2024 H2 Maths Paper 1 Q8 — A Maclaurin Expansion by Repeated Differentiation

The approximationsbuilding the series

Look this far from the origin0.60

The function The Maclaurin approximation The binomial series

Part (a)

3 marks

Given $y=\cos\left(1-e^{2x}\right)$ , show that $\dfrac{d^{2}y}{dx^{2}}=k\left(\dfrac{dy}{dx}-2ye^{4x}\right)$ , finding $k$ .

Approach

Differentiate twice with the chain rule, then look at the result and hunt for the pieces you already have names for. The whole art of this part is not expanding everything out: $\frac{dy}{dx}$ and $y$ are already sitting inside $\frac{d^{2}y}{dx^{2}}$ , and the target tells you they are what you are meant to find.

Let $u=1-e^{2x}$ , so $y=\cos u$ and $\frac{du}{dx}=-2e^{2x}$ . Then

\frac{dy}{dx}=-\sin u\cdot\frac{du}{dx}=-\sin u\left(-2e^{2x}\right)=2e^{2x}\sin u

Differentiating again, by the product rule on $2e^{2x}$ and $\sin u$ :

\begin{aligned}\frac{d^{2}y}{dx^{2}}&=4e^{2x}\sin u+2e^{2x}\cos u\cdot\frac{du}{dx}\\[2pt]&=4e^{2x}\sin u+2e^{2x}\cos u\left(-2e^{2x}\right)\\[2pt]&=4e^{2x}\sin u-4e^{4x}\cos u\end{aligned}

Now recognise the pieces. $4e^{2x}\sin u=2\left(2e^{2x}\sin u\right)=2\frac{dy}{dx}$ , and $\cos u=y$ , so

\frac{d^{2}y}{dx^{2}}=2\frac{dy}{dx}-4ye^{4x}=2\left(\frac{dy}{dx}-2ye^{4x}\right)\quad\textbf{(shown)}

To check your answer:

At $x=0$ : $u=1-1=0$ , so $y=\cos0=1$ and $\frac{dy}{dx}=2(1)\sin0=0$ . The relation then predicts $\frac{d^{2}y}{dx^{2}}=2\left(0-2(1)(1)\right)=-4$ , which agrees with computing $4e^{0}\sin0-4e^{0}\cos0=-4$ directly ✓.

$\therefore\ k=2$ .

Where the marks go

Two chain rules and one product rule, and the inner derivative $-2e^{2x}$ is used twice — once with a minus that cancels the one from $\frac{d}{du}\cos u$ , and once that does not. Writing $u$ explicitly and keeping it as $u$ until the last line is what stops the signs drifting; substituting $1-e^{2x}$ back in at every step buries the structure the question wants you to see.

Part (b)

4 marks

By differentiating the result in part (a), find the first three non-zero terms of the Maclaurin expansion of $\cos\left(1-e^{2x}\right)$ .

Approach

Maclaurin needs the derivatives at $x=0$ , not as functions. So evaluate what you already have at $x=0$ , differentiate the relation once more to reach the third derivative, evaluate that too, and assemble. "By differentiating the result in part (a)" is an instruction, not a suggestion — it is the cheap route to $y'''$ .

At $x=0$ , using $u=1-e^{0}=0$ :

\begin{aligned}y(0)&=\cos0=1\\y'(0)&=2e^{0}\sin0=0\\y''(0)&=2\left(y'(0)-2y(0)e^{0}\right)=2\left(0-2(1)(1)\right)=-4\end{aligned}

Differentiating the relation $y''=2\left(y'-2ye^{4x}\right)$ , with the product rule on $ye^{4x}$ :

y'''=2\left(y''-2\left[y'e^{4x}+4ye^{4x}\right]\right)

and at $x=0$ :

y'''(0)=2\left(-4-2\left[0+4(1)(1)\right]\right)=2(-4-8)=-24

The Maclaurin series is $y=y(0)+y'(0)x+\frac{y''(0)}{2!}x^{2}+\frac{y'''(0)}{3!}x^{3}+\cdots$ , so

y=1+0\cdot x-\frac{4}{2}x^{2}-\frac{24}{6}x^{3}+\cdots=1-2x^{2}-4x^{3}+\cdots

To check your answer:

Expand from standard series instead. $e^{2x}=1+2x+2x^{2}+\frac{4}{3}x^{3}+\cdots$ , so $u=1-e^{2x}=-2x-2x^{2}-\frac{4}{3}x^{3}-\cdots$ , giving $u^{2}=4x^{2}+8x^{3}+\cdots$ . Then $\cos u\approx1-\frac{u^{2}}{2}=1-2x^{2}-4x^{3}+\cdots$ ✓ — the same three terms by a completely different route.

$\therefore\ \cos\left(1-e^{2x}\right)\approx1-2x^{2}-4x^{3}$ .

Where the marks go

"First three non-zero terms" — and the $x$ term is zero, so you must go as far as $x^{3}$ . Stopping at $1-2x^{2}$ because that is three terms written down loses two marks. The other loss is dividing by the wrong factorial: the coefficient of $x^{3}$ is $\frac{y'''(0)}{3!}=\frac{-24}{6}$ , not $\frac{-24}{3}$ .

Part (c)

2 marks

The first two non-zero terms of the expansion equal the first two non-zero terms of the series for $\dfrac{1}{\sqrt{a+bx^{2}}}$ . Using standard series from MF26, find $a$ and $b$ .

Approach

MF26 gives $(1+z)^{n}$ , which needs a leading 1 — so factor $a$ out of the square root first. Then expand, take two terms, and match against $1-2x^{2}$ .

Factoring $a$ out:

\frac{1}{\sqrt{a+bx^{2}}}=\frac{1}{\sqrt{a}}\left(1+\frac{b}{a}x^{2}\right)^{-1/2}

and expanding with $n=-\frac{1}{2}$ , $z=\frac{b}{a}x^{2}$ :

=\frac{1}{\sqrt{a}}\left[1-\frac{1}{2}\cdot\frac{b}{a}x^{2}+\cdots\right]=\frac{1}{\sqrt{a}}-\frac{b}{2a^{3/2}}x^{2}+\cdots

Matching the first two non-zero terms with $1-2x^{2}$ :

\frac{1}{\sqrt{a}}=1\ \Longrightarrow\ a=1

-\frac{b}{2a^{3/2}}=-2\ \Longrightarrow\ -\frac{b}{2}=-2\ \Longrightarrow\ b=4

To check your answer:

Expanding the answer back out:

\frac{1}{\sqrt{1+4x^{2}}}=\left(1+4x^{2}\right)^{-1/2}=1-\frac{1}{2}\left(4x^{2}\right)+\cdots=1-2x^{2}+\cdots

✓, matching the first two non-zero terms of part (b). The two curves are drawn together in the model — they agree to the eye near the origin and part company as $\left|x\right|$ grows, which is exactly what "first two terms" promises and no more.

$\therefore\ a=1$ and $b=4$ .

Note — why the third terms need not agree

They do not: the expansion of $\left(1+4x^{2}\right)^{-1/2}$ is even, so its next term is in $x^{4}$ , while $\cos\left(1-e^{2x}\right)$ has a $-4x^{3}$ . The question only ever claims agreement to two terms, and the model shows the two curves separating precisely where that claim runs out.

What this question was really testing

Recognising your own expressions. Part (a) is two routine differentiations followed by the real step: spotting $\frac{dy}{dx}$ and $y$ inside the result instead of expanding them away.
Why a relation beats a formula. Differentiating $y''=2\left(y'-2ye^{4x}\right)$ to get $y'''$ is a two-line job; differentiating $\cos\left(1-e^{2x}\right)$ three times is not.
Non-zero terms. The $x$ coefficient vanishes, so "first three non-zero terms" reaches $x^{3}$ . Read the words, count the surviving terms.
Factoring before a binomial expansion. MF26's $(1+z)^{n}$ needs a leading 1, so $\sqrt{a+bx^{2}}$ must become $\sqrt{a}\sqrt{1+\frac{b}{a}x^{2}}$ first.

A series built from a relation, not from the function

Part (a)

Part (b)

Part (c)

What this question was really testing

This is how we mark, and how we teach.