FP2 June 2015 Q7

EdexcelOld spec11 marksTaylor Series

7. \[y = \tan^2 x, \qquad -\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\]

(a) Show that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6\sec^4 x - 4\sec^2 x\) (4)

(b) Hence show that \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = 8\sec^2 x\tan x\left(A\sec^2 x + B\right)\), where \(A\) and \(B\) are constants to be found. (3)

(c) Find the Taylor series expansion of \(\tan^2 x\), in ascending powers of \(\left(x - \dfrac{\pi}{3}\right)\), up to and including the term in \(\left(x - \dfrac{\pi}{3}\right)^3\) (4)

Mark scheme (a)

Scheme	Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\tan x\sec^2 x\)	B1
OR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\tan x\left(1 + \tan^2 x\right)\)
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\sec^4 x + 4\tan^2 x\sec^2 x\) OR \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2\sec^2 x + 2\times3\tan^2 x\sec^2 x\)	M1 A1
\(= 2\sec^4 x + 4\left(\sec^2 x - 1\right)\sec^2 x\) OR \(= 2\sec^2 x + 6\left(\sec^2 x - 1\right)\sec^2 x\)
\(= 6\sec^4 x - 4\sec^2 x\) *	A1cso
	(4)

Notes

B1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\tan x\sec^2 x\)

M1: attempting the second derivative, inc using the product rule or \(\sec^2\theta = \tan^2\theta + 1\) Must start from the result given in (a)

A1: a correct second derivative in any form

A1cso: for a correct result following completely correct working \(\sec^2\theta = \tan^2\theta + 1\) must be d seen or used

Mark scheme (b)

Scheme	Marks
\(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = 24\sec^3 x\sec x\tan x - 8\sec^2 x\tan x\)	M1A1
\(= 8\sec^2 x\tan x\left(3\sec^2 x - 1\right)\)	A1cso
	(3)

Notes

M1: attempting the third derivative, inc using the chain rule

A1: a correct derivative

A1: a completely correct final result

Mark scheme (c)

Scheme	Marks
\(y_{\frac{\pi}{3}} = \left(\sqrt{3}\right)^2\ (= 3) \qquad \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_{\frac{\pi}{3}} = 2\sqrt{3}\times\left(\dfrac{2}{1}\right)^2\ \left(= 8\sqrt{3}\right)\)	B1(both)
\(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{\frac{\pi}{3}} = 6\times2^4 - 4\times2^2 = 80\)
\(\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_{\frac{\pi}{3}} = 8\times4\times\sqrt{3}\left(3\times2^2 - 1\right) = 352\sqrt{3}\)	M1(attempt both)
\(\tan^2 x = y_{\frac{\pi}{3}} + \left(x - \dfrac{\pi}{3}\right)\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_{\frac{\pi}{3}} + \dfrac{1}{2!}\left(x - \dfrac{\pi}{3}\right)^2\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{\frac{\pi}{3}} + \dfrac{1}{3!}\left(x - \dfrac{\pi}{3}\right)^3\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_{\frac{\pi}{3}}\)
\(= 3 + 8\sqrt{3}\left(x - \dfrac{\pi}{3}\right) + 40\left(x - \dfrac{\pi}{3}\right)^2 + \dfrac{176}{3}\sqrt{3}\left(x - \dfrac{\pi}{3}\right)^3\)	M1A1
	(4)
	(11 marks)

Notes

B1: \(y_{\frac{\pi}{3}} = \left(\sqrt{3}\right)^2\) or 3 and \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_{\frac{\pi}{3}} = 2\sqrt{3}\times\left(\dfrac{2}{1}\right)^2\) or \(8\sqrt{3}\)

M1: obtaining values for second and third derivatives at \(\dfrac{\pi}{3}\) (need not be correct but must be obtained from their derivatives)

M1: using a correct Taylor’s expansion using \(\left(x - \dfrac{\pi}{3}\right)\) and their derivatives. (2! or 2, 3! or 6 must be seen or implied by the work shown) This mark is not dependent.

A1: for a correct final answer Must start \(\tan^2 x = \ldots\) or \(y = \ldots\) \(\mathrm{f}(x)\) scores A0 unless defined as \(\tan^2 x\) or \(y\) here or earlier. Accept equivalents eg awrt 610 (609.6…) \(\sqrt{371712}\) But no factorials in this final answer.