| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Determine if function is increasing/decreasing |
| Difficulty | Moderate -0.3 This is a straightforward chain rule application with standard follow-up tasks. Part (i) requires basic chain rule differentiation; part (ii) involves finding a tangent at x=0 (routine substitution); part (iii) requires solving dy/dx > 0, leading to a cubic inequality. While multi-part, each step uses standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = \frac{1}{8} × 3 × (2x-3)^2 × 2 - 4\) | B2,1, B1 [3] | Everything but the "\(×2\)". For the "\(×2\)", even if B0 given above. |
| (ii) \(x = 0\), \(y = -\frac{27}{8}\), \(y + \frac{27}{6} = 5x \rightarrow 2y + 9 = 10x\) | B1, M1 A1 [3] | For correct \(y\) value. Must be using calculus for \(m\). co. (ok unsimplified). |
| (iii) \((2x - 3)^2 - 4 (> 0)\) \(\rightarrow x = 2\frac{1}{2}\) or \(\frac{1}{2}\) \(\rightarrow x > 2\frac{1}{2}\), \(x < \frac{1}{2}\). | M1, DM1, A1 [3] | Links \(\frac{dy}{dx}\) with 0. Method for quadratic – lead to 2 answers. Correct set of values. |
$y = \frac{1}{8}(2x-3)^3 - 4x$
(i) $\frac{dy}{dx} = \frac{1}{8} × 3 × (2x-3)^2 × 2 - 4$ | B2,1, B1 [3] | Everything but the "$×2$". For the "$×2$", even if B0 given above.
(ii) $x = 0$, $y = -\frac{27}{8}$, $y + \frac{27}{6} = 5x \rightarrow 2y + 9 = 10x$ | B1, M1 A1 [3] | For correct $y$ value. Must be using calculus for $m$. co. (ok unsimplified).
(iii) $(2x - 3)^2 - 4 (> 0)$ $\rightarrow x = 2\frac{1}{2}$ or $\frac{1}{2}$ $\rightarrow x > 2\frac{1}{2}$, $x < \frac{1}{2}$. | M1, DM1, A1 [3] | Links $\frac{dy}{dx}$ with 0. Method for quadratic – lead to 2 answers. Correct set of values.The equation of a curve is $y = \frac{1}{6}(2x - 3)^3 - 4x$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$. [3]
\item Find the equation of the tangent to the curve at the point where the curve intersects the $y$-axis. [3]
\item Find the set of values of $x$ for which $\frac{1}{6}(2x - 3)^3 - 4x$ is an increasing function of $x$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2010 Q10 [9]}}