| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - logarithmic functions |
| Difficulty | Standard +0.3 This is a standard multi-part calculus question requiring differentiation using chain rule, finding a normal line equation, and calculating an area using integration. While it involves multiple steps (7 marks total), each technique is routine for P1 level: differentiating a composite function, finding intersection points, determining gradient of normal, and integrating. The question is slightly easier than average because it's methodical rather than requiring problem-solving insight, though the integration and area calculation add some computational work. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = [0] + \left[(2x+1)^{-3}\right] \times [+16]\) | B2,1,0 | OE. Full marks for 3 correct components. Withhold one mark for each error or omission |
| \(\int y\,dx = [x] + \left[(2x+1)^{-1}\right] \times [+2]\ (+c)\) | B2,1,0 | OE. Full marks for 3 correct components. Withhold one mark for each error or omission |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(A\), \(x = \frac{1}{2}\) | B1 | Ignore extra answer \(x = -1.5\) |
| \(\dfrac{dy}{dx} = 2 \rightarrow\) Gradient of normal \(= -\frac{1}{2}\) | \*M1 | With their positive \(x\) at \(A\) and their \(\dfrac{dy}{dx}\), uses \(m_1 m_2 = -1\) |
| Equation of normal: \(y - 0 = -\frac{1}{2}(x - \frac{1}{2})\) or \(y - 0 = -\frac{1}{2}(0 - \frac{1}{2})\) or \(0 = -\frac{1}{2}x\cdot\frac{1}{2} + c\) | DM1 | Use of their \(x\) at \(A\) and their normal gradient |
| \(B\ (0,\ \frac{1}{4})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\displaystyle\int_0^{\frac{1}{2}} 1 - \dfrac{4}{(2x+1)^2}\,dx\) | \*M1 | \(\int y\,dx\) SOI with \(0\) and their positive \(x\) coordinate of \(A\) |
| \([\frac{1}{2} + 1] - [0 + 2] = (-\frac{1}{2})\) | DM1 | Substitutes both \(0\) and their \(\frac{1}{2}\) into their \(\int y\,dx\) and subtracts |
| Area of triangle above \(x\)-axis \(= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{4}\ \left(= \dfrac{1}{16}\right)\) | B1 | |
| Total area of shaded region \(= \dfrac{9}{16}\) | A1 | OE (including AWRT 0.563) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\displaystyle\int_{-3}^{0} \dfrac{1}{(1-y)^{\frac{1}{2}}} - \frac{1}{2}\,dy\) | \*M1 | \(\int x\,dy\) SOI. Where \(x\) is of the form \(k\!\left(1-y\right)^{-\frac{1}{2}} + c\) with \(0\) and their negative \(y\) intercept of curve |
| \([-2] - \left[-4 + \frac{3}{2}\right] = (\frac{1}{2})\) | DM1 | Substitutes both \(0\) and their \(-3\) into their \(\int x\,dy\) and subtracts |
| Area of triangle \(= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{4}\ \left(= \dfrac{1}{16}\right)\) | B1 | |
| Total area \(= \dfrac{9}{16}\) | A1 | OE (including AWRT 0.563) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\displaystyle\int_0^{\frac{1}{2}} -\frac{1}{2}x + \frac{1}{4} - y\,dx\) | \*M1 | \(\int\)(their normal curve) with \(0\) and their positive \(x\) coordinate of \(A\) |
| Curve \([\frac{1}{2} + 1] - [0 + 2] = (-\frac{1}{2})\) | DM1 | Substitutes both \(0\) and their \(\frac{1}{2}\) into their \(\int y\,dx\) and subtracts |
| \(\displaystyle\int_0^{\frac{1}{2}} -\frac{1}{2}x + \frac{1}{4}\,dx = \dfrac{-x^2}{4} + \dfrac{x}{4} = \left[\dfrac{-1}{16} + \dfrac{1}{8}\right] - [0] = \dfrac{1}{16}\) | B1 | Substitutes both \(0\) and \(\frac{1}{2}\) into the correct integral and subtracts |
| Total area \(= \dfrac{9}{16}\) | A1 | OE (including AWRT 0.563) |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = [0] + \left[(2x+1)^{-3}\right] \times [+16]$ | B2,1,0 | OE. Full marks for 3 correct components. Withhold one mark for each error or omission |
| $\int y\,dx = [x] + \left[(2x+1)^{-1}\right] \times [+2]\ (+c)$ | B2,1,0 | OE. Full marks for 3 correct components. Withhold one mark for each error or omission |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $A$, $x = \frac{1}{2}$ | B1 | Ignore extra answer $x = -1.5$ |
| $\dfrac{dy}{dx} = 2 \rightarrow$ Gradient of normal $= -\frac{1}{2}$ | \*M1 | With their positive $x$ at $A$ and their $\dfrac{dy}{dx}$, uses $m_1 m_2 = -1$ |
| Equation of normal: $y - 0 = -\frac{1}{2}(x - \frac{1}{2})$ or $y - 0 = -\frac{1}{2}(0 - \frac{1}{2})$ or $0 = -\frac{1}{2}x\cdot\frac{1}{2} + c$ | DM1 | Use of their $x$ at $A$ and their normal gradient |
| $B\ (0,\ \frac{1}{4})$ | A1 | |
---
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\int_0^{\frac{1}{2}} 1 - \dfrac{4}{(2x+1)^2}\,dx$ | \*M1 | $\int y\,dx$ SOI with $0$ and their positive $x$ coordinate of $A$ |
| $[\frac{1}{2} + 1] - [0 + 2] = (-\frac{1}{2})$ | DM1 | Substitutes both $0$ and their $\frac{1}{2}$ into their $\int y\,dx$ and subtracts |
| Area of triangle above $x$-axis $= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{4}\ \left(= \dfrac{1}{16}\right)$ | B1 | |
| Total area of shaded region $= \dfrac{9}{16}$ | A1 | OE (including AWRT 0.563) |
**Alternative method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\int_{-3}^{0} \dfrac{1}{(1-y)^{\frac{1}{2}}} - \frac{1}{2}\,dy$ | \*M1 | $\int x\,dy$ SOI. Where $x$ is of the form $k\!\left(1-y\right)^{-\frac{1}{2}} + c$ with $0$ and their negative $y$ intercept of curve |
| $[-2] - \left[-4 + \frac{3}{2}\right] = (\frac{1}{2})$ | DM1 | Substitutes both $0$ and their $-3$ into their $\int x\,dy$ and subtracts |
| Area of triangle $= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{4}\ \left(= \dfrac{1}{16}\right)$ | B1 | |
| Total area $= \dfrac{9}{16}$ | A1 | OE (including AWRT 0.563) |
**Alternative method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\displaystyle\int_0^{\frac{1}{2}} -\frac{1}{2}x + \frac{1}{4} - y\,dx$ | \*M1 | $\int$(their normal curve) with $0$ and their positive $x$ coordinate of $A$ |
| Curve $[\frac{1}{2} + 1] - [0 + 2] = (-\frac{1}{2})$ | DM1 | Substitutes both $0$ and their $\frac{1}{2}$ into their $\int y\,dx$ and subtracts |
| $\displaystyle\int_0^{\frac{1}{2}} -\frac{1}{2}x + \frac{1}{4}\,dx = \dfrac{-x^2}{4} + \dfrac{x}{4} = \left[\dfrac{-1}{16} + \dfrac{1}{8}\right] - [0] = \dfrac{1}{16}$ | B1 | Substitutes both $0$ and $\frac{1}{2}$ into the correct integral and subtracts |
| Total area $= \dfrac{9}{16}$ | A1 | OE (including AWRT 0.563) |10\\
\includegraphics[max width=\textwidth, alt={}, center]{567c3d72-c633-4ae0-8605-f63f93d718c4-18_979_679_262_731}
The diagram shows part of the curve $y = 1 - \frac { 4 } { ( 2 x + 1 ) ^ { 2 } }$. The curve intersects the $x$-axis at $A$. The normal to the curve at $A$ intersects the $y$-axis at $B$.\\
(i) Obtain expressions for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\int y \mathrm {~d} x$.\\
(ii) Find the coordinates of $B$.\\
(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 Q10 [12]}}