| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about x=1: direct function expansion |
| Difficulty | Standard +0.8 This is a Further Maths question requiring two derivatives of a composite function involving ln and square root, then applying Taylor series about x=1 (not the origin). The differentiation is moderately challenging with quotient/chain rules, and students must evaluate at x=1 and construct the series correctly. More demanding than standard C3/C4 work but routine for F2. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = \dfrac{1}{2}(4+\ln x)^{-\frac{1}{2}}\times\dfrac{1}{x}\) | M1 A1 | M1: Chain rule, look for \(\frac{K}{x}(4+\ln x)^{-\frac{1}{2}}\); A1: Correct derivative |
| \(\dfrac{d^2y}{dx^2} = \dfrac{1}{2}\cdot\dfrac{0-\left(\sqrt{4+\ln x}+x\times\frac{1}{2}(4+\ln x)^{-\frac{1}{2}}\times\frac{1}{x}\right)}{x^2(4+\ln x)}\) or \(-\dfrac{1}{4x}(4+\ln x)^{-\frac{3}{2}}\times\dfrac{1}{x}-\dfrac{1}{x^2}\times\dfrac{1}{2}(4+\ln x)^{-\frac{1}{2}}\) | M1 | Product or quotient rule and chain rule; correct form for their \(\frac{dy}{dx}\) up to slips in coefficients |
| \(= \dfrac{\ldots}{4x^2(4+\ln x)^{\frac{3}{2}}}\) | M1 | Attempts to simplify to correct denominator; must be correct work for second derivative |
| \(= -\dfrac{9+2\ln x}{4x^2(4+\ln x)^{\frac{3}{2}}}\) | A1* | Given answer; needs suitable intermediate line with formation of correct common denominator between two fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^2 = 4+\ln x \Rightarrow 2y\dfrac{dy}{dx} = \dfrac{1}{x}\) | M1 A1 | Implicit differentiation |
| \(\Rightarrow 2y\dfrac{d^2y}{dx^2}+2\left(\dfrac{dy}{dx}\right)^2 = -\dfrac{1}{x^2}\) | M1 | Differentiates again implicitly |
| \(\Rightarrow \dfrac{d^2y}{dx^2} = -\dfrac{1}{2yx^2}-\dfrac{2}{8x^2y^3} = \dfrac{-2y^2-1}{4x^2y^3}\) | M1 | Makes \(\frac{d^2y}{dx^2}\) subject, single fraction with denominator \(kx^2y^3\) |
| \(= -\dfrac{9+2\ln x}{4x^2(4+\ln x)^{\frac{3}{2}}}\) | A1* | Correct second derivative, no errors in working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y_{x=1}=2,\quad \left.\dfrac{dy}{dx}\right | _{x=1}=\dfrac{1}{4},\quad \left.\dfrac{d^2y}{dx^2}\right | _{x=1}=-\dfrac{9}{32}\) |
| \(y = 2+\dfrac{1}{4}(x-1)-\dfrac{1}{2!}\times\dfrac{9}{32}(x-1)^2+\ldots\) | M1 | Applies Taylor's theorem with their values |
| \(= 2+\dfrac{1}{4}(x-1)-\dfrac{9}{64}(x-1)^2+\ldots\) | A1 | Correct expression (don't be concerned if \(y=\) is missing) |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = \dfrac{1}{2}(4+\ln x)^{-\frac{1}{2}}\times\dfrac{1}{x}$ | M1 A1 | M1: Chain rule, look for $\frac{K}{x}(4+\ln x)^{-\frac{1}{2}}$; A1: Correct derivative |
| $\dfrac{d^2y}{dx^2} = \dfrac{1}{2}\cdot\dfrac{0-\left(\sqrt{4+\ln x}+x\times\frac{1}{2}(4+\ln x)^{-\frac{1}{2}}\times\frac{1}{x}\right)}{x^2(4+\ln x)}$ or $-\dfrac{1}{4x}(4+\ln x)^{-\frac{3}{2}}\times\dfrac{1}{x}-\dfrac{1}{x^2}\times\dfrac{1}{2}(4+\ln x)^{-\frac{1}{2}}$ | M1 | Product or quotient rule and chain rule; correct form for their $\frac{dy}{dx}$ up to slips in coefficients |
| $= \dfrac{\ldots}{4x^2(4+\ln x)^{\frac{3}{2}}}$ | M1 | Attempts to simplify to correct denominator; must be correct work for second derivative |
| $= -\dfrac{9+2\ln x}{4x^2(4+\ln x)^{\frac{3}{2}}}$ | A1* | Given answer; needs suitable intermediate line with formation of correct common denominator between two fractions |
## Alt(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2 = 4+\ln x \Rightarrow 2y\dfrac{dy}{dx} = \dfrac{1}{x}$ | M1 A1 | Implicit differentiation |
| $\Rightarrow 2y\dfrac{d^2y}{dx^2}+2\left(\dfrac{dy}{dx}\right)^2 = -\dfrac{1}{x^2}$ | M1 | Differentiates again implicitly |
| $\Rightarrow \dfrac{d^2y}{dx^2} = -\dfrac{1}{2yx^2}-\dfrac{2}{8x^2y^3} = \dfrac{-2y^2-1}{4x^2y^3}$ | M1 | Makes $\frac{d^2y}{dx^2}$ subject, single fraction with denominator $kx^2y^3$ |
| $= -\dfrac{9+2\ln x}{4x^2(4+\ln x)^{\frac{3}{2}}}$ | A1* | Correct second derivative, no errors in working |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y_{x=1}=2,\quad \left.\dfrac{dy}{dx}\right|_{x=1}=\dfrac{1}{4},\quad \left.\dfrac{d^2y}{dx^2}\right|_{x=1}=-\dfrac{9}{32}$ | M1 | Evaluates $y$, $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$ at $x=1$; if substitution not seen, accept stated values following attempts |
| $y = 2+\dfrac{1}{4}(x-1)-\dfrac{1}{2!}\times\dfrac{9}{32}(x-1)^2+\ldots$ | M1 | Applies Taylor's theorem with their values |
| $= 2+\dfrac{1}{4}(x-1)-\dfrac{9}{64}(x-1)^2+\ldots$ | A1 | Correct expression (don't be concerned if $y=$ is missing) |
---5.
$$y = \sqrt { 4 + \ln x } \quad x > \frac { 1 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \frac { 9 + 2 \ln x } { 4 x ^ { 2 } ( 4 + \ln x ) ^ { \frac { 3 } { 2 } } }$$
\item Hence, or otherwise, determine the Taylor series expansion about $x = 1$ for $y$, in ascending powers of ( $x - 1$ ), up to and including the term in $( x - 1 ) ^ { 2 }$, giving each coefficient in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2022 Q5 [8]}}