| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding maximum/minimum on curve |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard differentiation (chain rule), integration by substitution (recognizing the form), finding a stationary point by setting dy/dx = 0, and computing an area using definite integration. All techniques are routine for P1 level with no novel insights 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 derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \left[\frac{1}{2}(4x+1)^{-\frac{1}{2}}\right][\times 4]\left[-\frac{9}{2}(4x+1)^{-\frac{3}{2}}\right][\times 4]\) | B1B1B1 | B1 B1 for each, without \(\times 4\). B1 for \(\times 4\) twice. SC: If no other marks awarded, B1 for both powers of \((4x+1)\) correct |
| \(\left(\frac{2}{\sqrt{4x+1}} - \frac{18}{(\sqrt{4x+1})^3}\text{ or }\frac{8x-16}{(4x+1)^{\frac{3}{2}}}\right)\) | ||
| \(\int y\,\mathrm{d}x = \left[\frac{(4x+1)^{\frac{3}{2}}}{\frac{3}{2}}\right][\div 4] + \left[\frac{9(4x+1)^{\frac{1}{2}}}{\frac{1}{2}}\right][\div 4](+C)\) | B1B1B1 | B1 B1 for each, without \(\div 4\). B1 for \(\div 4\) twice. \(+C\) not required. SC: B1 for both powers of \((4x+1)\) correct |
| \(\left(\frac{(\sqrt{4x+1})^3}{6} + \frac{9}{2}(\sqrt{4x+1})(+C)\right)\) | ||
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \rightarrow \frac{2}{\sqrt{4x+1}} - \frac{18}{(4x+1)^{\frac{3}{2}}} = 0\) | M1 | Sets \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to 0 and attempts to solve |
| \(4x + 1 = 9\) or \((4x+1)^2 = 81\) | A1 | Must be from correct differential |
| \(x = 2\), \(y = 6\) or M is \((2, 6)\) only | A1 | Both values required. Must be from correct differential |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Realises area is \(\int y\,\mathrm{d}x\) and attempt to use their 2 and sight of 0 | *M1 | Needs to use their integral and to see '*their 2*' substituted |
| Uses limits 0 to 2 correctly \(\rightarrow [4.5 + 13.5] - [\frac{1}{6} + 4.5]\) (\(= 13\frac{1}{3}\)) | DM1 | Uses both 0 and '*their 2*' and subtracts. Condone wrong way round |
| (Area \(=\)) \(1\frac{1}{3}\) or \(1.33\) | A1 | Must be from a correct differential and integral |
| 3 | \(13\frac{1}{3}\) or \(1\frac{1}{3}\) with little or no working scores M1DM0A0 |
## Question 11(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \left[\frac{1}{2}(4x+1)^{-\frac{1}{2}}\right][\times 4]\left[-\frac{9}{2}(4x+1)^{-\frac{3}{2}}\right][\times 4]$ | B1B1B1 | B1 B1 for each, without $\times 4$. B1 for $\times 4$ twice. SC: If no other marks awarded, B1 for both powers of $(4x+1)$ correct |
| $\left(\frac{2}{\sqrt{4x+1}} - \frac{18}{(\sqrt{4x+1})^3}\text{ or }\frac{8x-16}{(4x+1)^{\frac{3}{2}}}\right)$ | | |
| $\int y\,\mathrm{d}x = \left[\frac{(4x+1)^{\frac{3}{2}}}{\frac{3}{2}}\right][\div 4] + \left[\frac{9(4x+1)^{\frac{1}{2}}}{\frac{1}{2}}\right][\div 4](+C)$ | B1B1B1 | B1 B1 for each, without $\div 4$. B1 for $\div 4$ twice. $+C$ not required. SC: B1 for both powers of $(4x+1)$ correct |
| $\left(\frac{(\sqrt{4x+1})^3}{6} + \frac{9}{2}(\sqrt{4x+1})(+C)\right)$ | | |
| | **6** | |
---
## Question 11(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \rightarrow \frac{2}{\sqrt{4x+1}} - \frac{18}{(4x+1)^{\frac{3}{2}}} = 0$ | M1 | Sets $\frac{\mathrm{d}y}{\mathrm{d}x}$ to 0 and attempts to solve |
| $4x + 1 = 9$ or $(4x+1)^2 = 81$ | A1 | Must be from correct differential |
| $x = 2$, $y = 6$ **or** M is $(2, 6)$ only | A1 | Both values required. Must be from correct differential |
| | **3** | |
---
## Question 11(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Realises area is $\int y\,\mathrm{d}x$ **and** attempt to use their 2 and sight of 0 | *M1 | Needs to use their integral and to see '*their 2*' substituted |
| Uses limits 0 to 2 correctly $\rightarrow [4.5 + 13.5] - [\frac{1}{6} + 4.5]$ ($= 13\frac{1}{3}$) | DM1 | Uses both 0 and '*their 2*' and subtracts. Condone wrong way round |
| (Area $=$) $1\frac{1}{3}$ **or** $1.33$ | A1 | Must be from a correct differential and integral |
| | **3** | $13\frac{1}{3}$ or $1\frac{1}{3}$ with little or no working scores M1DM0A0 |11\\
\includegraphics[max width=\textwidth, alt={}, center]{ed5b77ae-6eac-4e73-bc43-613433abd3e1-16_723_942_260_598}
The diagram shows part of the curve $y = \sqrt { } ( 4 x + 1 ) + \frac { 9 } { \sqrt { } ( 4 x + 1 ) }$ and the minimum point $M$.\\
(i) Find expressions for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\int y \mathrm {~d} x$.\\
(ii) Find the coordinates of $M$.\\
The shaded region is bounded by the curve, the $y$-axis and the line through $M$ parallel to the $x$-axis.\\
(iii) Find, showing all necessary working, 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.\\
\hfill \mbox{\textit{CAIE P1 2019 Q11 [12]}}