| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of tangent line |
| Difficulty | Moderate -0.8 This is a straightforward differentiation question requiring basic power rule (rewriting √x as x^{1/2}), substituting x=4 to find gradient, and using y-y₁=m(x-x₁) for the tangent. Part (c) requires showing dy/dx≠0, which is routine algebraic manipulation. Easier than average A-level as it's purely procedural with no problem-solving insight needed. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Writes \(\sqrt{x}\) as \(x^{\frac{1}{2}}\) | B1 | PI by derivative with \(kx^{-\frac{1}{2}}\) |
| \(y = \frac{x^2}{8} + 4x^{\frac{1}{2}}\) → differentiates with at least one term correct | M1 | |
| \(\frac{dy}{dx} = \frac{x}{4} + 2x^{-\frac{1}{2}}\) | A1 | Correct expression for \(\frac{dy}{dx}\); ACF ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Obtains gradient of 2; substitutes \(x = 4\) into their \(\frac{dy}{dx}\) | M1 | |
| \(\frac{dy}{dx} = 2\), \(y - 10 = 2(x-4)\) i.e. \(y = 2x + 2\) | A1 | Does not need to be fully simplified; ACF ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equates their \(\frac{dy}{dx}\) to zero: \(\frac{x}{4} + 2x^{-\frac{1}{2}} = 0\) | M1 | OE |
| As \(x > 0\), \(\frac{2}{\sqrt{x}} > 0\) and \(\frac{x}{4} > 0\), therefore \(\frac{x}{4} + \frac{2}{\sqrt{x}} > 0\); equation has no solutions so the curve has no stationary points | R1 | Completes reasoned argument by correctly manipulating equation to obtain \(x^{\frac{3}{2}} = -8\) or \(x^{\frac{1}{2}} = -2\) and states \(x=4\) is a solution and deduces \(\frac{x}{4} + \frac{2}{\sqrt{x}} = 2 \neq 0\); OR obtains \(x^2 = -8\sqrt{x}\) or \(x^{\frac{3}{2}} = -8\) or \(x^{\frac{1}{2}} = -2\) and deduces no solutions referencing \(\sqrt{x} > 0\) |
## Question 4:
**Part 4(a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Writes $\sqrt{x}$ as $x^{\frac{1}{2}}$ | B1 | PI by derivative with $kx^{-\frac{1}{2}}$ |
| $y = \frac{x^2}{8} + 4x^{\frac{1}{2}}$ → differentiates with at least one term correct | M1 | |
| $\frac{dy}{dx} = \frac{x}{4} + 2x^{-\frac{1}{2}}$ | A1 | Correct expression for $\frac{dy}{dx}$; ACF ISW |
**Part 4(b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Obtains gradient of 2; substitutes $x = 4$ into their $\frac{dy}{dx}$ | M1 | |
| $\frac{dy}{dx} = 2$, $y - 10 = 2(x-4)$ i.e. $y = 2x + 2$ | A1 | Does not need to be fully simplified; ACF ISW |
**Part 4(c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equates their $\frac{dy}{dx}$ to zero: $\frac{x}{4} + 2x^{-\frac{1}{2}} = 0$ | M1 | OE |
| As $x > 0$, $\frac{2}{\sqrt{x}} > 0$ and $\frac{x}{4} > 0$, therefore $\frac{x}{4} + \frac{2}{\sqrt{x}} > 0$; equation has no solutions so the curve has no stationary points | R1 | Completes reasoned argument by correctly manipulating equation to obtain $x^{\frac{3}{2}} = -8$ or $x^{\frac{1}{2}} = -2$ **and** states $x=4$ is a solution **and** deduces $\frac{x}{4} + \frac{2}{\sqrt{x}} = 2 \neq 0$; OR obtains $x^2 = -8\sqrt{x}$ or $x^{\frac{3}{2}} = -8$ or $x^{\frac{1}{2}} = -2$ **and** deduces no solutions referencing $\sqrt{x} > 0$ |
---4 A curve has equation
$$y = \frac { x ^ { 2 } } { 8 } + 4 \sqrt { x }$$
4
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$\\
4
\item The point $P$ with coordinates $( 4,10 )$ lies on the curve.\\
Find an equation of the tangent to the curve at the point $P$\\
□\\
4
\item Show that the curve has no stationary points.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q4 [7]}}