| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a multi-part question involving standard calculus techniques: finding stationary points by differentiation, finding a normal line equation, and calculating an area bounded by a curve and a normal. All components are routine A-level procedures with no novel insight required, though the multi-step nature and area calculation add modest complexity beyond the most basic questions. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 3x^2 - 12x + 9\) | B1 | co (can be given in part (ii)) |
| Solves \(\frac{dy}{dx} = 0\) | M1 | Attempt to solve \(dy/dx = 0\) |
| \(\rightarrow A(1, 4)\), \(B(3, 0)\) | A1 [3] | Both needed |
| (ii) If \(x = 2\), \(m = -3\) | M1 | Use of \(m_1m_2 = -1\). needs calculus |
| Normal has \(m = \frac{1}{3}\) | ||
| Eqn \(y - 2 = \frac{1}{3}(x - 2)\) or \(3y = x + 4\) | M1 A1 [3] | Correct form of equation – needs calculus. A1 any form |
| (iii) area under curve – integrate \(y\) | ||
| \(\rightarrow \frac{1}{4}x^4 - 2x^3 + \frac{9}{2}x^2\) | B2,1 | For the 3 terms. –1 for each error |
| Limits 2 to "his 3" \(\rightarrow \frac{3}{4}\) (0.75) | M1 | Using 2 to "his 3" with integration |
| Area of trapezium \(= \frac{1}{2} \times 1 \times (2 + 2\frac{3}{4})\) | M1 | Any correct method for trapezium |
| \(= 2\frac{1}{8}\) | ||
| Subtract \(\rightarrow\) shaded area of \(1\frac{5}{12}\) | A1 [5] | co |
(i) $\frac{dy}{dx} = 3x^2 - 12x + 9$ | B1 | co (can be given in part (ii))
| |
Solves $\frac{dy}{dx} = 0$ | M1 | Attempt to solve $dy/dx = 0$
$\rightarrow A(1, 4)$, $B(3, 0)$ | A1 [3] | Both needed
(ii) If $x = 2$, $m = -3$ | M1 | Use of $m_1m_2 = -1$. needs calculus
Normal has $m = \frac{1}{3}$ | |
Eqn $y - 2 = \frac{1}{3}(x - 2)$ or $3y = x + 4$ | M1 A1 [3] | Correct form of equation – needs calculus. A1 any form
(iii) area under curve – integrate $y$ | |
$\rightarrow \frac{1}{4}x^4 - 2x^3 + \frac{9}{2}x^2$ | B2,1 | For the 3 terms. –1 for each error
Limits 2 to "his 3" $\rightarrow \frac{3}{4}$ (0.75) | M1 | Using 2 to "his 3" with integration
| |
Area of trapezium $= \frac{1}{2} \times 1 \times (2 + 2\frac{3}{4})$ | M1 | Any correct method for trapezium
$= 2\frac{1}{8}$ | |
Subtract $\rightarrow$ shaded area of $1\frac{5}{12}$ | A1 [5] | co11\\
\includegraphics[max width=\textwidth, alt={}, center]{3b527397-7781-41e9-8218-57277cc977bf-4_686_805_950_669}
The diagram shows the curve $y = x ^ { 3 } - 6 x ^ { 2 } + 9 x$ for $x \geqslant 0$. The curve has a maximum point at $A$ and a minimum point on the $x$-axis at $B$. The normal to the curve at $C ( 2,2 )$ meets the normal to the curve at $B$ at the point $D$.\\
(i) Find the coordinates of $A$ and $B$.\\
(ii) Find the equation of the normal to the curve at $C$.\\
(iii) Find the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2009 Q11 [11]}}