| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about x=1: direct function expansion |
| Difficulty | Standard +0.3 This is a structured, multi-part question that guides students through Taylor series expansion with clear scaffolding. Parts (a) and (b) involve routine differentiation of a composite function, part (c) applies the given formula mechanically by evaluating derivatives at x=1, and part (d) uses the series to evaluate a limit—a standard application. While it requires careful algebraic manipulation and understanding of Taylor series, the question provides the formula and breaks the problem into manageable steps with no novel insight required. It's slightly easier than average for Further Maths FP1. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = (1+\ln x)^2 \Rightarrow \frac{dy}{dx} = k(1+\ln x)\times\frac{1}{x}\) or \(y = 1 + 2\ln x + (\ln x)^2 \Rightarrow \frac{dy}{dx} = \frac{A}{x} + B\ln x \times\frac{1}{x}\) | M1 | Attempts first derivative including use of chain rule; may expand first |
| \(\frac{dy}{dx} = 2(1+\ln x)\times\frac{1}{x}\) or \(\frac{dy}{dx} = \frac{2}{x} + 2\ln x\times\frac{1}{x}\) | A1 | Correct first derivative, need not be simplified |
| \(\frac{d^2y}{dx^2} = \frac{k\left(\frac{1}{x}\right)\times x - k(1+\ln x)\times 1}{x^2}\) or \(\frac{k}{x}x^{-1} + k(1+\ln x)(-x^{-2})\) or \(\frac{d^2y}{dx^2} = -\frac{A}{x^2} + \frac{\frac{B}{x}\times x - 2\ln x \times 1}{x^2}\) | M1 | Attempts second derivative using quotient or product rule |
| \(\frac{d^2y}{dx^2} = \frac{\left(\frac{2}{x}\right)\times x - 2(1+\ln x)}{x^2}\) leading to \(\frac{d^2y}{dx^2} = -\frac{2\ln x}{x^2}\) | A1* | Correct result achieved from correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^3y}{dx^3} = \frac{\pm\frac{C}{x}\times x^2 \pm Dx\ln x}{x^4}\) or \(\frac{d^3y}{dx^3} = (-2\ln x)(-Cx^{-3}) + \left(-\frac{D}{x}\right)(x^{-2})\) | M1 | Applies quotient or product rule; if formula quoted it must be correct |
| \(\frac{d^3y}{dx^3} = -\frac{\frac{2}{x}\times x^2 - 4x\ln x}{x^4}\) or \(-\frac{2}{x^3} + \frac{4\ln x}{x^3}\) | A1 | Correct third derivative, any form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y(1)=1,\ y'(1)=2,\ y''(1)=0,\ y'''(1)=-2\) | M1 | Find value of derivatives at \(x=1\) |
| \([y] = 1 + 2(x-1) + \frac{0}{2!}(x-1)^2 + \frac{-2}{3!}(x-1)^3 + ...\) | M1 | Applies Taylor series expansion |
| \([y] = 1 + 2(x-1) - \frac{1}{3}(x-1)^3 + ...\) or \([y] = -1 + 2x - \frac{1}{3}(x-1)^3 + ...\) | A1 | Correct series, may be unsimplified; must be using correct third derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2x-1-(1+\ln x)^2}{(x-1)^3} = \frac{2x-1-1-2(x-1)+\frac{1}{3}(x-1)^3+...}{(x-1)^3} = \frac{\frac{1}{3}(x-1)^3+...}{(x-1)^3}\) | M1 | Applies series to the limit and cancels terms in numerator to leave term in \((x-1)^3\) and above |
| \(\frac{\frac{1}{3}(x-1)^3+...}{(x-1)^3} = \frac{1}{3}\) | M1 | Simplifies and realises \((x-1)^3\) cancels to achieve a constant |
| Hence \(\lim_{x\to1}\frac{2x-1-(1+\ln x)^2}{(x-1)^3} = \frac{1}{3}\), as all remaining terms are zero in the limit as they are multiples of \((x-1)^k\), which tends to 0 | A1 | Correct limit deduced with reasoning given why remaining terms disappear |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = (1+\ln x)^2 \Rightarrow \frac{dy}{dx} = k(1+\ln x)\times\frac{1}{x}$ or $y = 1 + 2\ln x + (\ln x)^2 \Rightarrow \frac{dy}{dx} = \frac{A}{x} + B\ln x \times\frac{1}{x}$ | M1 | Attempts first derivative including use of chain rule; may expand first |
| $\frac{dy}{dx} = 2(1+\ln x)\times\frac{1}{x}$ or $\frac{dy}{dx} = \frac{2}{x} + 2\ln x\times\frac{1}{x}$ | A1 | Correct first derivative, need not be simplified |
| $\frac{d^2y}{dx^2} = \frac{k\left(\frac{1}{x}\right)\times x - k(1+\ln x)\times 1}{x^2}$ or $\frac{k}{x}x^{-1} + k(1+\ln x)(-x^{-2})$ or $\frac{d^2y}{dx^2} = -\frac{A}{x^2} + \frac{\frac{B}{x}\times x - 2\ln x \times 1}{x^2}$ | M1 | Attempts second derivative using quotient or product rule |
| $\frac{d^2y}{dx^2} = \frac{\left(\frac{2}{x}\right)\times x - 2(1+\ln x)}{x^2}$ leading to $\frac{d^2y}{dx^2} = -\frac{2\ln x}{x^2}$ | A1* | Correct result achieved from correct work |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^3y}{dx^3} = \frac{\pm\frac{C}{x}\times x^2 \pm Dx\ln x}{x^4}$ or $\frac{d^3y}{dx^3} = (-2\ln x)(-Cx^{-3}) + \left(-\frac{D}{x}\right)(x^{-2})$ | M1 | Applies quotient or product rule; if formula quoted it must be correct |
| $\frac{d^3y}{dx^3} = -\frac{\frac{2}{x}\times x^2 - 4x\ln x}{x^4}$ or $-\frac{2}{x^3} + \frac{4\ln x}{x^3}$ | A1 | Correct third derivative, any form |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y(1)=1,\ y'(1)=2,\ y''(1)=0,\ y'''(1)=-2$ | M1 | Find value of derivatives at $x=1$ |
| $[y] = 1 + 2(x-1) + \frac{0}{2!}(x-1)^2 + \frac{-2}{3!}(x-1)^3 + ...$ | M1 | Applies Taylor series expansion |
| $[y] = 1 + 2(x-1) - \frac{1}{3}(x-1)^3 + ...$ or $[y] = -1 + 2x - \frac{1}{3}(x-1)^3 + ...$ | A1 | Correct series, may be unsimplified; must be using correct third derivative |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2x-1-(1+\ln x)^2}{(x-1)^3} = \frac{2x-1-1-2(x-1)+\frac{1}{3}(x-1)^3+...}{(x-1)^3} = \frac{\frac{1}{3}(x-1)^3+...}{(x-1)^3}$ | M1 | Applies series to the limit and cancels terms in numerator to leave term in $(x-1)^3$ and above |
| $\frac{\frac{1}{3}(x-1)^3+...}{(x-1)^3} = \frac{1}{3}$ | M1 | Simplifies and realises $(x-1)^3$ cancels to achieve a constant |
| Hence $\lim_{x\to1}\frac{2x-1-(1+\ln x)^2}{(x-1)^3} = \frac{1}{3}$, as all remaining terms are zero in the limit as they are multiples of $(x-1)^k$, which tends to 0 | A1 | Correct limit deduced with reasoning given why remaining terms disappear |
---\begin{enumerate}
\item The Taylor series expansion of $f ( x )$ about $x = a$ is given by
\end{enumerate}
$$f ( x ) = f ( a ) + ( x - a ) f ^ { \prime } ( a ) + \frac { ( x - a ) ^ { 2 } } { 2 ! } f ^ { \prime \prime } ( a ) + \ldots + \frac { ( x - a ) ^ { r } } { r ! } f ^ { ( r ) } ( a ) + \ldots$$
Given that
$$y = ( 1 + \ln x ) ^ { 2 } \quad x > 0$$
(a) show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \frac { 2 \ln x } { x ^ { 2 } }$\\
(b) Hence find $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$\\
(c) Determine the Taylor series expansion about $x = 1$ of
$$( 1 + \ln x ) ^ { 2 }$$
in ascending powers of ( $x - 1$ ), up to and including the term in $( x - 1 ) ^ { 3 }$\\
Give each coefficient in simplest form.\\
(d) Use this series expansion to evaluate
$$\lim _ { x \rightarrow 1 } \frac { 2 x - 1 - ( 1 + \ln x ) ^ { 2 } } { ( x - 1 ) ^ { 3 } }$$
explaining your reasoning clearly.
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\hfill \mbox{\textit{Edexcel FP1 2021 Q6 [12]}}