Further Mathematics · Pearson Edexcel 9FM0 · Option FP1 · Topic 2

Further calculus

Taylor series, using series to find limits, Leibnitz's theorem for repeated differentiation, l'Hôpital's rule and the Weierstrass substitution.

5 sub-topics
Taught in Year 13
59 practice questions
5 animated lessons

Practice questions (59) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

2.1Taylor seriesY13
2.2Series expansions to evaluate limitsY13
2.3Leibnitz's theoremY13
2.4L'Hôpital's ruleY13
2.5Weierstrass substitutionY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 59 questions in this topic, free, no account needed.

Further calculus

Not needed for Year 12Easy

The Taylor series · Year 13 · 1 mark

Which is the Taylor series of $f(x)$ about $x = a$ ?
(1)

Show worked solutionHide worked solution

A Maclaurin series is the special case $a = 0$ . $C$ : $f(a)$ ${}+ f'(a)(x - a)$ ${}+ \dfrac{f''(a)}{2!}(x - a)^2$ ${}+ \ldots$

Total1

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

How did it go?

Report a problem with this question

Further calculus

Not needed for Year 12Medium

Taylor series · Year 13 · 6 marks

(a)
Find the Taylor series expansion of $\sqrt{x}$ in ascending powers of $\left(x - 4\right)$ , up to and including the term in $\left(x - 4\right)^3$ .
(4)
(b)
Hence find an approximation for $\sqrt{4.1}$ , giving your answer to 5 decimal places.
(2)

Show worked solutionHide worked solution

Part (a)4 marks

The Taylor series about $x = a$ . $f(x)$ ${}= f(a)$ ${}+ f'(a)(x - a)$ ${}+ \frac{f''(a)}{2!}(x - a)^2$ ${}+ \frac{f'''(a)}{3!}(x - a)^3$
Differentiate three times and substitute. $f^{(0)}(4)$ ${}= 2, \ f^{(1)}(4)$ ${}= \frac{1}{4}, \ f^{(2)}(4)$ ${}= - \frac{1}{32}, \ f^{(3)}(4)$ ${}= \frac{3}{256}$
$2$ ${}+ \frac{1}{4}\left(x - 4\right)$ ${}- \frac{1}{64}\left(x - 4\right)^{2}$ ${}+ \frac{1}{512}\left(x - 4\right)^{3}$

Part (b)2 marks

Choose $x$ so that $x - a$ is small. $x - 4 = 0.1$
The calculator value is 2.02485. $\sqrt{4.1} \approx 2.02485$

Total6

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

How did it go?

Report a problem with this question

Further calculus

Not needed for Year 12Hard

Leibnitz's theorem · Year 13 · 5 marks

y = x^2\sin 3x

.

(a)
Use Leibnitz's theorem to find $\dfrac{\mathrm{d}^{6}y}{\mathrm{d}x^{6}}$ .
(4)
(b)
Hence find the value of $\dfrac{\mathrm{d}^{6}y}{\mathrm{d}x^{6}}$ when $x = 0$ .
(1)

Check your answer before you look at the solution

Show worked solutionHide worked solution

Part (a)4 marks

Take $u = x^{2}$ , whose third and higher derivatives are zero. $\frac{\mathrm{d}^{6}}{\mathrm{d}x^{6}}(uv)$ ${}= \sum_{k = 0}^{6} \binom{6}{k} u^{(k)} v^{(6 - k)}, \ u$ ${}= x^2$
Only three terms survive. $\left(x^{2}\right)\left(- 729 \sin 3 x\right)$ ${}+ 6\left(2 x\right)\left(243 \cos 3 x\right)$ ${}+ 15\left(2\right)\left(81 \sin 3 x\right)$
$= - 729 x^{2} \sin 3 x$ ${}+ 2916 x \cos 3 x$ ${}+ 2430 \sin 3 x$

Part (b)1 mark

$0$

Total5

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

How did it go?

Report a problem with this question