| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 This is a straightforward quotient rule application with basic algebraic simplification. The question requires applying the quotient rule to a rational function involving √x, then simplifying to match the given form. While it involves multiple steps and careful algebra, it's a standard technique with no conceptual challenges—slightly easier than average for A-level. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01d Proof by contradiction |
| VIIV SIHI NI III HM IONOO | VIAV SIHI NI III M M I O N OO | VIUV SIHI NI IIIUM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{x-4}{2+\sqrt{x}} \Rightarrow \frac{dy}{dx} = \frac{2+\sqrt{x} - (x-4)\frac{1}{2}x^{-\frac{1}{2}}}{(2+\sqrt{x})^2}\) | M1, A1 | M1: Correct rule (quotient or product & chain) to achieve \(\frac{\alpha(2+\sqrt{x}) - \beta(x-4)x^{-\frac{1}{2}}}{(2+\sqrt{x})^2}\); A1: Correct derivative in any form, in terms of single variable |
| \(= \frac{2+\sqrt{x} - \frac{1}{2}\sqrt{x} + 2x^{-\frac{1}{2}}}{(2+\sqrt{x})^2} = \frac{2\sqrt{x} + \frac{1}{2}x + 2}{\sqrt{x}(2+\sqrt{x})^2}\) | M1 | Multiply numerator and denominator by \(\sqrt{x}\) and collect terms into single fraction |
| \(= \frac{x + 4\sqrt{x} + 4}{2\sqrt{x}(2+\sqrt{x})^2} = \frac{(2+\sqrt{x})^2}{2\sqrt{x}(2+\sqrt{x})^2} = \frac{1}{2\sqrt{x}}\) | A1 | Correct expression showing all key steps; \(\frac{dy}{dx}\) must be seen at least once |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{x-4}{2+\sqrt{x}} \Rightarrow y = \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{2+\sqrt{x}} = \sqrt{x} - 2\) | M1, A1 | M1: Uses difference of two squares; A1: \(y = \sqrt{x}-2\) or \(y = t-2\) |
| \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\) | M1, A1 | M1: Differentiates expression of form \(y = \sqrt{x} + b\); A1: Correct with \(\frac{dy}{dx}\) seen at least once |
# Question 14:
## Method 1 (Quotient Rule):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{x-4}{2+\sqrt{x}} \Rightarrow \frac{dy}{dx} = \frac{2+\sqrt{x} - (x-4)\frac{1}{2}x^{-\frac{1}{2}}}{(2+\sqrt{x})^2}$ | M1, A1 | M1: Correct rule (quotient or product & chain) to achieve $\frac{\alpha(2+\sqrt{x}) - \beta(x-4)x^{-\frac{1}{2}}}{(2+\sqrt{x})^2}$; A1: Correct derivative in any form, in terms of single variable |
| $= \frac{2+\sqrt{x} - \frac{1}{2}\sqrt{x} + 2x^{-\frac{1}{2}}}{(2+\sqrt{x})^2} = \frac{2\sqrt{x} + \frac{1}{2}x + 2}{\sqrt{x}(2+\sqrt{x})^2}$ | M1 | Multiply numerator and denominator by $\sqrt{x}$ and collect terms into single fraction |
| $= \frac{x + 4\sqrt{x} + 4}{2\sqrt{x}(2+\sqrt{x})^2} = \frac{(2+\sqrt{x})^2}{2\sqrt{x}(2+\sqrt{x})^2} = \frac{1}{2\sqrt{x}}$ | A1 | Correct expression showing all key steps; $\frac{dy}{dx}$ must be seen at least once |
## Method 2 (Simplification first):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{x-4}{2+\sqrt{x}} \Rightarrow y = \frac{(\sqrt{x}+2)(\sqrt{x}-2)}{2+\sqrt{x}} = \sqrt{x} - 2$ | M1, A1 | M1: Uses difference of two squares; A1: $y = \sqrt{x}-2$ or $y = t-2$ |
| $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ | M1, A1 | M1: Differentiates expression of form $y = \sqrt{x} + b$; A1: Correct with $\frac{dy}{dx}$ seen at least once |\begin{enumerate}
\item Given that
\end{enumerate}
$$y = \frac { x - 4 } { 2 + \sqrt { x } } \quad x > 0$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \mathrm {~A} \sqrt { \mathrm { x } } } \quad x > 0$$
where $A$ is a constant to be found.
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VIIV SIHI NI III HM IONOO & VIAV SIHI NI III M M I O N OO & VIUV SIHI NI IIIUM ION OC \\
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\hfill \mbox{\textit{Edexcel Paper 1 2021 Q14 [4]}}