Question 7 - A-Level Maths

Edexcel P3 2021 June — Question 7 5 marks

Exam Board	Edexcel
Module	P3 (Pure Mathematics 3)
Year	2021
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Implicit equations and differentiation
Type	Parametric form dy/dx
Difficulty	Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = 1/(dx/dy)), basic trigonometric differentiation, and algebraic manipulation to match the given form. While it involves multiple steps, each technique is standard for P3 level with no novel insight required, making it slightly easier than average.
Spec	1.05a Sine, cosine, tangent: definitions for all arguments 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.07s Parametric and implicit differentiation

7. Given that $$x = 6 \sin ^ { 2 } 2 y \quad 0 < y < \frac { \pi } { 4 }$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { A \sqrt { \left( B x - x ^ { 2 } \right) } }$$ where $A$ and $B$ are integers to be found.
WIHV SIHI NI III HM ION OC
VIAV SIHI NI III IM I ON OC
WARV SIHI NI IIIIM I I ON OC

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x = 6\sin^2 2y \Rightarrow \frac{dx}{dy} = 24\sin 2y \cos 2y$	M1 A1	Chain rule to achieve $k\sin 2y \cos 2y$; fully correct derivative
Attempts to use $\frac{dy}{dx} = 1 \div \frac{dx}{dy}$	M1	Condone slip on coefficient; do not condone slip on variable
Attempts to use both $\sin 2y = \sqrt{\frac{x}{6}}$ and $\cos 2y = \sqrt{1 - \frac{x}{6}}$	M1	Change fully to function of $x$; expect $\sin 2y = p\sqrt{x}$ and $\cos 2y = \sqrt{1-qx}$
$\frac{dy}{dx} = \frac{1}{24\sin 2y \cos 2y} = \frac{1}{24 \times \sqrt{\frac{x}{6}} \times \sqrt{1 - \frac{x}{6}}} = \frac{1}{4\sqrt{6x - x^2}}$	A1	CSO

Way 2:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x = 3 - 3\cos 4y \Rightarrow \frac{dx}{dy} = 12\sin 4y$	M1 A1	Differentiates to obtain $k\sin 4y$; fully correct
$\frac{dy}{dx} = 1 \div \frac{dx}{dy}$	M1
Uses $\sin 4y = \frac{\sqrt{6x - x^2}}{3}$	M1	Expect $\sin 4y = p\sqrt{6x-x^2}$ or equivalent
$\frac{dy}{dx} = \frac{1}{4\sqrt{6x-x^2}}$	A1	CSO

(5 marks)

## Question 7:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 6\sin^2 2y \Rightarrow \frac{dx}{dy} = 24\sin 2y \cos 2y$ | M1 A1 | Chain rule to achieve $k\sin 2y \cos 2y$; fully correct derivative |
| Attempts to use $\frac{dy}{dx} = 1 \div \frac{dx}{dy}$ | M1 | Condone slip on coefficient; do not condone slip on variable |
| Attempts to use both $\sin 2y = \sqrt{\frac{x}{6}}$ and $\cos 2y = \sqrt{1 - \frac{x}{6}}$ | M1 | Change fully to function of $x$; expect $\sin 2y = p\sqrt{x}$ and $\cos 2y = \sqrt{1-qx}$ |
| $\frac{dy}{dx} = \frac{1}{24\sin 2y \cos 2y} = \frac{1}{24 \times \sqrt{\frac{x}{6}} \times \sqrt{1 - \frac{x}{6}}} = \frac{1}{4\sqrt{6x - x^2}}$ | A1 | CSO |

**Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 3 - 3\cos 4y \Rightarrow \frac{dx}{dy} = 12\sin 4y$ | M1 A1 | Differentiates to obtain $k\sin 4y$; fully correct |
| $\frac{dy}{dx} = 1 \div \frac{dx}{dy}$ | M1 | |
| Uses $\sin 4y = \frac{\sqrt{6x - x^2}}{3}$ | M1 | Expect $\sin 4y = p\sqrt{6x-x^2}$ or equivalent |
| $\frac{dy}{dx} = \frac{1}{4\sqrt{6x-x^2}}$ | A1 | CSO |

**(5 marks)**

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7. Given that

$$x = 6 \sin ^ { 2 } 2 y \quad 0 < y < \frac { \pi } { 4 }$$

show that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { A \sqrt { \left( B x - x ^ { 2 } \right) } }$$

where $A$ and $B$ are integers to be found.\\

WIHV SIHI NI III HM ION OC\\
VIAV SIHI NI III IM I ON OC\\
WARV SIHI NI IIIIM I I ON OC

\hfill \mbox{\textit{Edexcel P3 2021 Q7 [5]}}

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