| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 12 |
| Topic | Curve Sketching |
| Type | Find unknown coefficients from roots |
| Difficulty | Standard +0.3 This is a multi-part question requiring standard techniques: using factor theorem and differentiation to find coefficients (routine), sketching a cubic (standard), and polynomial expansion with symmetry interpretation (slightly beyond routine but still accessible). The symmetry deduction in part (iii) adds mild conceptual interest but overall this is slightly easier than average A-level fare. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07n Stationary points: find maxima, minima using derivatives |
**(i)** Use of Factor Theorem: M1
giving $a + b = 2$ A1
Zero gradient at $x = -1 \Rightarrow 3 + 6 + a = 0$ A1
Solving to give $a = -9,\ b = 11$ A1 **[4]**
**(ii)** Attempt at finding roots of quadratic factor $(x^2 - 2x - 11)$: M1
Roots: $1,\ 1 \pm 2\sqrt{3}$ A1
Finding $x$-coordinates of stationary points: $-1, 3$ B1
Sketch of cubic incorporating above information and correct behaviour beyond extreme roots B1 **[4]**
**(iii)** $\mathrm{P}(1+x) = (1+x)^3 - 3(1+x)^2 - 9(1+x) + 11$ M1
$= x^3 - 12x$ A1
$\mathrm{P}(1-x) = (-x)^3 - 12(-x) = -(x^3 - 12x) = -\mathrm{P}(1+x)$ **AG** B1
The graph has antisymmetry about the line $x = 1$, or equivalent. B1 **[4]**7 A cubic polynomial is given by
$$\mathrm { P } ( x ) = x ^ { 3 } - 3 x ^ { 2 } + a x + b$$
where $a$ and $b$ are constants.\\
(i) If $\mathrm { P } ( x )$ is exactly divisible by $x - 1$, and has a local maximum at $x = - 1$, determine the values of $a$ and $b$.\\
(ii) Sketch the curve $y = \mathrm { P } ( x )$, marking the intercepts and the $x$-coordinates of the stationary points.\\
(iii) Expand and simplify $\mathrm { P } ( 1 + x )$, and deduce that $\mathrm { P } ( 1 + x ) = - \mathrm { P } ( 1 - x )$. Interpret this result graphically.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q7 [12]}}