FP1-2 Further calculus (Further Pure 1) — revision question pack

5 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FP1-2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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FP1-2.1 · Derivation and use of Taylor series.

Explanation

The Taylor expansion of $f$ about $x=a$ is $f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3+\cdots$ , and the equivalent shifted form is $f(a+h)=f(a)+hf'(a)+\dfrac{h^2}{2!}f''(a)+\cdots$ .
Taking $a=0$ recovers the Maclaurin series already met in A-level Mathematics.
In practice you build a short table of $f$ , $f'$ , $f''$ , $f'''$ , evaluate every entry at $x=a$ , and then assemble the series; keep the powers as $(x-a)^k$ throughout rather than expanding brackets, because the question almost always asks for the answer in ascending powers of $(x-a)$ .
Where a function is a product or composition of standard series it is usually far quicker to multiply or substitute known Maclaurin series than to differentiate repeatedly, and the two routes are a useful check on each other.

Worked example

Find the Taylor expansion of $\sin x$ in ascending powers of $(x-\pi)$ as far as the term in $(x-\pi)^3$ .

1. $f(x)=\sin x$ gives $f(\pi)=0$ .
2. $f'(x)=\cos x$ gives $f'(\pi)=-1$ ; $f''(x)=-\sin x$ gives $f''(\pi)=0$ ; $f'''(x)=-\cos x$ gives $f'''(\pi)=1$ .
3.Substituting into $f(\pi)+f'(\pi)(x-\pi)+\tfrac{f''(\pi)}{2!}(x-\pi)^2+\tfrac{f'''(\pi)}{3!}(x-\pi)^3$ gives $-(x-\pi)+\tfrac16(x-\pi)^3$ .

Answer: $\sin x=-(x-\pi)+\dfrac{(x-\pi)^3}{6}+\cdots$

Common mistakes

Don't fall into the trap of dividing by $n$ instead of $n!$ in the general term.
Don't fall into the trap of evaluating the derivatives at $x$ rather than at $x=a$ , leaving variables in the coefficients.
Don't fall into the trap of expanding the brackets when the question asked for ascending powers of $(x-a)$ .

Exam tip

Set out a two-column table of derivatives and their values at $x=a$ before writing a single term of the series; the marks follow the table.

Tier 1 · Easy

1.
Find the Taylor expansion of $\mathrm{e}^{x}$ in ascending powers of $(x-1)$ as far as the term in $(x-1)^2$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find the Taylor expansion of $\ln x$ in ascending powers of $(x-1)$ as far as the term in $(x-1)^3$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Find the Taylor expansion of $\cos x$ in ascending powers of $\left(x-\tfrac{\pi}{3}\right)$ as far as the term in $\left(x-\tfrac{\pi}{3}\right)^2$ .
(4)
(Total for Question 1 is 4 marks)
2.
Find the Taylor expansion of $\tan x$ in ascending powers of $\left(x-\tfrac{\pi}{4}\right)$ as far as the term in $\left(x-\tfrac{\pi}{4}\right)^2$ .
(4)
(Total for Question 2 is 4 marks)
3.
Use Taylor's series to expand $\ln(\cos x)$ in ascending powers of $x$ as far as the term in $x^4$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Show that the Maclaurin expansion of $\sec x$ as far as the term in $x^4$ is $1+\dfrac{x^2}{2}+\dfrac{5x^4}{24}$ .
(6)
(Total for Question 1 is 6 marks)
2.
Find the Maclaurin expansion of $\mathrm{e}^{\sin x}$ as far as the term in $x^3$ , and state the value of the coefficient of $x^3$ .
(6)
(Total for Question 2 is 6 marks)
3.
Find the Maclaurin expansion of $\mathrm{e}^{x}\sin x$ as far as the term in $x^3$ .
(5)
(Total for Question 3 is 5 marks)
4.
Find the Taylor expansion of $\sqrt{x}$ in ascending powers of $(x-4)$ as far as the term in $(x-4)^2$ , and use it to estimate $\sqrt{4.2}$ to $6$ decimal places.
(6)
(Total for Question 4 is 6 marks)
5.
Find the Taylor expansion of $\sin x$ in ascending powers of $\left(x-\tfrac{\pi}{2}\right)$ as far as the term in $\left(x-\tfrac{\pi}{2}\right)^4$ , and explain why only even powers appear.
(6)
(Total for Question 5 is 6 marks)

FP1-2.2 · Use of series expansions to find limits.

Explanation

To evaluate an indeterminate limit by series, replace each function by its Maclaurin expansion, keep enough terms to reach the lowest power that survives, and then cancel.
The five expansions that cover almost every question are $\sin x=x-\tfrac{x^3}{6}+\cdots$ , $\cos x=1-\tfrac{x^2}{2}+\tfrac{x^4}{24}-\cdots$ , $\tan x=x+\tfrac{x^3}{3}+\cdots$ , $\mathrm{e}^{x}=1+x+\tfrac{x^2}{2}+\cdots$ and $\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\cdots$ ; $\arctan x=x-\tfrac{x^3}{3}+\cdots$ is also worth knowing.
Decide in advance how many terms you need: if the denominator is $x^3$ you must expand every numerator to at least $x^3$ , and stopping a term early is the standard way to produce a wrong limit.
Differences of two singular expressions such as $\tfrac1{x^2}-\tfrac1{\sin^2x}$ should first be combined over a common denominator, then expanded.

Worked example

Use series expansions to evaluate $\lim_{x\to0}\dfrac{x-\arctan x}{x^3}$ .

1. $\arctan x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\cdots$ .
2.So $x-\arctan x=\dfrac{x^3}{3}-\dfrac{x^5}{5}+\cdots$ .
3.Dividing by $x^3$ gives $\dfrac13-\dfrac{x^2}{5}+\cdots$ , and letting $x\to0$ leaves $\dfrac13$ .

Answer: The limit is $\dfrac13$ .

Common mistakes

Don't fall into the trap of truncating a series one term too early, so the surviving power is missed and the limit comes out as $0$ or infinite.
Don't fall into the trap of expanding only the numerator and leaving the denominator unexpanded when it is not already a pure power.
Don't fall into the trap of writing $\lim$ signs on lines where the variable has already been cancelled away.

Exam tip

Look at the denominator first: its degree tells you exactly how many terms of every numerator expansion you must keep.

Tier 1 · Easy

1.
Use a series expansion to evaluate $\lim_{x\to0}\dfrac{1-\cos x}{x^2}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Use a series expansion to evaluate $\lim_{x\to0}\dfrac{\mathrm{e}^{3x^2}-1}{x^2}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Use series expansions to evaluate $\lim_{x\to0}\dfrac{x-\sin x}{x^3}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Use series expansions to evaluate $\lim_{x\to0}\dfrac{\tan x-\sin x}{x^3}$ .
(4)
(Total for Question 2 is 4 marks)
3.
Use series expansions to evaluate $\lim_{x\to0}\dfrac{\sin x-x\cos x}{x^3}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Use series expansions to evaluate $\lim_{x\to0}\dfrac{\ln(1+x)-x+\tfrac12x^2}{x^3}$ .
(5)
(Total for Question 1 is 5 marks)
2.
Use series expansions to evaluate $\lim_{x\to0}\dfrac{\cos x-\mathrm{e}^{-x^2/2}}{x^4}$ .
(6)
(Total for Question 2 is 6 marks)
3.
Use series expansions to evaluate $\lim_{x\to0}\left(\dfrac1{x^2}-\dfrac1{\sin^2x}\right)$ .
(6)
(Total for Question 3 is 6 marks)
4.
Use series expansions to evaluate $\lim_{x\to0}\dfrac{(1+x)^{1/x}-\mathrm{e}}{x}$ .
(6)
(Total for Question 4 is 6 marks)
5.
Given that $\lim_{x\to0}\dfrac{a\sin x+b\tan x-x}{x^3}$ is finite and equal to $\tfrac14$ , find the values of the constants $a$ and $b$ .
(6)
(Total for Question 5 is 6 marks)

FP1-2.3 · Leibnitz's theorem.

Explanation

Leibnitz's theorem gives the $n$ th derivative of a product: $\dfrac{\mathrm{d}^n}{\mathrm{d}x^n}(uv)=\sum_{r=0}^{n}\binom{n}{r}u^{(n-r)}v^{(r)}$ , where $u^{(k)}$ means the $k$ th derivative.
It is the differentiation analogue of the binomial expansion, and the binomial coefficients are the same ones.
The theorem is most useful when one factor is a polynomial, because that factor's derivatives stop after finitely many terms and the sum truncates: a quadratic factor leaves only three terms whatever the value of $n$ .
Its second use is to differentiate a differential relation $n$ times, producing a recurrence between successive derivatives at a point; this is the standard route to a Maclaurin series for functions such as $\arcsin x$ and $\mathrm{e}^{\arctan x}$ .
Take care to attach the derivative counts to the correct factor: $u^{(n-r)}$ pairs with $v^{(r)}$ , not the other way round.

Worked example

Use Leibnitz's theorem to find the third derivative of $x^2\mathrm{e}^{x}$ .

1.Take $u=\mathrm{e}^{x}$ and $v=x^2$ , so $u^{(k)}=\mathrm{e}^{x}$ for every $k$ and $v'=2x$ , $v''=2$ , $v'''=0$ .
2. $\dfrac{\mathrm{d}^3}{\mathrm{d}x^3}(uv)=\binom30\mathrm{e}^{x}x^2+\binom31\mathrm{e}^{x}(2x)+\binom32\mathrm{e}^{x}(2)$ .
3.This is $\mathrm{e}^{x}\left(x^2+6x+6\right)$ .

Answer: $\dfrac{\mathrm{d}^3}{\mathrm{d}x^3}\left(x^2\mathrm{e}^{x}\right)=\mathrm{e}^{x}\left(x^2+6x+6\right)$ .

Common mistakes

Don't fall into the trap of swapping the derivative orders so that the polynomial is differentiated $n-r$ times instead of $r$ times.
Don't fall into the trap of continuing the sum past the point where the polynomial factor has been differentiated to zero, or stopping it too early.
Don't fall into the trap of using $\binom{n}{r}$ with the wrong $n$ after differentiating a relation such as $(1-x^2)y''=xy'$ .

Exam tip

Write the sum out with its binomial coefficients before substituting anything; getting the Leibnitz structure down first makes the algebra that follows routine.

Tier 1 · Easy

1.
Use Leibnitz's theorem to find the fourth derivative of $x\sin x$ .
(3)
(Total for Question 1 is 3 marks)
2.
Use Leibnitz's theorem to find the fourth derivative of $x^3\mathrm{e}^{-x}$ , and evaluate it at $x=1$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Use Leibnitz's theorem to show that the $n$ th derivative of $x^2\mathrm{e}^{x}$ is $\mathrm{e}^{x}\left(x^2+2nx+n(n-1)\right)$ .
(4)
(Total for Question 1 is 4 marks)
2.
Use Leibnitz's theorem to find the sixth derivative of $x^2\sin x$ .
(5)
(Total for Question 2 is 5 marks)
3.
Use Leibnitz's theorem to find the fifth derivative of $x^3\ln x$ , and hence write down its value at $x=1$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Use Leibnitz's theorem to find the $n$ th derivative of $x^2\mathrm{e}^{ax}$ , where $a$ is a non-zero constant and $n\ge2$ .
(5)
(Total for Question 1 is 5 marks)
2.
Show that the $n$ th derivative of $x^2\ln x$ is $\dfrac{2(-1)^{n-1}(n-3)!}{x^{\,n-2}}$ for $n\ge3$ .
(6)
(Total for Question 2 is 6 marks)
3.
Given $y=\arcsin x$ , show that $\left(1-x^2\right)y''=xy'$ , and use Leibnitz's theorem to prove that $\left(1-x^2\right)y^{(n+2)}-(2n+1)x\,y^{(n+1)}-n^2y^{(n)}=0$ .
(6)
(Total for Question 3 is 6 marks)
4.
Given $y=\mathrm{e}^{\arctan x}$ , show that $\left(1+x^2\right)y'=y$ , and use Leibnitz's theorem to prove that $\left(1+x^2\right)y^{(n+1)}+\left(2nx-1\right)y^{(n)}+n(n-1)y^{(n-1)}=0$ .
(6)
(Total for Question 4 is 6 marks)
5.
Use the result $\left(1-x^2\right)y^{(n+2)}-(2n+1)x\,y^{(n+1)}-n^2y^{(n)}=0$ for $y=\arcsin x$ to find the Maclaurin expansion of $\arcsin x$ as far as the term in $x^5$ .
(6)
(Total for Question 5 is 6 marks)

FP1-2.4 · L'Hospital's Rule.

Explanation

If $f(a)=g(a)=0$ , or if both $|f|$ and $|g|$ tend to infinity, and $g'$ is non-zero near $a$ , then $\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$ whenever the second limit exists.
The rule may be applied repeatedly while the quotient remains indeterminate, but you must check the $\tfrac00$ or $\tfrac{\infty}{\infty}$ condition again before every application; applying it to a quotient that is not indeterminate is a guaranteed way to get the wrong answer.
Indeterminate forms of the types $0\times\infty$ , $\infty-\infty$ , $0^0$ , $1^{\infty}$ and $\infty^0$ must first be rewritten: products become quotients by moving one factor to the denominator, differences are combined over a common denominator, and powers are handled by taking logarithms, finding the limit of $\ln y$ and exponentiating at the end.

Worked example

Evaluate $\lim_{x\to0}\dfrac{2\sin x-\sin 2x}{x-\sin x}$ .

1.Both numerator and denominator vanish at $x=0$ , so differentiate: $\dfrac{2\cos x-2\cos 2x}{1-\cos x}$ , still $\tfrac00$ .
2.Differentiate again: $\dfrac{-2\sin x+4\sin 2x}{\sin x}$ , still $\tfrac00$ .
3.Differentiate once more: $\dfrac{-2\cos x+8\cos 2x}{\cos x}$ , which at $x=0$ is $\dfrac{-2+8}{1}=6$ .

Answer: The limit is $6$ .

Common mistakes

Don't fall into the trap of differentiating the quotient with the quotient rule instead of differentiating numerator and denominator separately.
Don't fall into the trap of applying the rule again once the quotient has stopped being indeterminate.
Don't fall into the trap of forgetting to exponentiate at the end after taking logarithms for a $1^{\infty}$ or $0^0$ form.

Exam tip

Before you differentiate, make sure the expression is a quotient in indeterminate form — for a product or difference, the first method mark is for rewriting it as $\dfrac{f(x)}{g(x)}$ with $\dfrac{f(a)}{g(a)}=\tfrac00$ or $\tfrac{\infty}{\infty}$ (June 2022 9FM₀/3A Q₈(ii), M₁). Annotating ' $\tfrac00$ ' on later lines earns no mark of its own, but it stops the classic over-application error.

Tier 1 · Easy

1.
Use L'Hospital's Rule to evaluate $\lim_{x\to0}\dfrac{\mathrm{e}^{2x}-1}{\sin x}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Use L'Hospital's Rule to evaluate $\lim_{x\to1}\dfrac{\ln x}{x-1}$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Use L'Hospital's Rule to evaluate $\lim_{x\to0^{+}}x\ln x$ .
(4)
(Total for Question 1 is 4 marks)
2.
Use L'Hospital's Rule to evaluate $\lim_{x\to\frac{\pi}{2}}\dfrac{1-\sin x}{\cos^2x}$ .
(4)
(Total for Question 2 is 4 marks)
3.
Use L'Hospital's Rule to evaluate $\lim_{x\to0}\dfrac{\tan x-x}{x-\sin x}$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Use L'Hospital's Rule to evaluate $\lim_{x\to0}\left(\dfrac1x-\dfrac1{\mathrm{e}^{x}-1}\right)$ .
(5)
(Total for Question 1 is 5 marks)
2.
Use L'Hospital's Rule to evaluate $\lim_{x\to\infty}\left(1+\dfrac3x\right)^{x}$ .
(5)
(Total for Question 2 is 5 marks)
3.
Use L'Hospital's Rule to evaluate $\lim_{x\to0^{+}}x^{x}$ .
(5)
(Total for Question 3 is 5 marks)
4.
Use L'Hospital's Rule to evaluate $\lim_{x\to0}\left(\dfrac1{x^2}-\dfrac{\cot x}{x}\right)$ .
(6)
(Total for Question 4 is 6 marks)
5.
Explain why L'Hospital's Rule cannot be used directly to evaluate $\lim_{x\to0}\dfrac{x^2+\cos x}{x+1}$ , and state the value of the limit.
(4)
(Total for Question 5 is 4 marks)

FP1-2.5 · The Weierstrass substitution for integration.

Explanation

The Weierstrass, or tangent half-angle, substitution turns any rational function of $\sin x$ and $\cos x$ into a rational function of $t$ .
Put $t=\tan\tfrac{x}{2}$ , so that $\sin x=\dfrac{2t}{1+t^2}$ , $\cos x=\dfrac{1-t^2}{1+t^2}$ and, differentiating $t$ with respect to $x$ , $\dfrac{\mathrm{d}t}{\mathrm{d}x}=\tfrac12\sec^2\tfrac{x}{2}=\tfrac12\left(1+t^2\right)$ , giving $\mathrm{d}x=\dfrac{2\,\mathrm{d}t}{1+t^2}$ .
That factor of $\dfrac{2}{1+t^2}$ almost always cancels against the $1+t^2$ produced by the trigonometric terms, leaving a straightforward rational integral.
For a definite integral, change the limits with the substitution rather than converting back: $x=0$ gives $t=0$ and $x=\tfrac{\pi}{2}$ gives $t=1$ .
The substitution is invalid across $x=\pi$ , where $t$ becomes infinite, so an integral spanning that point must be split.

Worked example

Use the substitution $t=\tan\tfrac{x}{2}$ to find $\displaystyle\int\frac{\mathrm{d}x}{1+\cos x}$ .

1. $1+\cos x=1+\dfrac{1-t^2}{1+t^2}=\dfrac{2}{1+t^2}$ .
2.With $\mathrm{d}x=\dfrac{2\,\mathrm{d}t}{1+t^2}$ the integral becomes $\displaystyle\int\frac{1+t^2}{2}\cdot\frac{2\,\mathrm{d}t}{1+t^2}=\int\mathrm{d}t$ .
3.So the integral is $t+c$ .

Answer: $\displaystyle\int\frac{\mathrm{d}x}{1+\cos x}=\tan\frac{x}{2}+c$ .

Common mistakes

Don't fall into the trap of forgetting to replace $\mathrm{d}x$ by $\dfrac{2\,\mathrm{d}t}{1+t^2}$ .
Don't fall into the trap of converting back to $x$ in a definite integral instead of changing the limits, and then using the wrong branch of $\arctan$ .
Don't fall into the trap of applying the substitution across $x=\pi$ , where $t$ is undefined.

Exam tip

Write the three substitution formulae and $\mathrm{d}x$ at the top of your working; every mark in these questions flows from getting that $\dfrac{2}{1+t^2}$ right.

Tier 1 · Easy

1.
Use the substitution $t=\tan\tfrac{x}{2}$ to show that $\displaystyle\int\operatorname{cosec}x\,\mathrm{d}x=\ln\left|\tan\frac{x}{2}\right|+c$ .
(4)
(Total for Question 1 is 4 marks)
2.
Use the substitution $t=\tan\tfrac{x}{2}$ to find $\displaystyle\int\frac{\mathrm{d}x}{1+\sin x}$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Use the substitution $t=\tan\tfrac{x}{2}$ to find $\displaystyle\int\frac{\mathrm{d}x}{5+4\cos x}$ .
(5)
(Total for Question 1 is 5 marks)
2.
Use the substitution $t=\tan\tfrac{x}{2}$ to evaluate $\displaystyle\int_{0}^{\pi/2}\frac{\mathrm{d}x}{1+\sin x+\cos x}$ , giving your answer in exact form.
(5)
(Total for Question 2 is 5 marks)
3.
Use the substitution $t=\tan\tfrac{x}{2}$ to evaluate $\displaystyle\int_{0}^{\pi/2}\frac{\mathrm{d}x}{5-3\cos x}$ , giving your answer in exact form.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Use the substitution $t=\tan\tfrac{x}{2}$ to evaluate $\displaystyle\int_{0}^{\pi/2}\frac{\mathrm{d}x}{2+\cos x}$ , giving your answer in exact form.
(6)
(Total for Question 1 is 6 marks)
2.
Use the substitution $t=\tan\tfrac{x}{2}$ to show that $\displaystyle\int\sec x\,\mathrm{d}x=\ln\left|\frac{1+t}{1-t}\right|+c$ , and deduce that this agrees with the standard result $\ln\left|\sec x+\tan x\right|+c$ .
(6)
(Total for Question 2 is 6 marks)
3.
Use the substitution $t=\tan\tfrac{x}{2}$ to evaluate $\displaystyle\int_{0}^{\pi/2}\frac{\mathrm{d}x}{4\cos x+3\sin x+5}$ .
(6)
(Total for Question 3 is 6 marks)
4.
Use the substitution $t=\tan\tfrac{x}{2}$ to find $\displaystyle\int\frac{\mathrm{d}x}{\sin x+\cos x}$ , giving your answer in logarithmic form.
(6)
(Total for Question 4 is 6 marks)
5.
Explain why the substitution $t=\tan\tfrac{x}{2}$ cannot be applied in a single step to $\displaystyle\int_{0}^{3\pi/2}\frac{\mathrm{d}x}{2+\cos x}$ , and describe how to evaluate that integral correctly.
(5)
(Total for Question 5 is 5 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FP1-2.1 · Derivation and use of Taylor series.

Tier 1 · Easy

Mark scheme for FP1-2.1 Tier 1 · Easy
Question	Scheme	Marks
1	$f(1)=f'(1)=f''(1)=\mathrm{e}$ $\mathrm{e}^{x}=\mathrm{e}+\mathrm{e}(x-1)+\dfrac{\mathrm{e}}{2}(x-1)^2+\cdots$	3
(3 marks)		3
Notes
Every derivative of $\mathrm{e}^{x}$ is $\mathrm{e}^{x}$ , so each value at $x=1$ is $\mathrm{e}$ and the coefficients are $\mathrm{e}/k!$ . Independent check: $\mathrm{e}^{x}=\mathrm{e}\cdot\mathrm{e}^{x-1}=\mathrm{e}\left(1+(x-1)+\tfrac{(x-1)^2}{2}+\cdots\right)$ , which is the same series.
2	$f(1)=0$ , $f'(1)=1$ , $f''(1)=-1$ , $f'''(1)=2$ $\ln x=(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}-\cdots$	4
(4 marks)		4
Notes
With $f'(x)=x^{-1}$ , $f''(x)=-x^{-2}$ and $f'''(x)=2x^{-3}$ , the values at $x=1$ are $1$ , $-1$ and $2$ , so the coefficients are $1$ , $-\tfrac12$ and $\tfrac26=\tfrac13$ . Independent check: substituting $u=x-1$ into the standard expansion $\ln(1+u)=u-\tfrac{u^2}{2}+\tfrac{u^3}{3}-\cdots$ gives the same result.

Tier 2 · Standard

Mark scheme for FP1-2.1 Tier 2 · Standard
Question	Scheme	Marks
1	$f\left(\tfrac{\pi}{3}\right)=\tfrac12$ , $f'\left(\tfrac{\pi}{3}\right)=-\tfrac{\sqrt3}{2}$ , $f''\left(\tfrac{\pi}{3}\right)=-\tfrac12$ $\cos x=\dfrac12-\dfrac{\sqrt3}{2}\left(x-\tfrac{\pi}{3}\right)-\dfrac14\left(x-\tfrac{\pi}{3}\right)^2+\cdots$	4
(4 marks)		4
Notes
Here $f'(x)=-\sin x$ and $f''(x)=-\cos x$ , so the values at $\tfrac{\pi}{3}$ are $\tfrac12$ , $-\tfrac{\sqrt3}{2}$ and $-\tfrac12$ , and the quadratic coefficient is $-\tfrac12\div2!=-\tfrac14$ . Independent check by the compound-angle route: $\cos\left(\tfrac{\pi}{3}+h\right)=\tfrac12\cos h-\tfrac{\sqrt3}{2}\sin h=\tfrac12\left(1-\tfrac{h^2}{2}\right)-\tfrac{\sqrt3}{2}h+\cdots$ , matching term by term.
2	$f\left(\tfrac{\pi}{4}\right)=1$ $f'(x)=\sec^2x$ so $f'\left(\tfrac{\pi}{4}\right)=2$ $f''(x)=2\sec^2x\tan x$ so $f''\left(\tfrac{\pi}{4}\right)=4$ $\tan x=1+2\left(x-\tfrac{\pi}{4}\right)+2\left(x-\tfrac{\pi}{4}\right)^2+\cdots$	4
(4 marks)		4
Notes
Differentiating, $f'(x)=\sec^2x$ and $f''(x)=2\sec^2x\tan x$ , giving $2$ and $4$ at $x=\tfrac{\pi}{4}$ ; the quadratic coefficient is $4\div2!=2$ . Independent check: $\tan\left(\tfrac{\pi}{4}+h\right)=\dfrac{1+\tan h}{1-\tan h}=(1+h)(1+h+h^2)+\cdots=1+2h+2h^2+\cdots$ .
3	$f(0)=0$ , $f'(x)=-\tan x$ so $f'(0)=0$ $f''(x)=-\sec^2x$ so $f''(0)=-1$ $f'''(x)=-2\sec^2x\tan x$ so $f'''(0)=0$ ; $f^{(4)}(0)=-2$ $\ln(\cos x)=-\dfrac{x^2}{2}-\dfrac{x^4}{12}-\cdots$	5
(5 marks)		5
Notes
Repeated differentiation gives $f''(0)=-1$ and $f^{(4)}(0)=-2$ , with the odd derivatives vanishing because $\ln(\cos x)$ is even; the coefficients are $-\tfrac1{2!}$ and $-\tfrac2{4!}=-\tfrac1{12}$ . Independent check by substitution: $\ln(\cos x)=\ln\left(1-\tfrac{x^2}{2}+\tfrac{x^4}{24}\right)=\left(-\tfrac{x^2}{2}+\tfrac{x^4}{24}\right)-\tfrac12\left(-\tfrac{x^2}{2}\right)^2+\cdots=-\tfrac{x^2}{2}-\tfrac{x^4}{12}+\cdots$ .

Tier 3 · Hard

Mark scheme for FP1-2.1 Tier 3 · Hard
Question	Scheme	Marks
1	$f(0)=1$ and $f'(x)=\sec x\tan x$ gives $f'(0)=0$ $f''(x)=\sec x\tan^2x+\sec^3x$ gives $f''(0)=1$ $f'''(0)=0$ by oddness of the third derivative $f^{(4)}(0)=5$ $\sec x=1+\dfrac{x^2}{2!}+\dfrac{5x^4}{4!}=1+\dfrac{x^2}{2}+\dfrac{5x^4}{24}$	6
(6 marks)		6
Notes
Differentiate four times, using $f''=\sec x\tan^2x+\sec^3x$ and its derivative, and evaluate at $x=0$ to get $1,0,1,0,5$ . Dividing by the factorials gives the stated series. Independent check by reciprocal series: $\sec x=\left(1-\tfrac{x^2}{2}+\tfrac{x^4}{24}\right)^{-1}=1+\left(\tfrac{x^2}{2}-\tfrac{x^4}{24}\right)+\left(\tfrac{x^2}{2}\right)^2+\cdots=1+\tfrac{x^2}{2}+\tfrac{5x^4}{24}+\cdots$ .
2	$\sin x=x-\dfrac{x^3}{6}+\cdots$ $\mathrm{e}^{u}=1+u+\dfrac{u^2}{2}+\dfrac{u^3}{6}+\cdots$ with $u=\sin x$ The $x^3$ terms are $-\dfrac16$ from $u$ and $+\dfrac16$ from $\dfrac{u^3}{6}$ $\mathrm{e}^{\sin x}=1+x+\dfrac{x^2}{2}+0\cdot x^3+\cdots$ , so the coefficient of $x^3$ is $0$	6
(6 marks)		6
Notes
Substitute $u=x-\tfrac{x^3}{6}$ into the exponential series: $u$ contributes $x-\tfrac{x^3}{6}$ , $\tfrac{u^2}{2}$ contributes $\tfrac{x^2}{2}$ and $\tfrac{u^3}{6}$ contributes $\tfrac{x^3}{6}$ , so the cubic terms cancel exactly. Independent check by differentiation: with $y=\mathrm{e}^{\sin x}$ , $y'=y\cos x$ , $y''=y(\cos^2x-\sin x)$ and $y'''=y(\cos^3x-3\sin x\cos x-\cos x)$ , giving $y'''(0)=1-0-1=0$ .
3	$\mathrm{e}^{x}=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots$ and $\sin x=x-\dfrac{x^3}{6}+\cdots$ Multiplying and keeping terms up to $x^3$ gives $x+x^2+\dfrac{x^3}{2}-\dfrac{x^3}{6}$ $\mathrm{e}^{x}\sin x=x+x^2+\dfrac{x^3}{3}+\cdots$	5
(5 marks)		5
Notes
Multiply the two standard series and collect powers: the $x^3$ contributions are $\tfrac{x^2}{2}\cdot x=\tfrac{x^3}{2}$ and $1\cdot\left(-\tfrac{x^3}{6}\right)$ , giving $\tfrac12-\tfrac16=\tfrac13$ . Independent check by Taylor's theorem: with $y=\mathrm{e}^{x}\sin x$ , $y(0)=0$ , $y'(0)=1$ , $y''=2\mathrm{e}^{x}\cos x$ so $y''(0)=2$ , and $y'''=2\mathrm{e}^{x}(\cos x-\sin x)$ so $y'''(0)=2$ ; the coefficients are $1$ , $\tfrac{2}{2!}=1$ and $\tfrac{2}{3!}=\tfrac13$ .
4	$f(4)=2$ , $f'(x)=\tfrac12x^{-1/2}$ so $f'(4)=\tfrac14$ $f''(x)=-\tfrac14x^{-3/2}$ so $f''(4)=-\tfrac1{32}$ $\sqrt{x}=2+\dfrac{x-4}{4}-\dfrac{(x-4)^2}{64}+\cdots$ At $x=4.2$ the estimate is $2+0.05-0.000625=2.049375$	6
(6 marks)		6
Notes
Differentiating twice gives $f'(4)=\tfrac14$ and $f''(4)=-\tfrac1{32}$ , so the quadratic coefficient is $-\tfrac1{64}$ . Substituting $x-4=0.2$ gives $2+0.05-0.000625=2.049375$ . Independent check: squaring, $2.049375^2=4.199938$ , and the true value $\sqrt{4.2}=2.0493902$ exceeds the estimate by $1.52\times10^{-5}$ , which matches the neglected cubic term $\tfrac{f'''(4)}{3!}(0.2)^3=\tfrac1{512}(0.008)=1.56\times10^{-5}$ .
5	$f\left(\tfrac{\pi}{2}\right)=1$ , $f'\left(\tfrac{\pi}{2}\right)=0$ , $f''\left(\tfrac{\pi}{2}\right)=-1$ , $f'''\left(\tfrac{\pi}{2}\right)=0$ , $f^{(4)}\left(\tfrac{\pi}{2}\right)=1$ $\sin x=1-\dfrac{1}{2}\left(x-\tfrac{\pi}{2}\right)^2+\dfrac{1}{24}\left(x-\tfrac{\pi}{2}\right)^4-\cdots$ Writing $h=x-\tfrac{\pi}{2}$ gives $\sin x=\cos h$ , which is an even function of $h$ Hence every odd-power coefficient is zero	6
(6 marks)		6
Notes
The derivatives of $\sin x$ cycle as $\cos x,-\sin x,-\cos x,\sin x$ , giving values $0,-1,0,1$ at $\tfrac{\pi}{2}$ , so the coefficients are $-\tfrac1{2!}$ and $\tfrac1{4!}$ . Independent check: with $h=x-\tfrac{\pi}{2}$ , $\sin x=\sin\left(\tfrac{\pi}{2}+h\right)=\cos h=1-\tfrac{h^2}{2}+\tfrac{h^4}{24}-\cdots$ , which reproduces the series and shows directly that only even powers of $h$ can occur.

FP1-2.2 · Use of series expansions to find limits.

Tier 1 · Easy

Mark scheme for FP1-2.2 Tier 1 · Easy
Question	Scheme	Marks
1	$1-\cos x=\dfrac{x^2}{2}-\dfrac{x^4}{24}+\cdots$ Dividing by $x^2$ gives $\dfrac12-\dfrac{x^2}{24}+\cdots$ The limit is $\dfrac12$	3
(3 marks)		3
Notes
Substitute $\cos x=1-\tfrac{x^2}{2}+\tfrac{x^4}{24}-\cdots$ , so the numerator is $\tfrac{x^2}{2}+O(x^4)$ and the quotient tends to $\tfrac12$ . Independent check numerically: at $x=0.01$ the quotient is $0.4999958$ .
2	$\mathrm{e}^{3x^2}=1+3x^2+\dfrac{9x^4}{2}+\cdots$ So the quotient is $3+\dfrac{9x^2}{2}+\cdots$ The limit is $3$	3
(3 marks)		3
Notes
Put $u=3x^2$ into $\mathrm{e}^{u}=1+u+\tfrac{u^2}{2}+\cdots$ ; the constant terms cancel and the leading surviving term is $3x^2$ . Independent check numerically: at $x=0.01$ the quotient is $3.000450$ .

Tier 2 · Standard

Mark scheme for FP1-2.2 Tier 2 · Standard
Question	Scheme	Marks
1	$x-\sin x=\dfrac{x^3}{6}-\dfrac{x^5}{120}+\cdots$ The quotient is $\dfrac16-\dfrac{x^2}{120}+\cdots$ The limit is $\dfrac16$	3
(3 marks)		3
Notes
Use $\sin x=x-\tfrac{x^3}{6}+\tfrac{x^5}{120}-\cdots$ , so the $x$ terms cancel and the leading term of the numerator is $\tfrac{x^3}{6}$ . Independent check numerically: at $x=0.05$ the quotient is $0.1666458$ .
2	$\tan x=x+\dfrac{x^3}{3}+\cdots$ and $\sin x=x-\dfrac{x^3}{6}+\cdots$ $\tan x-\sin x=\dfrac{x^3}{3}+\dfrac{x^3}{6}+\cdots=\dfrac{x^3}{2}+\cdots$ The limit is $\dfrac12$	4
(4 marks)		4
Notes
Subtract the two expansions: the linear terms cancel and the cubic coefficients add as $\tfrac13+\tfrac16=\tfrac12$ . Independent check by factorising: $\tan x-\sin x=\sin x\left(\sec x-1\right)\approx x\cdot\tfrac{x^2}{2}=\tfrac{x^3}{2}$ , and numerically at $x=0.05$ the quotient is $0.500313$ .
3	$\sin x=x-\dfrac{x^3}{6}+\cdots$ and $x\cos x=x-\dfrac{x^3}{2}+\cdots$ The difference is $\left(-\dfrac16+\dfrac12\right)x^3+\cdots=\dfrac{x^3}{3}+\cdots$ The limit is $\dfrac13$	4
(4 marks)		4
Notes
Expand both terms to order $x^3$ ; the $x$ terms cancel and the cubic coefficients are $-\tfrac16$ and $+\tfrac12$ , giving $\tfrac13$ . Independent check numerically: at $x=0.05$ the quotient is $0.333250$ .

Tier 3 · Hard

Mark scheme for FP1-2.2 Tier 3 · Hard
Question	Scheme	Marks
1	$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots$ Adding $-x+\dfrac{x^2}{2}$ cancels the first two terms The numerator is $\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots$ The limit is $\dfrac13$	5
(5 marks)		5
Notes
The first two terms of the logarithm series are removed by the added polynomial, leaving $\tfrac{x^3}{3}$ as the leading behaviour. Independent check numerically: at $x=0.01$ the quotient is $0.330853$ , consistent with $\tfrac13-\tfrac{x}{4}=0.330833$ .
2	$\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots$ $\mathrm{e}^{-x^2/2}=1-\dfrac{x^2}{2}+\dfrac{x^4}{8}-\cdots$ The difference is $\left(\dfrac1{24}-\dfrac18\right)x^4+\cdots=-\dfrac{x^4}{12}+\cdots$ The limit is $-\dfrac1{12}$	6
(6 marks)		6
Notes
Both expansions agree up to $x^2$ , so the quartic terms decide the limit: $\tfrac1{24}-\tfrac{3}{24}=-\tfrac2{24}=-\tfrac1{12}$ . Independent check numerically: at $x=0.1$ the quotient is $-0.083139$ , approaching $-\tfrac1{12}=-0.083333$ as $x\to0$ .
3	$\sin^2x=x^2-\dfrac{x^4}{3}+\cdots=x^2\left(1-\dfrac{x^2}{3}+\cdots\right)$ $\dfrac1{\sin^2x}=\dfrac1{x^2}\left(1+\dfrac{x^2}{3}+\cdots\right)=\dfrac1{x^2}+\dfrac13+\cdots$ Hence $\dfrac1{x^2}-\dfrac1{\sin^2x}=-\dfrac13+\cdots$ The limit is $-\dfrac13$	6
(6 marks)		6
Notes
Square the sine series to get $\sin^2x=x^2-\tfrac{x^4}{3}+\cdots$ , factor out $x^2$ and use the binomial expansion of $(1-u)^{-1}$ with $u=\tfrac{x^2}{3}$ ; the singular parts cancel and the constant term $-\tfrac13$ remains. Independent check numerically: at $x=0.05$ the expression equals $-0.333500$ .
4	$\dfrac{\ln(1+x)}{x}=1-\dfrac{x}{2}+\dfrac{x^2}{3}-\cdots$ $(1+x)^{1/x}=\mathrm{e}^{1-\frac{x}{2}+\cdots}=\mathrm{e}\left(1-\dfrac{x}{2}+\cdots\right)$ So the numerator is $-\dfrac{\mathrm{e}x}{2}+\cdots$ The limit is $-\dfrac{\mathrm{e}}{2}$	6
(6 marks)		6
Notes
Write $(1+x)^{1/x}=\exp\left(\tfrac{\ln(1+x)}{x}\right)$ , expand the exponent as $1-\tfrac{x}{2}+O(x^2)$ , and then expand the exponential about $1$ : $\mathrm{e}\cdot\mathrm{e}^{-x/2+\cdots}=\mathrm{e}\left(1-\tfrac{x}{2}+\cdots\right)$ . Independent check numerically: at $x=0.001$ the quotient is $-1.35790$ , against $-\tfrac{\mathrm{e}}{2}=-1.35914$ .
5	$a\sin x+b\tan x-x=(a+b-1)x+\left(-\dfrac{a}{6}+\dfrac{b}{3}\right)x^3+\cdots$ A finite limit requires $a+b-1=0$ The cubic coefficient gives $-\dfrac{a}{6}+\dfrac{b}{3}=\dfrac14$ Solving simultaneously gives $b=\dfrac{5}{6}$ and $a=\dfrac16$	6
(6 marks)		6
Notes
Expand to order $x^3$ : $a\sin x=ax-\tfrac{a}{6}x^3$ and $b\tan x=bx+\tfrac{b}{3}x^3$ . For the quotient to be finite the coefficient of $x$ must vanish, giving $a+b=1$ ; equating the cubic coefficient to $\tfrac14$ gives $2b-a=\tfrac32$ . Substituting $a=1-b$ gives $3b=\tfrac52$ , so $b=\tfrac56$ and $a=\tfrac16$ . Independent check numerically with these values at $x=0.05$ : the quotient is $0.250282$ .

FP1-2.3 · Leibnitz's theorem.

Tier 1 · Easy

Mark scheme for FP1-2.3 Tier 1 · Easy
Question	Scheme	Marks
1	With $u=\sin x$ and $v=x$ only the terms $r=0$ and $r=1$ survive $\binom40\sin^{(4)}x\cdot x+\binom41\sin^{(3)}x\cdot1$ $=x\sin x-4\cos x$	3
(3 marks)		3
Notes
Since $v=x$ has $v'=1$ and $v''=0$ , the sum has two terms. The fourth derivative of $\sin x$ is $\sin x$ and the third is $-\cos x$ , so the answer is $x\sin x+4(-\cos x)$ . Independent check by direct differentiation: successive derivatives are $\sin x+x\cos x$ , $2\cos x-x\sin x$ , $-3\sin x-x\cos x$ , $-4\cos x+x\sin x$ .
2	$\mathrm{e}^{-x}\left(x^3-12x^2+36x-24\right)$ At $x=1$ this is $\mathrm{e}^{-1}(1-12+36-24)=\mathrm{e}^{-1}$	4
(4 marks)		4
Notes
With $u=\mathrm{e}^{-x}$ , $u^{(k)}=(-1)^{k}\mathrm{e}^{-x}$ , and $v=x^3$ contributes only $r=0,1,2,3$ . The four terms are $x^3$ , $4(-1)(3x^2)$ , $6(+1)(6x)$ and $4(-1)(6)$ , all multiplied by $\mathrm{e}^{-x}$ , giving $\mathrm{e}^{-x}(x^3-12x^2+36x-24)$ . Independent check: substituting $x=1$ into $1-12+36-24$ gives $1$ , so the value is $\mathrm{e}^{-1}=0.3679$ , which matches a numerical fourth difference of the function near $x=1$ .

Tier 2 · Standard

Mark scheme for FP1-2.3 Tier 2 · Standard
Question	Scheme	Marks
1	Only $r=0,1,2$ contribute because $v=x^2$ has $v'''=0$ $\binom{n}{0}\mathrm{e}^{x}x^2+\binom{n}{1}\mathrm{e}^{x}(2x)+\binom{n}{2}\mathrm{e}^{x}(2)$ $=\mathrm{e}^{x}\left(x^2+2nx+n(n-1)\right)$	4
(4 marks)		4
Notes
Every derivative of $\mathrm{e}^{x}$ is itself, so the sum is $\mathrm{e}^{x}\left[x^2+n(2x)+\tfrac{n(n-1)}{2}(2)\right]$ . Independent check at $n=3$ : the formula gives $\mathrm{e}^{x}(x^2+6x+6)$ , which agrees with the worked example obtained by direct differentiation.
2	Only $r=0,1,2$ contribute $\binom60(-\sin x)x^2+\binom61(\cos x)(2x)+\binom62(\sin x)(2)$ $=-x^2\sin x+12x\cos x+30\sin x$	5
(5 marks)		5
Notes
The derivatives of $\sin x$ cycle with period four, so the sixth is $-\sin x$ , the fifth is $\cos x$ and the fourth is $\sin x$ . With $\binom61=6$ and $\binom62=15$ the terms are $-x^2\sin x$ , $6\cos x\cdot2x=12x\cos x$ and $15\sin x\cdot2=30\sin x$ . Independent check numerically at $x=1$ : the expression gives $-0.8415+6.4836+25.2441=30.8863$ , matching a sixth-order finite difference of $x^2\sin x$ at $x=1$ .
3	$\left(\ln x\right)^{(k)}=(-1)^{k-1}(k-1)!\,x^{-k}$ The four surviving terms are $\dfrac{24}{x^2}$ , $-\dfrac{90}{x^2}$ , $\dfrac{120}{x^2}$ and $-\dfrac{60}{x^2}$ The fifth derivative is $-\dfrac{6}{x^2}$ At $x=1$ the value is $-6$	5
(5 marks)		5
Notes
Take $u=\ln x$ and $v=x^3$ , so only $r=0,1,2,3$ contribute. The terms are $1\cdot\tfrac{24}{x^5}\cdot x^3$ , $5\cdot\left(-\tfrac{6}{x^4}\right)\cdot3x^2$ , $10\cdot\tfrac{2}{x^3}\cdot6x$ and $10\cdot\left(-\tfrac1{x^2}\right)\cdot6$ , that is $24,-90,120,-60$ over $x^2$ , summing to $-6x^{-2}$ . Independent check: $x^3\ln x$ has $\tfrac{\mathrm{d}^3}{\mathrm{d}x^3}=6\ln x+11$ , whose next two derivatives are $6x^{-1}$ and $-6x^{-2}$ .

Tier 3 · Hard

Mark scheme for FP1-2.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\left(\mathrm{e}^{ax}\right)^{(k)}=a^{k}\mathrm{e}^{ax}$ Only $r=0,1,2$ contribute $\mathrm{e}^{ax}\left(a^{n}x^2+2na^{n-1}x+n(n-1)a^{n-2}\right)$	5
(5 marks)		5
Notes
The sum is $\binom{n}{0}a^{n}\mathrm{e}^{ax}x^2+\binom{n}{1}a^{n-1}\mathrm{e}^{ax}(2x)+\binom{n}{2}a^{n-2}\mathrm{e}^{ax}(2)$ , and $\binom{n}{2}\times2=n(n-1)$ . Independent check: putting $a=1$ recovers $\mathrm{e}^{x}\left(x^2+2nx+n(n-1)\right)$ , and putting $n=2$ , $a=2$ gives $\mathrm{e}^{2x}(4x^2+8x+2)$ , which is the direct second derivative of $x^2\mathrm{e}^{2x}$ .
2	$\left(\ln x\right)^{(k)}=(-1)^{k-1}(k-1)!\,x^{-k}$ The three surviving terms are $(-1)^{n-1}(n-1)!\,x^{2-n}$ , $2n(-1)^{n-2}(n-2)!\,x^{2-n}$ and $n(n-1)(-1)^{n-3}(n-3)!\,x^{2-n}$ Factorising $(-1)^{n-1}(n-3)!\,x^{2-n}$ leaves $(n-1)(n-2)-2n(n-2)+n(n-1)$ $=n^2-3n+2-2n^2+4n+n^2-n=2$ Hence the $n$ th derivative is $2(-1)^{n-1}(n-3)!\,x^{2-n}$	6
(6 marks)		6
Notes
Apply Leibnitz with $u=\ln x$ and $v=x^2$ ; only $r=0,1,2$ contribute, giving three terms in $x^{2-n}$ . Taking out the common factor $(-1)^{n-1}(n-3)!\,x^{2-n}$ leaves a quadratic in $n$ that simplifies to the constant $2$ . Independent check at $n=3$ : the formula gives $2x^{-1}$ , and differentiating $x^2\ln x$ three times gives $2x\ln x+x$ , then $2\ln x+3$ , then $2x^{-1}$ .
3	$y'=\left(1-x^2\right)^{-1/2}$ , so $\left(1-x^2\right)\left(y'\right)^2=1$ Differentiating gives $\left(1-x^2\right)y''=xy'$ Differentiating $n$ times by Leibnitz, the left side gives $\left(1-x^2\right)y^{(n+2)}-2nxy^{(n+1)}-n(n-1)y^{(n)}$ The right side gives $xy^{(n+1)}+ny^{(n)}$ Rearranging yields $\left(1-x^2\right)y^{(n+2)}-(2n+1)xy^{(n+1)}-n^2y^{(n)}=0$	6
(6 marks)		6
Notes
From $y'=(1-x^2)^{-1/2}$ we get $(1-x^2)(y')^2=1$ ; differentiating and dividing by $2y'$ gives $(1-x^2)y''=xy'$ . Applying Leibnitz $n$ times to each product uses only the first three derivatives of $1-x^2$ and the first two of $x$ . Collecting terms gives $-2nx-x=-(2n+1)x$ and $-n(n-1)-n=-n^2$ . Independent check at $n=1$ with $y=\arcsin x$ : $y''=x(1-x^2)^{-3/2}$ and $y'''=(1+2x^2)(1-x^2)^{-5/2}$ , and substituting into $(1-x^2)y'''-3xy''-y'$ gives $0$ identically.
4	$y'=\dfrac{\mathrm{e}^{\arctan x}}{1+x^2}$ , so $\left(1+x^2\right)y'=y$ Differentiating $n$ times by Leibnitz gives $\left(1+x^2\right)y^{(n+1)}+2nxy^{(n)}+n(n-1)y^{(n-1)}$ This equals $y^{(n)}$ Rearranging gives $\left(1+x^2\right)y^{(n+1)}+(2nx-1)y^{(n)}+n(n-1)y^{(n-1)}=0$	6
(6 marks)		6
Notes
The chain rule gives $y'=y/(1+x^2)$ , hence $(1+x^2)y'=y$ . Differentiating the left side $n$ times by Leibnitz uses $\left(1+x^2\right)'=2x$ and $\left(1+x^2\right)''=2$ , so only $r=0,1,2$ contribute; the right side becomes $y^{(n)}$ . Independent check at $n=1$ and $x=0$ : $y(0)=1$ , $y'(0)=1$ , and differentiating $(1+x^2)y'=y$ once gives $(1+x^2)y''+2xy'=y'$ , so $y''(0)=1$ , consistent with the stated relation.
5	At $x=0$ the relation gives $y^{(n+2)}(0)=n^2y^{(n)}(0)$ $y(0)=0$ , $y'(0)=1$ , $y''(0)=0$ $y'''(0)=1^2\times1=1$ and $y^{(5)}(0)=3^2\times1=9$ All even derivatives vanish $\arcsin x=x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+\cdots$	6
(6 marks)		6
Notes
Setting $x=0$ collapses the recurrence to $y^{(n+2)}(0)=n^2y^{(n)}(0)$ . Starting from $y'(0)=1$ gives $y'''(0)=1$ and $y^{(5)}(0)=9$ , while $y''(0)=0$ makes every even derivative zero. Dividing by factorials gives $\tfrac{1}{3!}=\tfrac16$ and $\tfrac{9}{5!}=\tfrac{3}{40}$ . Independent check by binomial integration: $\arcsin x=\int_0^x\left(1-u^2\right)^{-1/2}\mathrm{d}u=\int_0^x\left(1+\tfrac{u^2}{2}+\tfrac{3u^4}{8}+\cdots\right)\mathrm{d}u=x+\tfrac{x^3}{6}+\tfrac{3x^5}{40}+\cdots$ .

FP1-2.4 · L'Hospital's Rule.

Tier 1 · Easy

Mark scheme for FP1-2.4 Tier 1 · Easy
Question	Scheme	Marks
1	The quotient is of the form $\dfrac00$ Differentiating gives $\dfrac{2\mathrm{e}^{2x}}{\cos x}$ At $x=0$ this is $2$	3
(3 marks)		3
Notes
Both numerator and denominator vanish at $x=0$ , so differentiate each once to get $\dfrac{2\mathrm{e}^{2x}}{\cos x}$ , which is continuous at $0$ with value $2$ . Independent check by series: $\dfrac{2x+2x^2+\cdots}{x-\tfrac{x^3}{6}+\cdots}=\dfrac{2+2x+\cdots}{1-\tfrac{x^2}{6}+\cdots}\to2$ .
2	The quotient is of the form $\dfrac00$ as $x\to1$ Differentiating gives $\dfrac{1/x}{1}$ At $x=1$ this is $1$	3
(3 marks)		3
Notes
Since $\ln1=0$ and $1-1=0$ , differentiate to get $\dfrac1x$ , whose value at $x=1$ is $1$ . Independent check by series: putting $x=1+u$ gives $\dfrac{\ln(1+u)}{u}=\dfrac{u-\tfrac{u^2}{2}+\cdots}{u}\to1$ .

Tier 2 · Standard

Mark scheme for FP1-2.4 Tier 2 · Standard
Question	Scheme	Marks
1	Rewrite as $\dfrac{\ln x}{1/x}$ , which is of the form $\dfrac{-\infty}{\infty}$ Differentiating gives $\dfrac{1/x}{-1/x^2}=-x$ As $x\to0^{+}$ this tends to $0$	4
(4 marks)		4
Notes
The product form $0\times\infty$ is converted to a quotient by writing $x=\left(\tfrac1x\right)^{-1}$ ; one application of the rule reduces it to $-x$ , which tends to $0$ . Independent check numerically: at $x=10^{-4}$ , $x\ln x=-9.21\times10^{-4}$ , and at $x=10^{-6}$ it is $-1.38\times10^{-5}$ , decreasing in magnitude towards zero.
2	The quotient is of the form $\dfrac00$ Differentiating gives $\dfrac{-\cos x}{-2\cos x\sin x}=\dfrac1{2\sin x}$ At $x=\tfrac{\pi}{2}$ this is $\dfrac12$	4
(4 marks)		4
Notes
One application of the rule leaves $\dfrac{1}{2\sin x}$ after cancelling $\cos x$ , and substituting $x=\tfrac{\pi}{2}$ gives $\tfrac12$ . Independent check by algebra: $\cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x)$ , so the quotient is $\dfrac1{1+\sin x}$ , which tends to $\tfrac12$ .
3	Both terms vanish, so differentiate: $\dfrac{\sec^2x-1}{1-\cos x}$ , still $\dfrac00$ Differentiate again: $\dfrac{2\sec^2x\tan x}{\sin x}$ Writing $\tan x=\dfrac{\sin x}{\cos x}$ gives $\dfrac{2\sec^3x\sin x}{\sin x}=2\sec^3x$ At $x=0$ this is $2$	5
(5 marks)		5
Notes
Two applications of the rule reduce the quotient to $\dfrac{2\sec^2x\tan x}{\sin x}$ ; cancelling $\sin x$ leaves $2\sec^3x$ , whose value at $0$ is $2$ . Independent check by series: $\tan x-x=\tfrac{x^3}{3}+\cdots$ and $x-\sin x=\tfrac{x^3}{6}+\cdots$ , so the ratio tends to $\tfrac{1/3}{1/6}=2$ .

Tier 3 · Hard

Mark scheme for FP1-2.4 Tier 3 · Hard
Question	Scheme	Marks
1	Combining over a common denominator gives $\dfrac{\mathrm{e}^{x}-1-x}{x\left(\mathrm{e}^{x}-1\right)}$ , of the form $\dfrac00$ Differentiating gives $\dfrac{\mathrm{e}^{x}-1}{\mathrm{e}^{x}-1+x\mathrm{e}^{x}}$ , still $\dfrac00$ Differentiating again gives $\dfrac{\mathrm{e}^{x}}{2\mathrm{e}^{x}+x\mathrm{e}^{x}}$ At $x=0$ this is $\dfrac12$	5
(5 marks)		5
Notes
The $\infty-\infty$ form must first be combined over $x\left(\mathrm{e}^{x}-1\right)$ ; two applications of the rule then give $\dfrac{\mathrm{e}^{x}}{(2+x)\mathrm{e}^{x}}=\dfrac1{2+x}$ , which tends to $\tfrac12$ . Independent check by series: the numerator is $\tfrac{x^2}{2}+\cdots$ and the denominator is $x^2+\cdots$ , so the ratio tends to $\tfrac12$ .
2	Let $y=\left(1+\dfrac3x\right)^{x}$ , so $\ln y=\dfrac{\ln\left(1+3/x\right)}{1/x}$ This is of the form $\dfrac00$ as $x\to\infty$ Differentiating gives $\dfrac{-3/x^2\div\left(1+3/x\right)}{-1/x^2}=\dfrac{3}{1+3/x}$ So $\ln y\to3$ and the limit is $\mathrm{e}^{3}$	5
(5 marks)		5
Notes
Take logarithms to convert the $1^{\infty}$ form into a $\tfrac00$ quotient, apply the rule once, and simplify; the $x^{-2}$ factors cancel, leaving $\dfrac{3}{1+3/x}\to3$ , so $y\to\mathrm{e}^{3}$ . Independent check numerically: at $x=10^{6}$ , $\left(1+3\times10^{-6}\right)^{10^{6}}=20.0854$ , against $\mathrm{e}^{3}=20.0855$ .
3	Let $y=x^{x}$ , so $\ln y=x\ln x=\dfrac{\ln x}{1/x}$ This is of the form $\dfrac{-\infty}{\infty}$ Differentiating gives $-x$ , so $\ln y\to0$ Hence $y\to\mathrm{e}^{0}=1$	5
(5 marks)		5
Notes
The $0^0$ form is handled by logarithms: $\ln y=x\ln x$ , which the rule shows tends to $0$ , so $y\to1$ . Independent check numerically: $0.01^{0.01}=0.9550$ , $0.0001^{0.0001}=0.99908$ and $10^{-8}$ raised to itself is $0.99999982$ .
4	Combining gives $\dfrac{\sin x-x\cos x}{x^2\sin x}$ , of the form $\dfrac00$ Differentiating gives $\dfrac{x\sin x}{2x\sin x+x^2\cos x}$ Cancelling $x$ gives $\dfrac{\sin x}{2\sin x+x\cos x}$ , still $\dfrac00$ Differentiating gives $\dfrac{\cos x}{3\cos x-x\sin x}$ At $x=0$ this is $\dfrac13$	6
(6 marks)		6
Notes
Write the difference over the common denominator $x^2\sin x$ ; the numerator differentiates to $x\sin x$ and the denominator to $2x\sin x+x^2\cos x$ , and after cancelling $x$ one further application gives $\dfrac{\cos x}{3\cos x-x\sin x}\to\tfrac13$ . Independent check by series: $\sin x-x\cos x=\tfrac{x^3}{3}+\cdots$ and $x^2\sin x=x^3+\cdots$ , so the ratio tends to $\tfrac13$ .
5	At $x=0$ the numerator is $1$ and the denominator is $1$ The quotient is therefore not of the form $\dfrac00$ or $\dfrac{\infty}{\infty}$ Substituting directly gives the limit $1$ Differentiating numerator and denominator would give $\dfrac{2x-\sin x}{1}\to0$ , which is wrong	4
(4 marks)		4
Notes
L'Hospital's Rule requires an indeterminate form; here both parts are continuous and non-zero at $x=0$ , so the limit is obtained by direct substitution and equals $\tfrac11=1$ . Independent check numerically: at $x=0.01$ the quotient is $\dfrac{0.0001+0.99995}{1.01}=0.99015$ , and at $x=0.0001$ it is $0.99990$ , confirming the limit $1$ rather than the $0$ that misapplying the rule would suggest.

FP1-2.5 · The Weierstrass substitution for integration.

Tier 1 · Easy

Mark scheme for FP1-2.5 Tier 1 · Easy
Question	Scheme	Marks
1	$\operatorname{cosec}x=\dfrac{1+t^2}{2t}$ and $\mathrm{d}x=\dfrac{2\,\mathrm{d}t}{1+t^2}$ The integral becomes $\displaystyle\int\frac{\mathrm{d}t}{t}$ $=\ln\|t\|+c=\ln\left\|\tan\dfrac{x}{2}\right\|+c$	4
(4 marks)		4
Notes
The $1+t^2$ from $\operatorname{cosec}x$ cancels against the one in $\mathrm{d}x$ and the twos cancel, leaving $\int t^{-1}\,\mathrm{d}t$ . Independent check by differentiation: $\dfrac{\mathrm{d}}{\mathrm{d}x}\ln\left\|\tan\tfrac{x}{2}\right\|=\dfrac{\tfrac12\sec^2\frac{x}{2}}{\tan\frac{x}{2}}=\dfrac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\operatorname{cosec}x$ .
2	$1+\sin x=\dfrac{1+2t+t^2}{1+t^2}=\dfrac{(1+t)^2}{1+t^2}$ The integral becomes $\displaystyle\int\frac{2\,\mathrm{d}t}{(1+t)^2}$ $=-\dfrac{2}{1+t}+c=-\dfrac{2}{1+\tan\frac{x}{2}}+c$	4
(4 marks)		4
Notes
The numerator $1+t^2$ cancels with the denominator of $\mathrm{d}x$ , leaving $\int2(1+t)^{-2}\,\mathrm{d}t=-2(1+t)^{-1}+c$ . Independent check by differentiation: $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(-\dfrac{2}{1+t}\right)=\dfrac{2}{(1+t)^2}\cdot\dfrac{1+t^2}{2}=\dfrac{1+t^2}{(1+t)^2}=\dfrac{1}{1+\sin x}$ .

Tier 2 · Standard

Mark scheme for FP1-2.5 Tier 2 · Standard
Question	Scheme	Marks
1	$5+4\cos x=\dfrac{5\left(1+t^2\right)+4\left(1-t^2\right)}{1+t^2}=\dfrac{9+t^2}{1+t^2}$ The integral becomes $\displaystyle\int\frac{2\,\mathrm{d}t}{9+t^2}$ $=\dfrac23\arctan\dfrac{t}{3}+c$ $=\dfrac23\arctan\left(\dfrac13\tan\dfrac{x}{2}\right)+c$	5
(5 marks)		5
Notes
After substitution the $1+t^2$ factors cancel and the integrand is $\dfrac{2}{t^2+9}$ , a standard arctangent form with $a=3$ . Independent check by differentiation: $\dfrac{\mathrm{d}}{\mathrm{d}x}\left[\tfrac23\arctan\tfrac{t}{3}\right]=\tfrac23\cdot\dfrac{1/3}{1+t^2/9}\cdot\dfrac{1+t^2}{2}=\dfrac{1+t^2}{9+t^2}=\dfrac1{5+4\cos x}$ .
2	$1+\sin x+\cos x=\dfrac{\left(1+t^2\right)+2t+\left(1-t^2\right)}{1+t^2}=\dfrac{2+2t}{1+t^2}$ The integral becomes $\displaystyle\int_{0}^{1}\frac{\mathrm{d}t}{1+t}$ $=\left[\ln(1+t)\right]_{0}^{1}$ $=\ln2$	5
(5 marks)		5
Notes
The limits transform as $x=0\Rightarrow t=0$ and $x=\tfrac{\pi}{2}\Rightarrow t=1$ , and after cancelling $1+t^2$ the integrand is $\dfrac1{1+t}$ . Independent check numerically: evaluating the original integral by Simpson's rule with four strips on $\left[0,\tfrac{\pi}{2}\right]$ gives $0.69335$ , against $\ln2=0.69315$ .
3	$5-3\cos x=\dfrac{5\left(1+t^2\right)-3\left(1-t^2\right)}{1+t^2}=\dfrac{2+8t^2}{1+t^2}$ The integral becomes $\displaystyle\int_{0}^{1}\frac{2\,\mathrm{d}t}{2+8t^2}=\int_{0}^{1}\frac{\mathrm{d}t}{1+4t^2}$ $=\left[\tfrac12\arctan 2t\right]_{0}^{1}$ $=\dfrac12\arctan2$	5
(5 marks)		5
Notes
After substitution the integrand is $\dfrac{1}{1+4t^2}$ , whose antiderivative is $\tfrac12\arctan2t$ ; the limits are $t=0$ and $t=1$ . Independent check numerically: $\tfrac12\arctan2=0.553574$ , and a Simpson's rule estimate of the original integral with four strips gives $0.553475$ .

Tier 3 · Hard

Mark scheme for FP1-2.5 Tier 3 · Hard
Question	Scheme	Marks
1	$2+\cos x=\dfrac{2\left(1+t^2\right)+\left(1-t^2\right)}{1+t^2}=\dfrac{3+t^2}{1+t^2}$ The integral becomes $\displaystyle\int_{0}^{1}\frac{2\,\mathrm{d}t}{3+t^2}$ $=\left[\dfrac{2}{\sqrt3}\arctan\dfrac{t}{\sqrt3}\right]_{0}^{1}=\dfrac{2}{\sqrt3}\cdot\dfrac{\pi}{6}$ $=\dfrac{\pi\sqrt3}{9}$	6
(6 marks)		6
Notes
After substitution the integrand is $\dfrac{2}{t^2+3}$ with limits $t=0$ and $t=1$ ; since $\arctan\tfrac1{\sqrt3}=\tfrac{\pi}{6}$ , the value is $\dfrac{2}{\sqrt3}\cdot\dfrac{\pi}{6}=\dfrac{\pi}{3\sqrt3}=\dfrac{\pi\sqrt3}{9}$ . Independent check numerically: $\dfrac{\pi\sqrt3}{9}=0.604600$ , and Simpson's rule with four strips on the original integral gives $0.604620$ .
2	$\sec x=\dfrac{1+t^2}{1-t^2}$ , so the integral becomes $\displaystyle\int\frac{2\,\mathrm{d}t}{1-t^2}$ $\dfrac{2}{1-t^2}=\dfrac{1}{1-t}+\dfrac{1}{1+t}$ Integrating gives $-\ln\|1-t\|+\ln\|1+t\|+c=\ln\left\|\dfrac{1+t}{1-t}\right\|+c$ Since $\sec x+\tan x=\dfrac{1+t^2+2t}{1-t^2}=\dfrac{(1+t)^2}{(1-t)(1+t)}=\dfrac{1+t}{1-t}$ , the two answers are identical	6
(6 marks)		6
Notes
Substituting and cancelling gives $\int\dfrac{2\,\mathrm{d}t}{1-t^2}$ , which partial fractions turn into two logarithms. The deduction uses the t-formula identity $\sec x+\tan x=\dfrac{1+t}{1-t}$ proved from the same substitution. Independent check by differentiation: $\dfrac{\mathrm{d}}{\mathrm{d}x}\ln\left\|\dfrac{1+t}{1-t}\right\|=\left(\dfrac1{1+t}+\dfrac1{1-t}\right)\cdot\dfrac{1+t^2}{2}=\dfrac{2}{1-t^2}\cdot\dfrac{1+t^2}{2}=\sec x$ .
3	The denominator becomes $\dfrac{4\left(1-t^2\right)+6t+5\left(1+t^2\right)}{1+t^2}=\dfrac{t^2+6t+9}{1+t^2}$ $t^2+6t+9=(t+3)^2$ The integral becomes $\displaystyle\int_{0}^{1}\frac{2\,\mathrm{d}t}{(t+3)^2}$ $=\left[-\dfrac{2}{t+3}\right]_{0}^{1}=-\dfrac12+\dfrac23$ $=\dfrac16$	6
(6 marks)		6
Notes
The quadratic in $t$ is a perfect square, so the integral is elementary: $\int_0^1 2(t+3)^{-2}\,\mathrm{d}t=\left[-2(t+3)^{-1}\right]_0^1=\tfrac23-\tfrac12=\tfrac16$ . Independent check numerically: Simpson's rule with four strips on the original integral gives $0.166698$ , against $\tfrac16=0.166667$ .
4	$\sin x+\cos x=\dfrac{2t+1-t^2}{1+t^2}$ The integral becomes $\displaystyle\int\frac{2\,\mathrm{d}t}{1+2t-t^2}=\int\frac{2\,\mathrm{d}t}{2-(t-1)^2}$ Using $\displaystyle\int\frac{\mathrm{d}u}{a^2-u^2}=\frac1{2a}\ln\left\|\frac{a+u}{a-u}\right\|$ with $a=\sqrt2$ , $u=t-1$ $=\dfrac1{\sqrt2}\ln\left\|\dfrac{\sqrt2-1+t}{\sqrt2+1-t}\right\|+c$	6
(6 marks)		6
Notes
Completing the square gives $1+2t-t^2=2-(t-1)^2$ , a standard partial-fraction logarithm form. Independent check by differentiation: at $x=0.3$ the stated antiderivative has derivative $0.799452$ and $\dfrac1{\sin0.3+\cos0.3}=0.799452$ ; the same agreement holds at $x=0.7$ , where both equal $0.709693$ , and at $x=1.1$ , where both equal $0.743603$ .
5	$t=\tan\tfrac{x}{2}$ is undefined at $x=\pi$ , where $\tfrac{x}{2}=\tfrac{\pi}{2}$ On $\left[0,\tfrac{3\pi}{2}\right]$ the substitution is therefore discontinuous, and $t\to+\infty$ as $x\to\pi^{-}$ and $t\to-\infty$ as $x\to\pi^{+}$ Split the integral as $\displaystyle\int_{0}^{\pi}+\int_{\pi}^{3\pi/2}$ Evaluate each part as an improper $t$ -integral, over $[0,\infty)$ and $(-\infty,-1]$ respectively The original integrand is continuous throughout, so the two parts add to a finite value	5
(5 marks)		5
Notes
The substitution is a bijection only on $(-\pi,\pi)$ ; crossing $x=\pi$ sends $t$ through infinity, so a single antiderivative in $t$ jumps by a constant there and a blind application loses that jump. Splitting at $x=\pi$ and taking the two improper $t$ -integrals separately restores the correct value. Independent check: $\displaystyle\int_{0}^{3\pi/2}\frac{\mathrm{d}x}{2+\cos x}=\frac{2}{\sqrt3}\left(\pi-\arctan\frac{1}{\sqrt3}\right)=3.0230$ , whereas naively evaluating $\left[\tfrac{2}{\sqrt3}\arctan\tfrac{t}{\sqrt3}\right]$ between $t=0$ and $t=-1$ gives $-0.6046$ , which is not even positive.