Question 4 - A-Level Maths

Edexcel C3 2007 January — Question 4 11 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2007
Session	January
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Product & Quotient Rules
Type	Find stationary points coordinates
Difficulty	Standard +0.3 Part (i) requires quotient rule differentiation and solving a quadratic to find stationary points—standard C3 technique with straightforward algebra. Part (ii) applies chain rule to an exponential composite function with substitution—routine but requires careful execution. Both are textbook-style exercises with no novel insight needed.
Spec	1.06a Exponential function: a^x and e^x graphs and properties 1.07n Stationary points: find maxima, minima using derivatives 1.07q Product and quotient rules: differentiation 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

4. (i) The curve $C$ has equation $$y = \frac { x } { 9 + x ^ { 2 } }$$ Use calculus to find the coordinates of the turning points of $C$.
(ii) Given that $$y = \left( 1 + \mathrm { e } ^ { 2 x } \right) ^ { \frac { 3 } { 2 } }$$ find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $x = \frac { 1 } { 2 } \ln 3$.

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Answer	Marks	Guidance
(i) $\frac{dy}{dx} = \frac{(9+x^2) - x(2x)}{(9+x^2)^2} = \frac{9-x^2}{(9+x^2)^2}$; $\frac{dy}{dx} = 0 \Rightarrow 9 - x^2 = 0 \Rightarrow x = \pm 3$; $\left(3, \frac{1}{6}\right), \left(-3, -\frac{1}{6}\right)$	M1, A1, M1, A1, A1, A1	Final two A marks depend on second M only (6 marks)
(ii) $\frac{dy}{dx} = \frac{3}{2}(1+e^{2x})^{\frac{1}{2}} \times 2e^{2x}$; $x = \frac{1}{2}\ln 3 \Rightarrow \frac{dy}{dx} = \frac{3}{2}(1+e^{\ln 3})^{\frac{1}{2}} \times 2e^{\ln 3} = 3 \times 4^{\frac{1}{2}} \times 3 = 18$	M1, A1, A1, M1, A1	(5 marks)

(i) $\frac{dy}{dx} = \frac{(9+x^2) - x(2x)}{(9+x^2)^2} = \frac{9-x^2}{(9+x^2)^2}$; $\frac{dy}{dx} = 0 \Rightarrow 9 - x^2 = 0 \Rightarrow x = \pm 3$; $\left(3, \frac{1}{6}\right), \left(-3, -\frac{1}{6}\right)$ | M1, A1, M1, A1, A1, A1 | Final two A marks depend on second M only (6 marks)

(ii) $\frac{dy}{dx} = \frac{3}{2}(1+e^{2x})^{\frac{1}{2}} \times 2e^{2x}$; $x = \frac{1}{2}\ln 3 \Rightarrow \frac{dy}{dx} = \frac{3}{2}(1+e^{\ln 3})^{\frac{1}{2}} \times 2e^{\ln 3} = 3 \times 4^{\frac{1}{2}} \times 3 = 18$ | M1, A1, A1, M1, A1 | (5 marks)

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4. (i) The curve $C$ has equation

$$y = \frac { x } { 9 + x ^ { 2 } }$$

Use calculus to find the coordinates of the turning points of $C$.\\
(ii) Given that

$$y = \left( 1 + \mathrm { e } ^ { 2 x } \right) ^ { \frac { 3 } { 2 } }$$

find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $x = \frac { 1 } { 2 } \ln 3$.\\

\hfill \mbox{\textit{Edexcel C3 2007 Q4 [11]}}

This paper (8 questions)

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Q1 7 Q2 8 Q3 9 Q4 11 Q5 8 Q6 13 Q7 13 Q8 6

Answer	Marks	Guidance
(i) \(\frac{dy}{dx} = \frac{(9+x^2) - x(2x)}{(9+x^2)^2} = \frac{9-x^2}{(9+x^2)^2}\); \(\frac{dy}{dx} = 0 \Rightarrow 9 - x^2 = 0 \Rightarrow x = \pm 3\); \(\left(3, \frac{1}{6}\right), \left(-3, -\frac{1}{6}\right)\)	M1, A1, M1, A1, A1, A1	Final two A marks depend on second M only (6 marks)
(ii) \(\frac{dy}{dx} = \frac{3}{2}(1+e^{2x})^{\frac{1}{2}} \times 2e^{2x}\); \(x = \frac{1}{2}\ln 3 \Rightarrow \frac{dy}{dx} = \frac{3}{2}(1+e^{\ln 3})^{\frac{1}{2}} \times 2e^{\ln 3} = 3 \times 4^{\frac{1}{2}} \times 3 = 18\)	M1, A1, A1, M1, A1	(5 marks)