11 \Rightarrow kx^2 + 2(3k-2)x + (9k-11) = 0$
For creating a quadratic in $x$: M1
$\Delta = 4(9k^2-12k+4) - 4k(9k-11)$, for considering the discriminant: M1
Real roots $\Leftrightarrow \Delta \geq 0$ gives $k \leq 4$, from use of correct condition: A1
$y(=k) = 4 \Rightarrow 4x^2 + 20x + 25 = 0$, for substituting back and finding $x$: M1
$(2x+5)^2 = 0 \Rightarrow x = -2.5$ **cao**: A1 **[5]**
**(ii)** Crossing points at $\left(0, \frac{11}{9}\right)$ & $\left(-\frac{11}{4}, 0\right)$: B1, B1
Asymptotes $x = -3$ & $y = 0$: B1, B1
All these MUST be stated clearly or shown on graph **&** be in approx correct positions, for shape relative to $y=0$ or $x=-3$: B1; All correct: B1 **[6]**7 A curve has equation $y = \frac { 4 x + 11 } { ( x + 3 ) ^ { 2 } }$.\\
(i) Show that the curve meets the line $y = k$ if and only if $k \leq 4$, and deduce the coordinates of the turning point on the curve.\\
(ii) Sketch the curve, stating the coordinates of the points where it cuts the axes, and showing clearly its asymptotes and the turning point.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q7 [6]}}