Question 10 - A-Level Maths

Pre-U Pre-U 9794/1 2018 June — Question 10 12 marks

Exam Board	Pre-U
Module	Pre-U 9794/1 (Pre-U Mathematics Paper 1)
Year	2018
Session	June
Marks	12
Topic	Differential equations
Type	Separable variables - partial fractions
Difficulty	Standard +0.8 This question requires separating variables, decomposing 1/(y-y³) into partial fractions (three terms), integrating both sides involving logarithms, then rearranging to make y the subject (requiring algebraic manipulation with exponentials). The limit analysis in part (ii) adds further challenge. While the techniques are A-level standard, the multi-step algebraic manipulation and making y explicit pushes this moderately above average difficulty.
Spec	1.02y Partial fractions: decompose rational functions 1.08k Separable differential equations: dy/dx = f(x)g(y)

Using partial fractions, find the general solution of the differential equation $$2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = y - y ^ { 3 } \text { for } 0 < y < 1$$ giving your solution in the form $y = \mathrm { f } ( x )$.
Determine $\lim _ { x \rightarrow - \infty } \mathrm { f } ( x )$ and $\lim _ { x \rightarrow + \infty } \mathrm { f } ( x )$.

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Question 10:

(i) $\int \frac{2}{y - y^3}\,\text{d}y = \int \text{d}x$ M1 (Some intention to separate variables)

$\frac{A}{y} + \frac{B}{1+y} + \frac{C}{1-y}$ or $\frac{A}{y} + \frac{By+C}{1-y^2}$ or $\pm\frac{A}{y} + \frac{By+C}{y^2-1}$ M1 (Set up partial fractions in a reasonably correct form)

For example $2, 2, 0$ or $1, 1, 0$ or $2, -1, 1$ or $1, \frac{1}{2}, -\frac{1}{2}$ A1 (Obtain at least one correct value)

[Correct triplet for their partial fraction format] A1

$2\ln y - \ln(1+y) - \ln(1-y) = x + c$

Or $2\ln y - \ln(1-y^2) = x + C$

Or $\ln y - \frac{1}{2}\ln(1-y^2) = \frac{1}{2}x + C$ M1 (Obtain $k\ln y \pm m\ln(1+y) \pm n\ln(1-y) = \pm px (+c)$)

$\frac{y^2}{(1+y)(1-y)} = \text{e}^x\text{e}^c$ M1 (Use at least one log law correctly)

[Approximately correct expression without logs] M1 (Allow a coefficient slip but not omission of $+c$)

$y = \sqrt{\frac{A\text{e}^x}{1+A\text{e}^x}}$ M1 (Obtain their reasonable expression in the form $y = \text{f}(x)$ by a reasonable method)

[Correct expression for $y$] A1 (Do not accept $\text{e}^{x+c}$ for this mark)

(ii) As $x$ becomes large and negative, $A\text{e}^x$ tends to $0$ and $\lim_{x\to-\infty} f(x) = 0$ B1 (Dep — requires a correct $y$)

Since $A\text{e}^x \neq 0$, $y = \sqrt{\dfrac{A\text{e}^x}{\dfrac{1}{A\text{e}^x} + \dfrac{A\text{e}^x}{A\text{e}^x}}}$ M1 (Dep — Accept minimal or no working)

Hence $\lim_{x\to+\infty} f(x) = 1$ A1

Total: 12 marks

**Question 10:**

**(i)** $\int \frac{2}{y - y^3}\,\text{d}y = \int \text{d}x$ **M1** (Some intention to separate variables)

$\frac{A}{y} + \frac{B}{1+y} + \frac{C}{1-y}$ or $\frac{A}{y} + \frac{By+C}{1-y^2}$ or $\pm\frac{A}{y} + \frac{By+C}{y^2-1}$ **M1** (Set up partial fractions in a reasonably correct form)

For example $2, 2, 0$ or $1, 1, 0$ or $2, -1, 1$ or $1, \frac{1}{2}, -\frac{1}{2}$ **A1** (Obtain at least one correct value)

[Correct triplet for their partial fraction format] **A1**

$2\ln y - \ln(1+y) - \ln(1-y) = x + c$
Or $2\ln y - \ln(1-y^2) = x + C$
Or $\ln y - \frac{1}{2}\ln(1-y^2) = \frac{1}{2}x + C$ **M1** (Obtain $k\ln y \pm m\ln(1+y) \pm n\ln(1-y) = \pm px (+c)$)

$\frac{y^2}{(1+y)(1-y)} = \text{e}^x\text{e}^c$ **M1** (Use at least one log law correctly)

[Approximately correct expression without logs] **M1** (Allow a coefficient slip but not omission of $+c$)

$y = \sqrt{\frac{A\text{e}^x}{1+A\text{e}^x}}$ **M1** (Obtain their reasonable expression in the form $y = \text{f}(x)$ by a reasonable method)

[Correct expression for $y$] **A1** (Do not accept $\text{e}^{x+c}$ for this mark)

**(ii)** As $x$ becomes large and negative, $A\text{e}^x$ tends to $0$ and $\lim_{x\to-\infty} f(x) = 0$ **B1** (Dep — requires a correct $y$)

Since $A\text{e}^x \neq 0$, $y = \sqrt{\dfrac{A\text{e}^x}{\dfrac{1}{A\text{e}^x} + \dfrac{A\text{e}^x}{A\text{e}^x}}}$ **M1** (Dep — Accept minimal or no working)

Hence $\lim_{x\to+\infty} f(x) = 1$ **A1**

**Total: 12 marks**

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10 (i) Using partial fractions, find the general solution of the differential equation

$$2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = y - y ^ { 3 } \text { for } 0 < y < 1$$

giving your solution in the form $y = \mathrm { f } ( x )$.\\
(ii) Determine $\lim _ { x \rightarrow - \infty } \mathrm { f } ( x )$ and $\lim _ { x \rightarrow + \infty } \mathrm { f } ( x )$.

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2018 Q10 [12]}}

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Q1 4 Q2 7 Q3 4 Q4 5 Q5 10 Q6 8 Q7 8 Q8 7 Q9 12 Q10 12

Pre-U Pre-U 9794/1 2018 June — Question 10 12 marks

Question 10:

(i) \(\int \frac{2}{y - y^3}\,\text{d}y = \int \text{d}x\) M1 (Some intention to separate variables)

\(\frac{A}{y} + \frac{B}{1+y} + \frac{C}{1-y}\) or \(\frac{A}{y} + \frac{By+C}{1-y^2}\) or \(\pm\frac{A}{y} + \frac{By+C}{y^2-1}\) M1 (Set up partial fractions in a reasonably correct form)

For example \(2, 2, 0\) or \(1, 1, 0\) or \(2, -1, 1\) or \(1, \frac{1}{2}, -\frac{1}{2}\) A1 (Obtain at least one correct value)

[Correct triplet for their partial fraction format] A1

\(2\ln y - \ln(1+y) - \ln(1-y) = x + c\)

Or \(2\ln y - \ln(1-y^2) = x + C\)

Or \(\ln y - \frac{1}{2}\ln(1-y^2) = \frac{1}{2}x + C\) M1 (Obtain \(k\ln y \pm m\ln(1+y) \pm n\ln(1-y) = \pm px (+c)\))

\(\frac{y^2}{(1+y)(1-y)} = \text{e}^x\text{e}^c\) M1 (Use at least one log law correctly)

[Approximately correct expression without logs] M1 (Allow a coefficient slip but not omission of \(+c\))

\(y = \sqrt{\frac{A\text{e}^x}{1+A\text{e}^x}}\) M1 (Obtain their reasonable expression in the form \(y = \text{f}(x)\) by a reasonable method)

[Correct expression for \(y\)] A1 (Do not accept \(\text{e}^{x+c}\) for this mark)

(ii) As \(x\) becomes large and negative, \(A\text{e}^x\) tends to \(0\) and \(\lim_{x\to-\infty} f(x) = 0\) B1 (Dep — requires a correct \(y\))

Since \(A\text{e}^x \neq 0\), \(y = \sqrt{\dfrac{A\text{e}^x}{\dfrac{1}{A\text{e}^x} + \dfrac{A\text{e}^x}{A\text{e}^x}}}\) M1 (Dep — Accept minimal or no working)

Hence \(\lim_{x\to+\infty} f(x) = 1\) A1

Total: 12 marks

This paper (10 questions)