Standard +0.8 This question combines partial fractions with separable differential equations and requires integration of trigonometric functions. Part (a) is routine, but part (b) requires recognizing the separation of variables, integrating cot x (less common), applying initial conditions, and manipulating the result into the specified form involving sec²x. The multi-step nature and the need to work with less familiar trigonometric integrals elevates this above average difficulty.
7. (a) Express \(\frac { 2 } { 4 - y ^ { 2 } }\) in partial fractions.
(b) Hence obtain the solution of
$$2 \cot x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 4 - y ^ { 2 } \right)$$
for which \(y = 0\) at \(x = \frac { \pi } { 3 }\), giving your answer in the form \(\sec ^ { 2 } x = \mathrm { g } ( y )\).
*(If no working seen, but candidate writes down correct partial fraction then award all three marks. If working is seen but one of A or B is incorrect then M0A0A0.)*
Either \(\pm a\ln(\lambda - y)\) or \(\pm b\ln(\lambda+y)\); their \(\int\frac{1}{\cot x}\,dx =\) LHS correct with ft for their \(A\) and \(B\) and no error with the "2" with or without \(+c\)
Using the log laws correctly to obtain a single log term on both sides of the equation
\(\frac{2+y}{2-y} = \frac{\sec^2 x}{4}\); Hence \(\sec^2 x = \frac{8+4y}{2-y}\)
A1 aef
\(\sec^2 x = \frac{8+4y}{2-y}\)
[8 marks] — Total: 11 marks
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{4-y^2} \equiv \frac{2}{(2-y)(2+y)} \equiv \frac{A}{(2-y)} + \frac{B}{(2+y)}$; $2 \equiv A(2+y) + B(2-y)$ | M1 | Forming this identity. NB: A & B are not assigned in this question |
| Let $y=-2$: $2 = B(4) \Rightarrow B = \frac{1}{2}$; Let $y=2$: $2=A(4) \Rightarrow A=\frac{1}{2}$ | A1 | Either one of $A=\frac{1}{2}$ or $B=\frac{1}{2}$ |
| $\frac{\frac{1}{2}}{(2-y)} + \frac{\frac{1}{2}}{(2+y)}$ | A1 cao | $\frac{\frac{1}{2}}{(2-y)} + \frac{\frac{1}{2}}{(2+y)}$, aef |
**[3 marks]**
*(If no working seen, but candidate writes down correct partial fraction then award all three marks. If working is seen but one of A or B is incorrect then M0A0A0.)*
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{2}{4-y^2}\,dy = \int \frac{1}{\cot x}\,dx$ | B1 | Separates variables as shown. Can be implied. Ignore integral signs and the '2'. |
| $\int \frac{\frac{1}{2}}{(2-y)} + \frac{\frac{1}{2}}{(2+y)}\,dy = \int \tan x\,dx$ | | |
| $\therefore -\frac{1}{2}\ln(2-y) + \frac{1}{2}\ln(2+y) = \ln(\sec x) + c$ | B1 | $\ln(\sec x)$ or $-\ln(\cos x)$ |
| | M1 | Either $\pm a\ln(\lambda - y)$ or $\pm b\ln(\lambda+y)$; their $\int\frac{1}{\cot x}\,dx =$ LHS correct with ft for their $A$ and $B$ and no error with the "2" with or without $+c$ |
| | A1$\checkmark$ | |
| $y=0$, $x=\frac{\pi}{3}$: $-\frac{1}{2}\ln 2 + \frac{1}{2}\ln 2 = \ln\!\left(\frac{1}{\cos(\frac{\pi}{3})}\right) + c$ | M1* | Use of $y=0$ and $x=\frac{\pi}{3}$ in an integrated equation containing $c$ |
| $0 = \ln 2 + c \Rightarrow \underline{c = -\ln 2}$ | | |
| $\frac{1}{2}\ln\!\left(\frac{2+y}{2-y}\right) = \ln\!\left(\frac{\sec x}{2}\right)$ | M1 | Using either the quotient (or product) or power laws for logarithms correctly |
| $\ln\!\left(\frac{2+y}{2-y}\right) = 2\ln\!\left(\frac{\sec x}{2}\right) = \ln\!\left(\frac{\sec x}{2}\right)^2$ | dM1* | Using the log laws correctly to obtain a single log term on both sides of the equation |
| $\frac{2+y}{2-y} = \frac{\sec^2 x}{4}$; Hence $\sec^2 x = \frac{8+4y}{2-y}$ | A1 aef | $\sec^2 x = \frac{8+4y}{2-y}$ |
**[8 marks] — Total: 11 marks**
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7. (a) Express $\frac { 2 } { 4 - y ^ { 2 } }$ in partial fractions.\\
(b) Hence obtain the solution of
$$2 \cot x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 4 - y ^ { 2 } \right)$$
for which $y = 0$ at $x = \frac { \pi } { 3 }$, giving your answer in the form $\sec ^ { 2 } x = \mathrm { g } ( y )$.\\
\hfill \mbox{\textit{Edexcel C4 2008 Q7 [11]}}