| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of tangent |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring expansion of the cubic, differentiation using standard rules, finding tangent equations, solving simultaneous equations, and computing definite integrals. While it involves multiple steps, each technique is routine for P1 level with no novel insights required. Slightly above average difficulty due to the algebraic manipulation and multi-step nature, but well within standard A-level expectations. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = x^3 - 3x^2 + 2x\) \(\frac{dy}{dx} = 3x^2 - 6x + 2\) | M1 A1 | Attempt at differentiation. co. |
| At \(A(1,0)\), \(m = -1 \rightarrow y = -(x-1)\) At \(B(2,0)\), \(m = 2 \rightarrow y = 2(x-2)\) Sim equations \(\rightarrow x = \frac{2}{3}\) | M1 once M1 A1 [5] | Correct form of eqn of tangent - not normal. Solution of Sim Eqns - even if normals. |
| (ii) \(R_1 = \int (x^3 - 3x^2 + 2x) dx\) \(= \left[\frac{x^4}{4} - \frac{3x^3}{3} - \frac{2x^2}{2}\right] = \frac{1}{4}\) | M1 A1∨ DM1 | Attempt at integration. Integration correct for his cubic. Correct use of limits once. |
| \(R_2 = []^2 - []^1 = -\frac{1}{4}\) | A1 [4] | Both correct (allow ± in either/both cases) (0 to 2 → 0 + reason ok). Ignore errors over ± |
(i) $y = x^3 - 3x^2 + 2x$ $\frac{dy}{dx} = 3x^2 - 6x + 2$ | M1 A1 | Attempt at differentiation. co.
At $A(1,0)$, $m = -1 \rightarrow y = -(x-1)$ At $B(2,0)$, $m = 2 \rightarrow y = 2(x-2)$ Sim equations $\rightarrow x = \frac{2}{3}$ | M1 once M1 A1 [5] | Correct form of eqn of tangent - not normal. Solution of Sim Eqns - even if normals.
(ii) $R_1 = \int (x^3 - 3x^2 + 2x) dx$ $= \left[\frac{x^4}{4} - \frac{3x^3}{3} - \frac{2x^2}{2}\right] = \frac{1}{4}$ | M1 A1∨ DM1 | Attempt at integration. Integration correct for his cubic. Correct use of limits once.
$R_2 = []^2 - []^1 = -\frac{1}{4}$ | A1 [4] | Both correct (allow ± in either/both cases) (0 to 2 → 0 + reason ok). Ignore errors over ±
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\includegraphics[max width=\textwidth, alt={}, center]{dd2cb0ec-5df9-4d99-9e15-5ae1f1c07b96-3_490_665_1793_740}
The diagram shows the curve $y = x ( x - 1 ) ( x - 2 )$, which crosses the $x$-axis at the points $O ( 0,0 )$, $A ( 1,0 )$ and $B ( 2,0 )$.\\
(i) The tangents to the curve at the points $A$ and $B$ meet at the point $C$. Find the $x$-coordinate of $C$.\\
(ii) Show by integration that the area of the shaded region $R _ { 1 }$ is the same as the area of the shaded region $R _ { 2 }$.
\hfill \mbox{\textit{CAIE P1 2006 Q7 [9]}}