Edexcel F2 2015 June — Question 5 9 marks

Exam BoardEdexcel
ModuleF2 (Further Pure Mathematics 2)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeDifferentiation of reciprocal functions
DifficultyChallenging +1.2 This is a structured Further Maths question requiring differentiation of cot x, algebraic manipulation with trig identities, and Taylor series application. While it involves multiple steps and Further Maths content (making it harder than typical A-level), each part is guided and uses standard techniques: (a) differentiate twice using chain rule and Pythagorean identity, (b) differentiate the result from (a), (c) apply Taylor series formula with given derivatives. The scaffolding and routine nature of the calculus operations place it moderately above average difficulty.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07d Second derivatives: d^2y/dx^2 notation4.08a Maclaurin series: find series for function
  1. Given that \(y = \cot x\),
    1. show that
    $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \cot x + 2 \cot ^ { 3 } x$$
  2. Hence show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = p \cot ^ { 4 } x + q \cot ^ { 2 } x + r$$ where \(p , q\) and \(r\) are integers to be found.
  3. Find the Taylor series expansion of \(\cot x\) in ascending powers of \(\left( x - \frac { \pi } { 3 } \right)\) up to and including the term in \(\left( x - \frac { \pi } { 3 } \right) ^ { 3 }\).
Question 5:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dy}{dx} = -\cosec^2 x\)— Given
\(\frac{d^2y}{dx^2} = (-2\cosec x)(-\cosec x \cot x)\)M1A1 M1: Differentiates using chain rule or product/quotient rule; A1: Correct derivative
\(= 2\cosec^2 x \cot x = 2\cot x + 2\cot^3 x\) *A1cso* A1: Correct completion to printed answer; \(1 + \cot^2 x = \cosec^2 x\) or \(\cos^2 x + \sin^2 x = 1\) must be used; Full working must be shown
Alternative:
AnswerMarks Guidance
\(y = \frac{\cos x}{\sin x} \rightarrow \frac{dy}{dx} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x}\)— —
\(\frac{d^2y}{dx^2} = -(-2\sin^{-3}x\cos x) = \ldots\)M1A1 —
Correct completion to printed answerA1 See above
Total: (3)
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d^3y}{dx^3} = -2\cosec^2 x - 6\cot^2 x \cosec^2 x\)B1 Correct third derivative
\(= -2(1+\cot^2 x) - 6\cot^2 x(1+\cot^2 x)\)M1 Uses \(1 + \cot^2 x = \cosec^2 x\)
\(= -6\cot^4 x - 8\cot^2 x - 2\)A1 cso
Total: (3)
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f\!\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}},\ f'\!\left(\frac{\pi}{3}\right) = -\frac{4}{3},\ f''\!\left(\frac{\pi}{3}\right) = \frac{8}{3\sqrt{3}},\ f'''\!\left(\frac{\pi}{3}\right) = -\frac{16}{3}\)M1 Attempts all 4 values at \(\frac{\pi}{3}\); no working need be shown
\((y=)\frac{1}{\sqrt{3}} - \frac{4}{3}\!\left(x-\frac{\pi}{3}\right) + \frac{4}{3\sqrt{3}}\!\left(x-\frac{\pi}{3}\right)^2 - \frac{8}{9}\!\left(x-\frac{\pi}{3}\right)^3\)M1A1 M1: Correct application of Taylor using their values, must include up to \(\left(x-\frac{\pi}{3}\right)^3\) term; A1: Correct expression, must start \(y=\) or \(\cot x\); decimal equivalents allowed (min 3 sf)
Total: (3)
## Question 5:

**Part (a):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -\cosec^2 x$ | — | Given |
| $\frac{d^2y}{dx^2} = (-2\cosec x)(-\cosec x \cot x)$ | M1A1 | M1: Differentiates using chain rule or product/quotient rule; A1: Correct derivative |
| $= 2\cosec^2 x \cot x = 2\cot x + 2\cot^3 x$ * | A1cso* | A1: Correct completion to printed answer; $1 + \cot^2 x = \cosec^2 x$ or $\cos^2 x + \sin^2 x = 1$ must be used; Full working must be shown |

**Alternative:**
| $y = \frac{\cos x}{\sin x} \rightarrow \frac{dy}{dx} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x}$ | — | — |
| $\frac{d^2y}{dx^2} = -(-2\sin^{-3}x\cos x) = \ldots$ | M1A1 | — |
| Correct completion to printed answer | A1 | See above |

**Total: (3)**

---

**Part (b):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^3y}{dx^3} = -2\cosec^2 x - 6\cot^2 x \cosec^2 x$ | B1 | Correct third derivative |
| $= -2(1+\cot^2 x) - 6\cot^2 x(1+\cot^2 x)$ | M1 | Uses $1 + \cot^2 x = \cosec^2 x$ |
| $= -6\cot^4 x - 8\cot^2 x - 2$ | A1 | cso |

**Total: (3)**

---

**Part (c):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f\!\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}},\ f'\!\left(\frac{\pi}{3}\right) = -\frac{4}{3},\ f''\!\left(\frac{\pi}{3}\right) = \frac{8}{3\sqrt{3}},\ f'''\!\left(\frac{\pi}{3}\right) = -\frac{16}{3}$ | M1 | Attempts all 4 values at $\frac{\pi}{3}$; no working need be shown |
| $(y=)\frac{1}{\sqrt{3}} - \frac{4}{3}\!\left(x-\frac{\pi}{3}\right) + \frac{4}{3\sqrt{3}}\!\left(x-\frac{\pi}{3}\right)^2 - \frac{8}{9}\!\left(x-\frac{\pi}{3}\right)^3$ | M1A1 | M1: Correct application of Taylor using their values, must include up to $\left(x-\frac{\pi}{3}\right)^3$ term; A1: Correct expression, must start $y=$ or $\cot x$; decimal equivalents allowed (min 3 sf) |

**Total: (3)**

---
\begin{enumerate}
  \item Given that $y = \cot x$,\\
(a) show that
\end{enumerate}

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \cot x + 2 \cot ^ { 3 } x$$

(b) Hence show that

$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = p \cot ^ { 4 } x + q \cot ^ { 2 } x + r$$

where $p , q$ and $r$ are integers to be found.\\
(c) Find the Taylor series expansion of $\cot x$ in ascending powers of $\left( x - \frac { \pi } { 3 } \right)$ up to and including the term in $\left( x - \frac { \pi } { 3 } \right) ^ { 3 }$.

\hfill \mbox{\textit{Edexcel F2 2015 Q5 [9]}}
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