| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | March |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (extended problem with normals, stationary points, or further geometry) |
| Difficulty | Moderate -0.3 This is a structured multi-part integration question requiring standard techniques: finding stationary points by setting dy/dx=0, integrating using power rule (including negative powers), using a boundary condition to find the constant, solving a quartic equation, and finding area under a curve. All steps are routine P1/AS-level procedures with no novel problem-solving required, making it slightly easier than average but still requiring careful execution across multiple parts. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2x - \frac{2}{x^3} = 0\) | M1 | Set \(= 0\) |
| \(x^4 = 1 \Rightarrow x = 1\) at \(A\) cao | A1 | Allow 'spotted' \(x = 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = x^2 + \frac{1}{x^2}\ (+c)\) cao | B1 | |
| \(\frac{189}{16} = 16 + \frac{1}{16} + c\) | M1 | Sub \(\left(4,\ \frac{189}{16}\right)\). \(c\) must be present. Dep. on integration |
| \(c = -\frac{17}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 + \frac{1}{x^2} - \frac{17}{4} = 0 \Rightarrow 4x^4 - 17x^2 + 4\ (=0)\) | M1 | Multiply by \(4x^2\) (or similar) to transform into 3-term quartic |
| \((4x^2 - 1)(x^2 - 4)\ (=0)\) | M1 | Treat as quadratic in \(x^2\) and attempt solution or factorisation |
| \(x = \frac{1}{2},\ 2\) | A1A1 | Not necessary to distinguish. Ignore negative values. No working scores 0/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int(x^2 + x^{-2} - \frac{17}{4})\,dx = \frac{x^3}{3} - \frac{1}{x} - \frac{17x}{4}\) | B2,1,0 | Mark final integral |
| \(\left(\frac{8}{3} - \frac{1}{2} - \frac{17}{2}\right) - \left(\frac{1}{24} - 2 - \frac{17}{8}\right)\) | M1 | Apply *their* limits from (iii) (Seen). Dep. on integration of at least 1 term of \(y\) |
| Area \(= \frac{9}{4}\) | A1 | Mark final answer. \(\int y^2\) scores 0/4 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x - \frac{2}{x^3} = 0$ | **M1** | Set $= 0$ |
| $x^4 = 1 \Rightarrow x = 1$ at $A$ cao | **A1** | Allow 'spotted' $x = 1$ |
---
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = x^2 + \frac{1}{x^2}\ (+c)$ cao | **B1** | |
| $\frac{189}{16} = 16 + \frac{1}{16} + c$ | **M1** | Sub $\left(4,\ \frac{189}{16}\right)$. $c$ must be present. Dep. on integration |
| $c = -\frac{17}{4}$ | **A1** | |
---
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + \frac{1}{x^2} - \frac{17}{4} = 0 \Rightarrow 4x^4 - 17x^2 + 4\ (=0)$ | **M1** | Multiply by $4x^2$ (or similar) to transform into 3-term quartic |
| $(4x^2 - 1)(x^2 - 4)\ (=0)$ | **M1** | Treat as quadratic in $x^2$ and attempt solution or factorisation |
| $x = \frac{1}{2},\ 2$ | **A1A1** | Not necessary to distinguish. Ignore negative values. No working scores 0/4 |
---
## Question 10(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int(x^2 + x^{-2} - \frac{17}{4})\,dx = \frac{x^3}{3} - \frac{1}{x} - \frac{17x}{4}$ | **B2,1,0** | Mark final integral |
| $\left(\frac{8}{3} - \frac{1}{2} - \frac{17}{2}\right) - \left(\frac{1}{24} - 2 - \frac{17}{8}\right)$ | **M1** | Apply *their* limits from **(iii)** (Seen). Dep. on integration of at least 1 term of $y$ |
| Area $= \frac{9}{4}$ | **A1** | Mark final answer. $\int y^2$ scores 0/4 |10\\
\includegraphics[max width=\textwidth, alt={}, center]{f759ce41-708e-4fe7-80b9-adc2be2972ac-18_611_531_262_808}
The diagram shows the curve $y = \mathrm { f } ( x )$ defined for $x > 0$. The curve has a minimum point at $A$ and crosses the $x$-axis at $B$ and $C$. It is given that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x - \frac { 2 } { x ^ { 3 } }$ and that the curve passes through the point $\left( 4 , \frac { 189 } { 16 } \right)$.\\
(i) Find the $x$-coordinate of $A$.\\
(ii) Find $\mathrm { f } ( x )$.\\
(iii) Find the $x$-coordinates of $B$ and $C$.\\
(iv) Find, showing all necessary working, the area of the shaded region.\\
{www.cie.org.uk} after the live examination series.
}
\hfill \mbox{\textit{CAIE P1 2017 Q10 [13]}}