| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a standard multi-part calculus question requiring differentiation to find a tangent equation, then integration to find an area. All techniques are routine for P1 level: finding dy/dx, using point-slope form, setting up and evaluating a definite integral, and subtracting a triangle area. Slightly above average due to the multi-step nature and area calculation involving both integration and geometry, but no novel insight required. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 6 - 6x\) | B1 | |
| At \(x=2\), gradient \(= -6\) soi | B1\(\checkmark\) | |
| \(y - 9 = -6(x-2)\) oe — Expect \(y = -6x + 21\) | M1 | Line through \((2, 9)\) and with gradient *their* \(-6\) |
| When \(y=0\), \(x = 3\frac{1}{2}\) cao | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area under curve: \(\int 9 + 6x - 3x^2\,dx = 9x + 3x^2 - x^3\) | B2,1,0 | Allow unsimplified terms |
| \((27 + 27 - 27) - (18 + 12 - 8)\) | M1 | Apply limits 2, 3. Expect 5 |
| Area under tangent: \(\frac{1}{2} \times \frac{3}{2} \times 9\ \left(= \frac{27}{4}\right)\) | B1\(\checkmark\) | OR \(\int_2^{7/2}(-6x+21)\,dx \to \frac{27}{4}\). Ft on *their* \(-6x+21\) and/or *their* \(7/2\). |
| Area required \(\frac{27}{4} - 5 = \frac{7}{4}\) | A1 [5] |
## Question 10(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 6 - 6x$ | B1 | |
| At $x=2$, gradient $= -6$ soi | B1$\checkmark$ | |
| $y - 9 = -6(x-2)$ oe — Expect $y = -6x + 21$ | M1 | Line through $(2, 9)$ and with gradient *their* $-6$ |
| When $y=0$, $x = 3\frac{1}{2}$ cao | A1 [4] | |
## Question 10(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area under curve: $\int 9 + 6x - 3x^2\,dx = 9x + 3x^2 - x^3$ | B2,1,0 | Allow unsimplified terms |
| $(27 + 27 - 27) - (18 + 12 - 8)$ | M1 | Apply limits 2, 3. Expect 5 |
| Area under tangent: $\frac{1}{2} \times \frac{3}{2} \times 9\ \left(= \frac{27}{4}\right)$ | B1$\checkmark$ | OR $\int_2^{7/2}(-6x+21)\,dx \to \frac{27}{4}$. Ft on *their* $-6x+21$ and/or *their* $7/2$. |
| Area required $\frac{27}{4} - 5 = \frac{7}{4}$ | A1 [5] | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{8da9e73a-3126-471b-b904-25e3c156f6bf-3_682_1319_1525_413}
Points $A ( 2,9 )$ and $B ( 3,0 )$ lie on the curve $y = 9 + 6 x - 3 x ^ { 2 }$, as shown in the diagram. The tangent at $A$ intersects the $x$-axis at $C$. Showing all necessary working,\\
(i) find the equation of the tangent $A C$ and hence find the $x$-coordinate of $C$,\\
(ii) find the area of the shaded region $A B C$.\\[0pt]
[Question 11 is printed on the next page.]
\hfill \mbox{\textit{CAIE P1 2015 Q10 [9]}}