Question 7 - A-Level Maths

Edexcel C3 Specimen — Question 7 14 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Session	Specimen
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Differentiating Transcendental Functions
Type	Find gradient at a point - direct evaluation
Difficulty	Moderate -0.3 This is a straightforward multi-part differentiation question testing standard C3 techniques: basic differentiation of trig functions, implicit differentiation with chain rule, and product rule with a 'show that' format. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average.
Spec	1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07s Parametric and implicit differentiation

7. (i) Given that \(y = \tan x + 2 \cos x\), find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = \frac { \pi } { 4 }\).
(ii) Given that \(x = \tan \frac { 1 } { 2 } y\), prove that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { 1 + x ^ { 2 } }\).
(iii) Given that \(y = \mathrm { e } ^ { - x } \sin 2 x\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be expressed in the form \(R \mathrm { e } ^ { - x } \cos ( 2 x + \alpha )\). Find, to 3 significant figures, the values of \(R\) and \(\alpha\), where \(0 < \alpha < \frac { \pi } { 2 }\).

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Question 7:

Part (i):

Answer	Marks	Guidance
Answer/Working	Marks	Notes
\(\frac{dy}{dx} = \sec^2 x - 2\sin x\)	B1 B1
When \(x = \frac{1}{4}\pi\), \(\frac{dy}{dx} = 2 - \sqrt{2}\)	B1	(3)

Part (ii):

Answer	Marks	Guidance
Answer/Working	Marks	Notes
\(\frac{dx}{dy} = \frac{1}{2}\sec^2\frac{1}{2}y\)	B1
\(\frac{dy}{dx} = \frac{2}{\sec^2\!\left(\frac{y}{2}\right)} = \frac{2}{1+\tan^2\!\left(\frac{y}{2}\right)} = \frac{2}{1+x^2}\)	M1 M1 A1	(4)

Part (iii):

Answer	Marks	Guidance
Answer/Working	Marks	Notes
\(\frac{dy}{dx} = 2e^{-x}\cos 2x - e^{-x}\sin 2x = e^{-x}(2\cos 2x - \sin 2x)\)	M1 A1 A1
Method for \(R\): \(R = 2.24\) (allow \(\sqrt{5}\))	M1 A1
Method for \(\alpha\): \(\alpha = 0.464\)	M1 A1	(7)

## Question 7:

### Part (i):

| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dy}{dx} = \sec^2 x - 2\sin x$ | B1 B1 | |
| When $x = \frac{1}{4}\pi$, $\frac{dy}{dx} = 2 - \sqrt{2}$ | B1 | **(3)** |

### Part (ii):

| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dx}{dy} = \frac{1}{2}\sec^2\frac{1}{2}y$ | B1 | |
| $\frac{dy}{dx} = \frac{2}{\sec^2\!\left(\frac{y}{2}\right)} = \frac{2}{1+\tan^2\!\left(\frac{y}{2}\right)} = \frac{2}{1+x^2}$ | M1 M1 A1 | **(4)** |

### Part (iii):

| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dy}{dx} = 2e^{-x}\cos 2x - e^{-x}\sin 2x = e^{-x}(2\cos 2x - \sin 2x)$ | M1 A1 A1 | |
| Method for $R$: $R = 2.24$ (allow $\sqrt{5}$) | M1 A1 | |
| Method for $\alpha$: $\alpha = 0.464$ | M1 A1 | **(7)** |

Show LaTeX source

7. (i) Given that $y = \tan x + 2 \cos x$, find the exact value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $x = \frac { \pi } { 4 }$.\\
(ii) Given that $x = \tan \frac { 1 } { 2 } y$, prove that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { 1 + x ^ { 2 } }$.\\
(iii) Given that $y = \mathrm { e } ^ { - x } \sin 2 x$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x }$ can be expressed in the form $R \mathrm { e } ^ { - x } \cos ( 2 x + \alpha )$. Find, to 3 significant figures, the values of $R$ and $\alpha$, where $0 < \alpha < \frac { \pi } { 2 }$.

\hfill \mbox{\textit{Edexcel C3  Q7 [14]}}

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