C4 June 2008 Q7

EdexcelOld spec11 marksAlgebraic Fractions Integration

(a) Express \(\dfrac{2}{4 - y^2}\) in partial fractions. (3)

(b) Hence obtain the solution of\[2\cot x\,\frac{\mathrm{d}y}{\mathrm{d}x} = (4 - y^2)\]for which \(y = 0\) at \(x = \dfrac{\pi}{3}\), giving your answer in the form \(\sec^2 x = \mathrm{g}(y)\). (8)

Mark scheme (a)

Scheme	Marks
\(\dfrac{2}{4 - y^2} \equiv \dfrac{2}{(2-y)(2+y)} \equiv \dfrac{A}{(2-y)} + \dfrac{B}{(2+y)}\)
\(2 \equiv A(2+y) + B(2-y)\)	M1
Let \(y = -2\), \(2 = B(4) \Rightarrow B = \tfrac{1}{2}\)
Let \(y = 2\), \(2 = A(4) \Rightarrow A = \tfrac{1}{2}\)	A1
giving \(\underline{\dfrac{\frac{1}{2}}{(2-y)} + \dfrac{\frac{1}{2}}{(2+y)}}\)	A1 cao
	(3)

Notes

M1: Forming this identity. NB: A & B are not assigned in this question

A1: Either one of \(A = \tfrac{1}{2}\) or \(B = \tfrac{1}{2}\) A1 cao: \(\underline{\dfrac{\frac{1}{2}}{(2-y)} + \dfrac{\frac{1}{2}}{(2+y)}}\), aef

(If no working seen, but candidate writes down correct partial fraction then award all three marks. If no working is seen but one of \(A\) or \(B\) is incorrect then M0A0A0.)

Mark scheme (b)

Scheme	Marks
\(\displaystyle\int\frac{2}{4 - y^2}\,\mathrm{d}y = \int\frac{1}{\cot x}\,\mathrm{d}x\)	B1
\(\displaystyle\int\frac{\frac{1}{2}}{(2-y)} + \frac{\frac{1}{2}}{(2+y)}\,\mathrm{d}y = \int\tan x\,\mathrm{d}x\)
\(\therefore -\tfrac{1}{2}\ln(2 - y) + \tfrac{1}{2}\ln(2 + y) = \ln(\sec x)\ + (c)\)	B1 M1; A1ft
\(y = 0,\ x = \tfrac{\pi}{3} \Rightarrow -\tfrac{1}{2}\ln 2 + \tfrac{1}{2}\ln 2 = \ln\left(\tfrac{1}{\cos(\frac{\pi}{3})}\right) + c\)	M1*
\(\left\{0 = \ln 2 + c \Rightarrow \underline{c = -\ln 2}\right\}\)
\(-\tfrac{1}{2}\ln(2 - y) + \tfrac{1}{2}\ln(2 + y) = \ln(\sec x) - \ln 2\)
\(\dfrac{1}{2}\ln\left(\dfrac{2+y}{2-y}\right) = \ln\left(\dfrac{\sec x}{2}\right)\)	M1
\(\ln\left(\dfrac{2+y}{2-y}\right) = 2\ln\left(\dfrac{\sec x}{2}\right)\)
\(\ln\left(\dfrac{2+y}{2-y}\right) = \ln\left(\dfrac{\sec x}{2}\right)^2\)	dM1*
\(\dfrac{2+y}{2-y} = \dfrac{\sec^2 x}{4}\)
Hence, \(\underline{\sec^2 x = \dfrac{8 + 4y}{2 - y}}\)	A1 aef
	(8)
	(11 marks)

Notes

B1: Separates variables as shown. Can be implied. Ignore the integral signs, and the ‘2’.

B1: \(\ln(\sec x)\) or \(-\ln(\cos x)\)

M1: Either \(\pm a\ln(\lambda - y)\) or \(\pm b\ln(\lambda + y)\);

A1ft: their \(\displaystyle\int\frac{1}{\cot x}\,\mathrm{d}x =\) LHS correct with ft for their \(A\) and \(B\) and no error with the “2” with or without \(+\,c\)

M1*: Use of \(y = 0\) and \(x = \tfrac{\pi}{3}\) in an integrated equation containing c;

M1: Using either the quotient (or product) or power laws for logarithms CORRECTLY.

dM1*: Using the log laws correctly to obtain a single log term on both sides of the equation.

A1 aef: \(\underline{\sec^2 x = \dfrac{8 + 4y}{2 - y}}\)