| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line-Axis Bounded Region |
| Difficulty | Standard +0.3 This is a standard A-level integration question requiring finding intersection points by solving a quadratic (after substitution), verifying a stationary point using differentiation, and calculating area between curves. All techniques are routine P1/C1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4x^{\frac{1}{2}} - 2x = 3 - x \rightarrow x - 4x^{\frac{1}{2}} + 3\ (=0)\) | *M1 | 3-term quadratic. Can be expressed as e.g. \(u^2 - 4u + 3\ (=0)\) |
| \(\left(x^{\frac{1}{2}}-1\right)\left(x^{\frac{1}{2}}-3\right)(=0)\) or \((u-1)(u-3)(=0)\) | DM1 | Or quadratic formula or completing square |
| \(x^{\frac{1}{2}} = 1, 3\) | A1 | SOI |
| \(x = 1, 9\) | A1 | |
| Alternative method: | ||
| \(\left(4x^{\frac{1}{2}}\right)^2 = (3+x)^2\) | *M1 | Isolate \(x^{\frac{1}{2}}\) |
| \(16x = 9 + 6x + x^2 \rightarrow x^2 - 10x + 9\ (=0)\) | A1 | 3-term quadratic |
| \((x-1)(x-9)\ (=0)\) | DM1 | Or formula or completing square on a quadratic obtained by a correct method |
| \(x = 1, 9\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 2x^{1/2} - 2\) | *B1 | |
| \(\frac{dy}{dx}\) or \(2x^{1/2} - 2 = 0\) when \(x = 1\) hence \(B\) is a stationary point | DB1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area of correct triangle \(= \frac{1}{2}(9-3) \times 6\) | M1 | or \(\int_{3}^{9}(3-x)(dx) = \left[3x - \frac{1}{2}x^2\right] \rightarrow -18\) |
| \(\int(4x^{\frac{1}{2}} - 2x)(dx) = \left[\frac{4x^{\frac{3}{2}}}{\frac{3}{2}} - x^2\right]\) | B1 B1 | |
| \((72-81) - \left(\frac{64}{3} - 16\right)\) | M1 | Apply limits \(4 \rightarrow\) their \(9\) to an integrated expression |
| \(-14\frac{1}{3}\) | A1 | OE |
| Shaded region \(= 18 - 14\frac{1}{3} = 3\frac{2}{3}\) | A1 | OE |
| Total | 6 |
## Question 12(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4x^{\frac{1}{2}} - 2x = 3 - x \rightarrow x - 4x^{\frac{1}{2}} + 3\ (=0)$ | *M1 | 3-term quadratic. Can be expressed as e.g. $u^2 - 4u + 3\ (=0)$ |
| $\left(x^{\frac{1}{2}}-1\right)\left(x^{\frac{1}{2}}-3\right)(=0)$ or $(u-1)(u-3)(=0)$ | DM1 | Or quadratic formula or completing square |
| $x^{\frac{1}{2}} = 1, 3$ | A1 | SOI |
| $x = 1, 9$ | A1 | |
| **Alternative method:** | | |
| $\left(4x^{\frac{1}{2}}\right)^2 = (3+x)^2$ | *M1 | Isolate $x^{\frac{1}{2}}$ |
| $16x = 9 + 6x + x^2 \rightarrow x^2 - 10x + 9\ (=0)$ | A1 | 3-term quadratic |
| $(x-1)(x-9)\ (=0)$ | DM1 | Or formula or completing square on a quadratic obtained by a correct method |
| $x = 1, 9$ | A1 | |
| **Total** | **4** | |
## Question 12(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 2x^{1/2} - 2$ | *B1 | |
| $\frac{dy}{dx}$ or $2x^{1/2} - 2 = 0$ when $x = 1$ hence $B$ is a stationary point | DB1 | |
| **Total** | **2** | |
## Question 12(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of correct triangle $= \frac{1}{2}(9-3) \times 6$ | M1 | or $\int_{3}^{9}(3-x)(dx) = \left[3x - \frac{1}{2}x^2\right] \rightarrow -18$ |
| $\int(4x^{\frac{1}{2}} - 2x)(dx) = \left[\frac{4x^{\frac{3}{2}}}{\frac{3}{2}} - x^2\right]$ | B1 B1 | |
| $(72-81) - \left(\frac{64}{3} - 16\right)$ | M1 | Apply limits $4 \rightarrow$ their $9$ to an integrated expression |
| $-14\frac{1}{3}$ | A1 | OE |
| Shaded region $= 18 - 14\frac{1}{3} = 3\frac{2}{3}$ | A1 | OE |
| **Total** | **6** | |12\\
\includegraphics[max width=\textwidth, alt={}, center]{fdd6e942-b5bc-4369-8587-6de120459776-18_557_677_264_733}
The diagram shows a curve with equation $y = 4 x ^ { \frac { 1 } { 2 } } - 2 x$ for $x \geqslant 0$, and a straight line with equation $y = 3 - x$. The curve crosses the $x$-axis at $A ( 4,0 )$ and crosses the straight line at $B$ and $C$.
\begin{enumerate}[label=(\alph*)]
\item Find, by calculation, the $x$-coordinates of $B$ and $C$.
\item Show that $B$ is a stationary point on the curve.
\item Find the area of the shaded region.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q12 [12]}}