| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.3 This question requires setting a derivative equal to -1 to find a point, then integrating to find the original function using a boundary condition. While it involves multiple steps (solving an equation with fractional powers, integration with the reverse power rule, and substitution), these are all standard P1 techniques with no novel insight required. Slightly easier than average due to straightforward application of routine methods. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(3z - \dfrac{2}{z} = -1 \Rightarrow 3z^2 + z - 2 = 0\) | M1 | Express as 3-term quad. Accept \(x^{1/2}\) for \(z\); OR \(3x-1 = -\sqrt{x}\), \(9x^2-13x+4=0\) |
| \(x^{1/2}\) (or \(z\)) \(= 2/3\) or \(-1\) | A1 | |
| \(x = 4/9\) only | A1 [3] | M1, A1, A1 \(x = 4/9\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \dfrac{3x^{3/2}}{3/2} - \dfrac{2x^{1/2}}{1/2}\) \((+c)\) | B1B1 | |
| Sub \(x=4\), \(y=10\): \(10 = 16 - 8 + c \Rightarrow c = 2\) | M1A1 | \(c\) must be present |
| When \(x = \dfrac{4}{9}\), \(y = 2\!\left(\dfrac{4}{9}\right)^{3/2} - 4\!\left(\dfrac{4}{9}\right)^{1/2} + 2\) | M1 | Substituting \(x\) value from part (i) |
| \(-2/27\) | A1 [6] |
## Question 10:
### Part (i):
$3z - \dfrac{2}{z} = -1 \Rightarrow 3z^2 + z - 2 = 0$ | **M1** | Express as 3-term quad. Accept $x^{1/2}$ for $z$; OR $3x-1 = -\sqrt{x}$, $9x^2-13x+4=0$
$x^{1/2}$ (or $z$) $= 2/3$ or $-1$ | **A1** |
$x = 4/9$ only | **A1** [3] | **M1, A1, A1** $x = 4/9$
### Part (ii):
$f(x) = \dfrac{3x^{3/2}}{3/2} - \dfrac{2x^{1/2}}{1/2}$ $(+c)$ | **B1B1** |
Sub $x=4$, $y=10$: $10 = 16 - 8 + c \Rightarrow c = 2$ | **M1A1** | $c$ must be present
When $x = \dfrac{4}{9}$, $y = 2\!\left(\dfrac{4}{9}\right)^{3/2} - 4\!\left(\dfrac{4}{9}\right)^{1/2} + 2$ | **M1** | Substituting $x$ value from part (i)
$-2/27$ | **A1** [6] |
---10 A curve has equation $y = \mathrm { f } ( x )$ and it is given that $\mathrm { f } ^ { \prime } ( x ) = 3 x ^ { \frac { 1 } { 2 } } - 2 x ^ { - \frac { 1 } { 2 } }$. The point $A$ is the only point on the curve at which the gradient is - 1 .\\
(i) Find the $x$-coordinate of $A$.\\
(ii) Given that the curve also passes through the point $( 4,10 )$, find the $y$-coordinate of $A$, giving your answer as a fraction.
\hfill \mbox{\textit{CAIE P1 2016 Q10 [9]}}