Question 1 - A-Level Maths

Edexcel C3 2009 January — Question 1 10 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2009
Session	January
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Product & Quotient Rules
Type	Find gradient at point
Difficulty	Moderate -0.3 This is a straightforward application of the product rule (part a) and quotient rule (part b) with no additional complications. Part (a) requires differentiating x² × (5x-1)^(1/2) using product rule plus chain rule, then substituting x=2. Part (b) is a direct quotient rule application. Both are standard textbook exercises testing basic technique with minimal problem-solving, making this slightly easier than average.
Spec	1.07q Product and quotient rules: differentiation 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

(a) Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point where $x = 2$ on the curve with equation

$$y = x ^ { 2 } \sqrt { } ( 5 x - 1 )$$ (b) Differentiate $\frac { \sin 2 x } { x ^ { 2 } }$ with respect to $x$.

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(a)

Answer	Marks	Guidance
$\frac{d}{dx}\left(\sqrt{5x-1}\right) = \frac{d}{dx}\left((5x-1)^{\frac{1}{2}}\right)$	M1 A1
$= 5 \times \frac{1}{2}(5x-1)^{-\frac{1}{2}}$	M1 A1ft
$\frac{dy}{dx} = 2x\sqrt{5x-1} + \frac{5}{2}x^2(5x-1)^{-\frac{1}{2}}$	M1 A1ft
At $x = 2$: $\frac{dy}{dx} = 4\sqrt{9} + \frac{10}{\sqrt{9}} = 12 + \frac{10}{3} = \frac{46}{3}$	M1, Accept awrt 15.3	A1

(b)

Answer	Marks	Guidance
$\frac{d}{dx}\left(\frac{\sin 2x}{x^2}\right) = \frac{2x^2\cos 2x - 2x\sin 2x}{x^4}$	M1, A1+A1	A1

Alternative to (b):

Answer	Marks	Guidance
$\frac{d}{dx}(\sin 2x \times x^{-2}) = 2\cos 2x \times x^{-2} + \sin 2x \times (-2)x^{-3}$	M1 A1 + A1
$= 2x^{-2}\cos 2x - 2x^{-3}\sin 2x$ (= $\frac{2\cos 2x}{x^2} - \frac{2\sin 2x}{x^3}$)	A1	(4)

**(a)**

$\frac{d}{dx}\left(\sqrt{5x-1}\right) = \frac{d}{dx}\left((5x-1)^{\frac{1}{2}}\right)$ | M1 A1

$= 5 \times \frac{1}{2}(5x-1)^{-\frac{1}{2}}$ | M1 A1ft

$\frac{dy}{dx} = 2x\sqrt{5x-1} + \frac{5}{2}x^2(5x-1)^{-\frac{1}{2}}$ | M1 A1ft

At $x = 2$: $\frac{dy}{dx} = 4\sqrt{9} + \frac{10}{\sqrt{9}} = 12 + \frac{10}{3} = \frac{46}{3}$ | M1, Accept awrt 15.3 | A1 | (6)

**(b)**

$\frac{d}{dx}\left(\frac{\sin 2x}{x^2}\right) = \frac{2x^2\cos 2x - 2x\sin 2x}{x^4}$ | M1, A1+A1 | A1 | (4) [10]

**Alternative to (b):**

$\frac{d}{dx}(\sin 2x \times x^{-2}) = 2\cos 2x \times x^{-2} + \sin 2x \times (-2)x^{-3}$ | M1 A1 + A1

$= 2x^{-2}\cos 2x - 2x^{-3}\sin 2x$ (= $\frac{2\cos 2x}{x^2} - \frac{2\sin 2x}{x^3}$) | A1 | (4)

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Show LaTeX source

\begin{enumerate}
  \item (a) Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point where $x = 2$ on the curve with equation
\end{enumerate}

$$y = x ^ { 2 } \sqrt { } ( 5 x - 1 )$$

(b) Differentiate $\frac { \sin 2 x } { x ^ { 2 } }$ with respect to $x$.\\

\hfill \mbox{\textit{Edexcel C3 2009 Q1 [10]}}

This paper (8 questions)

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Q1 10 Q2 7 Q3 6 Q4 6 Q5 10 Q6 13 Q7 11 Q8 12

Answer	Marks	Guidance
\(\frac{d}{dx}\left(\sqrt{5x-1}\right) = \frac{d}{dx}\left((5x-1)^{\frac{1}{2}}\right)\)	M1 A1
\(= 5 \times \frac{1}{2}(5x-1)^{-\frac{1}{2}}\)	M1 A1ft
\(\frac{dy}{dx} = 2x\sqrt{5x-1} + \frac{5}{2}x^2(5x-1)^{-\frac{1}{2}}\)	M1 A1ft
At \(x = 2\): \(\frac{dy}{dx} = 4\sqrt{9} + \frac{10}{\sqrt{9}} = 12 + \frac{10}{3} = \frac{46}{3}\)	M1, Accept awrt 15.3	A1

Answer	Marks	Guidance
\(\frac{d}{dx}(\sin 2x \times x^{-2}) = 2\cos 2x \times x^{-2} + \sin 2x \times (-2)x^{-3}\)	M1 A1 + A1
\(= 2x^{-2}\cos 2x - 2x^{-3}\sin 2x\) (= \(\frac{2\cos 2x}{x^2} - \frac{2\sin 2x}{x^3}\))	A1	(4)