| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a straightforward application of the chain rule to find derivatives, followed by standard stationary point analysis. The chain rule application is routine (differentiating a linear function and a square root), and finding where dy/dx = 0 leads to a simple equation. The second derivative test is mechanical. Slightly easier than average due to the simple algebraic structure and standard procedure. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \{3\} + \left\{-4 \times \frac{1}{2}(3x+1)^{-\frac{1}{2}} \times 3\right\} = 3 - 6(3x+1)^{-\frac{1}{2}}\) | B1 B1 | Correct differentiation of \(3x+1\) and no other terms and correct differentiation of \(-4(3x+1)^{\frac{1}{2}}\). Accept unsimplified. |
| \(\frac{d^2y}{dx^2} = -\frac{1}{2} \times -6(3x+1)^{-\frac{3}{2}} \times 3\ [= 9(3x+1)^{-\frac{3}{2}}]\) | B1 | WWW. Accept unsimplified. Do not award if \(\frac{dy}{dx}\) is incorrect. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 0\) leading to \(3 - 6(3x+1)^{-\frac{1}{2}} = 0\) | M1 | Setting *their* \(\frac{dy}{dx} = 0\) |
| \((3x+1)^{\frac{1}{2}} = 2 \Rightarrow 3x+1 = 4\) leading to \(x = 1\) | A1 | CAO – do not ISW for a second answer |
| \(y = -4\) [coordinates \((1, -4)\)] | A1 | Condone inclusion of second value from a second answer |
| \(\frac{d^2y}{dx^2} = 9(3\times1+1)^{-\frac{3}{2}} = \frac{9}{8}\) or \(> 0\) so minimum | A1 | Some evidence of substitution needed but \(\frac{d^2y}{dx^2}\). Do not award if \(\frac{d^2y}{dx^2}\) is incorrect or wrongly evaluated. Accept correct consideration of gradients either side of \(x = 1\). |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \{3\} + \left\{-4 \times \frac{1}{2}(3x+1)^{-\frac{1}{2}} \times 3\right\} = 3 - 6(3x+1)^{-\frac{1}{2}}$ | B1 B1 | Correct differentiation of $3x+1$ and no other terms and correct differentiation of $-4(3x+1)^{\frac{1}{2}}$. Accept unsimplified. |
| $\frac{d^2y}{dx^2} = -\frac{1}{2} \times -6(3x+1)^{-\frac{3}{2}} \times 3\ [= 9(3x+1)^{-\frac{3}{2}}]$ | B1 | WWW. Accept unsimplified. Do not award if $\frac{dy}{dx}$ is incorrect. |
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## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 0$ leading to $3 - 6(3x+1)^{-\frac{1}{2}} = 0$ | M1 | Setting *their* $\frac{dy}{dx} = 0$ |
| $(3x+1)^{\frac{1}{2}} = 2 \Rightarrow 3x+1 = 4$ leading to $x = 1$ | A1 | CAO – do not ISW for a second answer |
| $y = -4$ [coordinates $(1, -4)$] | A1 | Condone inclusion of second value from a second answer |
| $\frac{d^2y}{dx^2} = 9(3\times1+1)^{-\frac{3}{2}} = \frac{9}{8}$ or $> 0$ so minimum | A1 | Some evidence of substitution needed but $\frac{d^2y}{dx^2}$. Do not award if $\frac{d^2y}{dx^2}$ is incorrect or wrongly evaluated. Accept correct consideration of gradients either side of $x = 1$. |
---9 The equation of a curve is $y = 3 x + 1 - 4 ( 3 x + 1 ) ^ { \frac { 1 } { 2 } }$ for $x > - \frac { 1 } { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Find the coordinates of the stationary point of the curve and determine its nature.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q9 [7]}}