| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.8 This is a straightforward multi-part calculus question requiring standard techniques: differentiation of x^n terms, finding stationary points by setting dy/dx=0, second derivative test, equation of normal line, and basic integration. All parts are routine applications with no novel problem-solving required, making it easier than average for A-level. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{dy}{dx} = 2 - \dfrac{16}{x^3}\) | B1 | For \(-16/x^3\) |
| \(\dfrac{d^2y}{dx^2} = \dfrac{48}{x^4}\) | B1, B1\(\sqrt{}\) [3] | For "2" and for "0"; for \(d/dx\) of his \(-16/x^3\) providing \(-\text{ve}\) power differentiated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{dy}{dx} = 0 \rightarrow x = 2\), \(y = 6\) | M1, A1 | Sets \(dy/dx\) to \(0\) + attempt at \(x\); needs both coordinates |
| \(\dfrac{d^2y}{dx^2}\) is \(+\text{ve}\) — Minimum | A1\(\sqrt{}\) [3] | Looks at sign. Correct conclusion for his \(x\) and his \(2^{\text{nd}}\) differential |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = -2\), \(m = 4\) | ||
| Perp gradient \(= -\frac{1}{4}\) | M1 | Uses \(m_1 m_2 = -1\) with \(dy/dx\) |
| \(y + 2 = -\frac{1}{4}(x + 2)\) | DM1 | Correct form of equation (not for tan) |
| Sets \(y\) to \(0 \rightarrow x = -10\) | A1 [3] | Co – nb answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area \(= \left[x^2 - \dfrac{8}{x}\right]\) | B1, B1 | For each term |
| Evaluated from \(1\) to \(2 \rightarrow 7\) | B1 [3] | Co. (\(-7 \Rightarrow 7\) gets B0) |
## Question 10(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = 2 - \dfrac{16}{x^3}$ | B1 | For $-16/x^3$ |
| $\dfrac{d^2y}{dx^2} = \dfrac{48}{x^4}$ | B1, B1$\sqrt{}$ [3] | For "2" and for "0"; for $d/dx$ of his $-16/x^3$ providing $-\text{ve}$ power differentiated |
## Question 10(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = 0 \rightarrow x = 2$, $y = 6$ | M1, A1 | Sets $dy/dx$ to $0$ + attempt at $x$; needs both coordinates |
| $\dfrac{d^2y}{dx^2}$ is $+\text{ve}$ — Minimum | A1$\sqrt{}$ [3] | Looks at sign. Correct conclusion for his $x$ and his $2^{\text{nd}}$ differential |
## Question 10(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = -2$, $m = 4$ | | |
| Perp gradient $= -\frac{1}{4}$ | M1 | Uses $m_1 m_2 = -1$ with $dy/dx$ |
| $y + 2 = -\frac{1}{4}(x + 2)$ | DM1 | Correct form of equation (not for tan) |
| Sets $y$ to $0 \rightarrow x = -10$ | A1 [3] | Co – nb answer given |
## Question 10(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $= \left[x^2 - \dfrac{8}{x}\right]$ | B1, B1 | For each term |
| Evaluated from $1$ to $2 \rightarrow 7$ | B1 [3] | Co. ($-7 \Rightarrow 7$ gets B0) |
---10 The equation of a curve is $y = 2 x + \frac { 8 } { x ^ { 2 } }$.\\
(i) Obtain expressions for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(ii) Find the coordinates of the stationary point on the curve and determine the nature of the stationary point.\\
(iii) Show that the normal to the curve at the point $( - 2 , - 2 )$ intersects the $x$-axis at the point $( - 10,0 )$.\\
(iv) Find the area of the region enclosed by the curve, the $x$-axis and the lines $x = 1$ and $x = 2$.
\hfill \mbox{\textit{CAIE P1 2007 Q10 [12]}}