| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2005 |
| Session | November |
| Marks | 12 |
| 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 -0.8 This is a straightforward multi-part integration question requiring only standard techniques: integrating x^(-3), using a point to find the constant, finding where dy/dx equals the negative reciprocal of -1/2, and computing a definite integral. All parts are routine applications of basic calculus with no problem-solving insight required, making it easier than average but not trivial due to the multi-step nature. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = \frac{16x^{-2}}{-2} + c\) | M1 A1 | Attempt to \(\int\). Allow A1 unsimplified. (ignore +c for these marks) |
| Use of \((1,4) \rightarrow y = -\frac{8}{x^2} + 12\) | M1A1 | Using coordinates in an integrated expression. Co. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{16}{x^3} = 2 \rightarrow x = 2\) | M1 DM1 | Use of \(m_1m_2=-1\) Link with \(\frac{dy}{dx} \rightarrow\) value for \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| ie \(2y + x = 22\) | M1 A1 | Using calculated \(x\) and \(y\) to find equation. Allow for correct 3 term linear equation |
| (iii) \(A = \int(\frac{-8}{x^2}+12)dx\) | M1 | Attempt at integration of \(y\) from (i) even if linear |
| \(= \frac{8}{x} + 12x\) | A1√ | Allow ft, providing original has two terms, one of which is a -ve power |
| \(= [\ ]\) at 2 \(- [\ ]\) at 1 | M1 | Correct use of correct limits in some integrated expression. |
| \(\rightarrow 8\) | A1 | [4] |
$\frac{dy}{dx} = \frac{16}{x^3}$, through $(1,4)$
(i) $y = \frac{16x^{-2}}{-2} + c$ | M1 A1 | Attempt to $\int$. Allow A1 unsimplified. (ignore +c for these marks)
Use of $(1,4) \rightarrow y = -\frac{8}{x^2} + 12$ | M1A1 | Using coordinates in an integrated expression. Co. | [4]
(ii) Normal has $m = -\frac{1}{2}$,
Perpendicular = 2
$\frac{16}{x^3} = 2 \rightarrow x = 2$ | M1 DM1 | Use of $m_1m_2=-1$ Link with $\frac{dy}{dx} \rightarrow$ value for $x$
$y = 10$
Eqn of normal $y - 10 = -\frac{1}{2}(x-2)$
ie $2y + x = 22$ | M1 A1 | Using calculated $x$ and $y$ to find equation. Allow for correct 3 term linear equation | [4]
(iii) $A = \int(\frac{-8}{x^2}+12)dx$ | M1 | Attempt at integration of $y$ from (i) even if linear
$= \frac{8}{x} + 12x$ | A1√ | Allow ft, providing original has two terms, one of which is a -ve power
$= [\ ]$ at 2 $- [\ ]$ at 1 | M1 | Correct use of correct limits in some integrated expression.
$\rightarrow 8$ | A1 | [4]10 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 16 } { x ^ { 3 } }$, and $( 1,4 )$ is a point on the curve.\\
(i) Find the equation of the curve.\\
(ii) A line with gradient $- \frac { 1 } { 2 }$ is a normal to the curve. Find the equation of this normal, giving your answer in the form $a x + b y = c$.\\
(iii) 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 2005 Q10 [12]}}