A2 June 2025 Q4

EdexcelCurrent spec8 marksTaylor Series

\[y = \sqrt{\left(15 + \mathrm{e}^{2x}\right)}\]

(a) Show that\[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{\mathrm{e}^{2x}\left(\mathrm{e}^{2x} + 30\right)}{\left(15 + \mathrm{e}^{2x}\right)^{\frac{3}{2}}}\] (3)

(b) Hence determine the Maclaurin series expansion for \(y\), in ascending powers of \(x\), up to and including the term in \(x^2\), giving each term in simplest form. (2)

(c) Use the series expansion for \(\sin x\) in ascending powers of \(x\) to determine, in simplest form, the first 2 non-zero terms in ascending powers of \(x\) of the series for \(\sin 3x\). (1)

(d) Use the answers to parts (b) and (c) to show that\[\lim_{x \to 0} \frac{\sqrt{\left(15 + \mathrm{e}^{2x}\right)} - 4}{\sin 3x} = \frac{1}{12}\] (2)

Mark scheme (a)

Scheme	Marks	AO
\(y = \left(15 + \mathrm{e}^{2x}\right)^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\left(15 + \mathrm{e}^{2x}\right)^{-\frac{1}{2}} \times 2\mathrm{e}^{2x}\) or \(y^2 = 15 + \mathrm{e}^{2x} \Rightarrow 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x}\)	B1	1.1b
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \dfrac{\left(15 + \mathrm{e}^{2x}\right)^{\frac{1}{2}} \times A\mathrm{e}^{2x} + B\left(15 + \mathrm{e}^{2x}\right)^{-\frac{1}{2}} \times \mathrm{e}^{2x} \times \mathrm{e}^{2x}}{15 + \mathrm{e}^{2x}}\) \(\left\{\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \dfrac{\left(15 + \mathrm{e}^{2x}\right)^{\frac{1}{2}} \times 2\mathrm{e}^{2x} - \frac{1}{2}\left(15 + \mathrm{e}^{2x}\right)^{-\frac{1}{2}} \times 2\mathrm{e}^{2x} \times \mathrm{e}^{2x}}{15 + \mathrm{e}^{2x}}\right\}\) or \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = A\mathrm{e}^{2x}\left(15 + \mathrm{e}^{2x}\right)^{-\frac{1}{2}} + B\mathrm{e}^{2x} \times \mathrm{e}^{2x}\left(15 + \mathrm{e}^{2x}\right)^{-\frac{3}{2}}\) \(\left\{\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 2\mathrm{e}^{2x}\left(15 + \mathrm{e}^{2x}\right)^{-\frac{1}{2}} - \dfrac{1}{2} \times 2\mathrm{e}^{2x} \times \mathrm{e}^{2x}\left(15 + \mathrm{e}^{2x}\right)^{-\frac{3}{2}}\right\}\) Or \(A\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + By\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = C\mathrm{e}^{2x}\) \(\left\{2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 2y\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 4\mathrm{e}^{2x}\right\}\)	M1	1.1b
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \dfrac{\left(15 + \mathrm{e}^{2x}\right) \times 2\mathrm{e}^{2x} - \mathrm{e}^{4x}}{\left(15 + \mathrm{e}^{2x}\right)^{\frac{3}{2}}} = \dfrac{\mathrm{e}^{2x}\left(\mathrm{e}^{2x} + 30\right)}{\left(15 + \mathrm{e}^{2x}\right)^{\frac{3}{2}}}\ *\)	A1*	2.1
	(3)

Notes

B1: Correct first derivative in any form, may square first and use implicit differentiation

M1: Differentiates again applying the quotient or product rule correctly to achieve the correct form or uses implicit differentiation to achieve a correct form

A1*: Correct proof with no errors. Must show sufficient working.
If uses the product rule they must show a common denominator
If uses the quotient rule there must be evidence to show the multiplication of \(\left(15 + \mathrm{e}^{2x}\right)^{\frac{1}{2}}\) top and bottom
If uses implicit differentiation we need to see the substitution for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(y\) and rearranging.

Mark scheme (b)

Scheme	Marks	AO
\(x = 0 \Rightarrow y = 4,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{4},\ \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \dfrac{31}{64}\) leading to \(\Rightarrow y = 4 + \dfrac{x}{4} + \dfrac{31}{64}\dfrac{x^2}{2!} + \ldots = 4 + \dfrac{x}{4} + \dfrac{31}{128}x^2\)	M1 A1	1.1b 1.1b
	(2)

Notes

M1: A full method to obtain the required expansion. I.e. finds the values up to the second derivative and uses a correct formula

A1: Correct simplified expansion with \(y =\) or \(\mathrm{f}(x) =\) seen somewhere in their solution

Scheme	Marks	AO
\(\sin 3x = 3x - \dfrac{9}{2}x^3 + \ldots\)	B1	2.2a
	(1)

Mark scheme (d)

Scheme	Marks	AO
\(\left\{\lim\limits_{x \to 0}\right\}\dfrac{\sqrt{\left(15 + \mathrm{e}^{2x}\right)} - 4}{\sin 3x} = \left\{\lim\limits_{x \to 0}\right\}\dfrac{4 + \frac{x}{4} + \frac{31}{128}x^2 - 4}{3x - \frac{9}{2}x^3} = \left\{\lim\limits_{x \to 0}\right\}\dfrac{\frac{1}{4} + \frac{31}{128}x}{3 - \frac{9}{2}x^2}\)	M1	2.1
\(\lim\limits_{x \to 0}\dfrac{\frac{1}{4} + \frac{31}{128}x}{3 - \frac{9}{2}x^2}\left\{= \dfrac{1}{4} \div 3\right\} = \dfrac{1}{12}\ *\) cso	A1*	1.1b
	(2)
	(8 marks)

Notes

M1: A rigorous argument using their answers to (b) and (c), showing the cancelling constant term and the division by \(x\) in the numerator and denominator to establish the limiting behaviour.

A1*: Correct proof with no errors including correct limiting notation seen in the final stage cso

Note: Using L’Hospital’s Rule with the functions \(\sqrt{\left(15 + \mathrm{e}^{2x}\right)} - 4\) and \(\sin 3x\) is M0A0

Using L’Hospital’s Rule with their expansions \(\dfrac{4 + \frac{x}{4} + \frac{31}{128}x^2 - 4}{3x - \frac{9}{2}x^3}\) leading to \(\dfrac{\frac{1}{4} + \frac{31}{64}x}{3 - \frac{27}{2}x^2}\) could score M1A1