Edexcel P1 2020 January — Question 9 6 marks

Exam BoardEdexcel
ModuleP1 (Pure Mathematics 1)
Year2020
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind stationary points coordinates
DifficultyStandard +0.3 This is a straightforward stationary point question requiring quotient rule (or rewriting as a product), setting derivative to zero, and solving. While it involves multiple steps and algebraic manipulation with surds, it follows a standard template with no conceptual surprises—slightly easier than average for A-level but not trivial due to the surd manipulation required.
Spec1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives

9. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. A curve has equation $$y = \frac { 4 x ^ { 2 } + 9 } { 2 \sqrt { x } } \quad x > 0$$ Find the \(x\) coordinate of the point on the curve at which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\)

Question 9:
AnswerMarks Guidance
AnswerMarks Guidance
\(\dfrac{4x^2 + 9}{2\sqrt{x}} = \dfrac{4x^2}{2\sqrt{x}} + \dfrac{9}{2\sqrt{x}} = 2x^{\frac{3}{2}} + \dfrac{9}{2}x^{-\frac{1}{2}}\)M1 Attempts to divide by \(2\sqrt{x}\). Award for one correct term including \(\dfrac{9}{2x^{\frac{1}{2}}}\). Allow if they combine with common denominator of 2, but indices must be processed.
\(2x^{\frac{3}{2}} + \dfrac{9}{2}x^{-\frac{1}{2}}\)A1 May be left unsimplified but indices must be processed
\(\dfrac{dy}{dx} = 3x^{\frac{1}{2}} - \dfrac{9}{4}x^{-\frac{3}{2}}\)M1 Attempts to differentiate. Award for one power decreasing by one, indices must be processed. Cannot award if just differentiating top and bottom of fraction separately.
\(\dfrac{dy}{dx} = 3x^{\frac{1}{2}} - \dfrac{9}{4}x^{-\frac{3}{2}}\)A1 May be left unsimplified but indices must be processed
\(\dfrac{dy}{dx} = 0 \Rightarrow x^2 = \dfrac{3}{4} \Rightarrow x = \dfrac{\sqrt{3}}{2}\)M1 Sets \(\dfrac{dy}{dx} = 0\) and proceeds to \(x^{\pm 2} = \ldots\) or \(x^{\pm 4} = \ldots\) following a derivative of form \(Ax^{\frac{1}{2}} - Bx^{-\frac{3}{2}},\ A,B > 0\)
\(x = \dfrac{\sqrt{3}}{2}\)A1 Or exact equivalent cso. A correct exact answer can imply final M1A1. Rounded answer with no working (e.g. awrt 0.87) is M0A0. Withhold final mark if \(-\dfrac{\sqrt{3}}{2}\) is not rejected.
## Question 9:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{4x^2 + 9}{2\sqrt{x}} = \dfrac{4x^2}{2\sqrt{x}} + \dfrac{9}{2\sqrt{x}} = 2x^{\frac{3}{2}} + \dfrac{9}{2}x^{-\frac{1}{2}}$ | M1 | Attempts to divide by $2\sqrt{x}$. Award for one correct term including $\dfrac{9}{2x^{\frac{1}{2}}}$. Allow if they combine with common denominator of 2, but indices must be processed. |
| $2x^{\frac{3}{2}} + \dfrac{9}{2}x^{-\frac{1}{2}}$ | A1 | May be left unsimplified but indices must be processed |
| $\dfrac{dy}{dx} = 3x^{\frac{1}{2}} - \dfrac{9}{4}x^{-\frac{3}{2}}$ | M1 | Attempts to differentiate. Award for one power decreasing by one, indices must be processed. Cannot award if just differentiating top and bottom of fraction separately. |
| $\dfrac{dy}{dx} = 3x^{\frac{1}{2}} - \dfrac{9}{4}x^{-\frac{3}{2}}$ | A1 | May be left unsimplified but indices must be processed |
| $\dfrac{dy}{dx} = 0 \Rightarrow x^2 = \dfrac{3}{4} \Rightarrow x = \dfrac{\sqrt{3}}{2}$ | M1 | Sets $\dfrac{dy}{dx} = 0$ and proceeds to $x^{\pm 2} = \ldots$ or $x^{\pm 4} = \ldots$ following a derivative of form $Ax^{\frac{1}{2}} - Bx^{-\frac{3}{2}},\ A,B > 0$ |
| $x = \dfrac{\sqrt{3}}{2}$ | A1 | Or exact equivalent cso. A correct exact answer can imply final M1A1. Rounded answer with no working (e.g. awrt 0.87) is M0A0. Withhold final mark if $-\dfrac{\sqrt{3}}{2}$ is not rejected. |
9. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

A curve has equation

$$y = \frac { 4 x ^ { 2 } + 9 } { 2 \sqrt { x } } \quad x > 0$$

Find the $x$ coordinate of the point on the curve at which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$\\

\hfill \mbox{\textit{Edexcel P1 2020 Q9 [6]}}