| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Related rates with substitution |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining basic integration (with simple substitution u=4x+1), standard related rates using the chain rule (dy/dt = dy/dx × dx/dt), and algebraic manipulation to show a product is constant. All techniques are routine for P1 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \dfrac{2}{3}(4x+1)^{\frac{3}{2}} \div 4\ (+C) \left(= \dfrac{(4x+1)^{\frac{3}{2}}}{6}\right)\) | B1 B1 | B1 without \(\div 4\). B1 for \(\div 4\) oe. Unsimplified OK |
| Uses \(x = 2\), \(y = 5\) | M1 | Uses \((2, 5)\) in an integral (indicated by an increase in power by 1) |
| \(\rightarrow c = \frac{1}{2}\) oe isw | A1 | No isw if candidate now goes on to produce a straight line equation |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}\) | ||
| \(\dfrac{dx}{dt} = 0.06 \div 3\) | M1 | Ignore notation. Must be \(0.06 \div 3\) for M1 |
| \(= 0.02\) oe | A1 | Correct answer with no working scores 2/2 |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{d^2y}{dx^2} = \frac{1}{2}(4x+1)^{-\frac{1}{2}} \times 4\) | B1 | |
| \(\dfrac{d^2y}{dx^2} \times \dfrac{dy}{dx} = \dfrac{2}{\sqrt{4x+1}} \times \sqrt{4x+1}\ (= 2)\) | B1FT | Must either show the algebraic product and state that it results in a constant or evaluate it as \('= 2'\). Must not evaluate at \(x = 2\). ft to apply only if \(\dfrac{d^2y}{dx^2}\) is of the form \(k(4x+1)^{-\frac{1}{2}}\) |
| Total | 2 |
## Question 9:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \dfrac{2}{3}(4x+1)^{\frac{3}{2}} \div 4\ (+C) \left(= \dfrac{(4x+1)^{\frac{3}{2}}}{6}\right)$ | B1 B1 | B1 without $\div 4$. B1 for $\div 4$ oe. Unsimplified OK |
| Uses $x = 2$, $y = 5$ | M1 | Uses $(2, 5)$ in an integral (indicated by an increase in power by 1) |
| $\rightarrow c = \frac{1}{2}$ oe isw | A1 | No isw if candidate now goes on to produce a straight line equation |
| **Total** | **4** | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$ | | |
| $\dfrac{dx}{dt} = 0.06 \div 3$ | M1 | Ignore notation. Must be $0.06 \div 3$ for M1 |
| $= 0.02$ oe | A1 | Correct answer with no working scores 2/2 |
| **Total** | **2** | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{d^2y}{dx^2} = \frac{1}{2}(4x+1)^{-\frac{1}{2}} \times 4$ | B1 | |
| $\dfrac{d^2y}{dx^2} \times \dfrac{dy}{dx} = \dfrac{2}{\sqrt{4x+1}} \times \sqrt{4x+1}\ (= 2)$ | B1FT | Must either show the algebraic product and state that it results in a constant or evaluate it as $'= 2'$. Must not evaluate at $x = 2$. ft to apply only if $\dfrac{d^2y}{dx^2}$ is of the form $k(4x+1)^{-\frac{1}{2}}$ |
| **Total** | **2** | |
---9 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { } ( 4 x + 1 )$ and $( 2,5 )$ is a point on the curve.\\
(i) Find the equation of the curve.\\
(ii) A point $P$ moves along the curve in such a way that the $y$-coordinate is increasing at a constant rate of 0.06 units per second. Find the rate of change of the $x$-coordinate when $P$ passes through $( 2,5 )$.\\
(iii) Show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \times \frac { \mathrm { d } y } { \mathrm {~d} x }$ is constant.\\
\hfill \mbox{\textit{CAIE P1 2018 Q9 [8]}}