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Maclaurin Series

Practice exam-style IB Math AA questions for Maclaurin Series, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

The function ff is defined by f(x)=e−2xf(x)=e^{-2x}.

A

the Maclaurin series for f(x)f(x) up to and including the term in x3x^3.

[2]
B

a polynomial approximation to e−15e^{-\frac{1}{5}}, giving your answer as a single fraction.

[2]
Question 2
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

Consider the function g(x)=1+3xg(x)=\sqrt{1+3x}.

A

the Maclaurin series for g(x)g(x) up to and including the term in x3x^3.

[3]
B

lim⁡x→01+3x−1−32xx2\displaystyle \lim_{x\to 0}\frac{\sqrt{1+3x}-1-\frac{3}{2}x}{x^2}.

[2]
Question 3
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the function f(x)=1−3xf(x)=\sqrt{1-3x}.

A

the Maclaurin series for f(x)f(x) up to and including the term in x3x^3.

[4]
B

the interval of convergence for this series.

[1]
Question 4
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=exsin⁡(2x)f(x)=e^x\sin(2x).

A

the Maclaurin series for f(x)f(x) up to and including the term in x4x^4.

[5]

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Question 5
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the function f(x)=arctan⁡(3x2)f(x)=\arctan(3x^2).

A

the Maclaurin series for f(x)f(x) up to and including the term in x6x^6.

[3]
B

the set of values of xx for which this series converges.

[1]
Question 6
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let

I(x)=∫0xln⁡(1+t)t dtI(x)=\int_0^x \frac{\ln(1+t)}{t}\,\mathrm{d}t
A

Using the Maclaurin series for ln⁡(1+t)\ln(1+t), the Maclaurin series for ln⁡(1+t)t\dfrac{\ln(1+t)}{t} up to and including the term in t3t^3.

[2]
B

the Maclaurin series for I(x)I(x) up to and including the term in x4x^4.

[3]
Question 7
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The geometric series is

11+x=1−x+x2−x3+x4−⋯\frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots
A

By differentiating term by term, the Maclaurin series for 1(1+x)2\dfrac{1}{(1+x)^2} up to and including the term in x4x^4.

[4]
B

the interval of convergence of this series.

[1]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A

lim⁡x→0ln⁡(1+2x)−sin⁡(2x)x2\lim_{x\to 0}\frac{\ln(1+2x)-\sin(2x)}{x^2}
[4]

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Question 9
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let f(x)=e2xln⁡(1+x)f(x)=e^{2x}\ln(1+x).

A

the Maclaurin series for f(x)f(x) up to and including the term in x4x^4.

[4]
B

Use your series to ∫00.2e2xln⁡(1+x) dx\displaystyle \int_0^{0.2} e^{2x}\ln(1+x)\,\mathrm{d}x.

[2]
Question 10
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let h(x)=arctan⁡x1+x2h(x)=\dfrac{\arctan x}{1+x^2}.

A

the Maclaurin series for h(x)h(x) up to and including the term in x5x^5.

[4]
B

Use your series to h(0.4)h(0.4).

[2]
Question 11
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let

I(x)=∫0x1−cos⁡(t2)t2 dtI(x)=\int_0^x \frac{1-\cos(t^2)}{t^2}\,\mathrm{d}t

where the integrand is defined by its limiting value at t=0t=0.

A

the Maclaurin series for I(x)I(x) up to and including the term in x11x^{11}.

[4]
B

Use your series to I(1)I(1).

[1]
Question 12
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let

f(x)=1ex+cos⁡xf(x)=\frac{1}{e^x+\cos x}
A

the Maclaurin series for f(x)f(x) up to and including the term in x3x^3.

[4]
B

f′′′(0)f'''(0).

[1]

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Question 13
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A function yy satisfies the differential equation

y′=y+xy'=y+x

with y(0)=2y(0)=2.

A

y′(0)y'(0), y′′(0)y''(0), y′′′(0)y'''(0) and y(4)(0)y^{(4)}(0).

[3]
B

the Maclaurin series for yy up to and including the term in x4x^4.

[2]
Question 14
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let

h(x)=ex(1+ax)−12h(x)=e^x(1+ax)^{-\frac{1}{2}}

where aa is a real constant. The coefficient of x2x^2 in the Maclaurin series for h(x)h(x) is 11.

A

the possible values of aa.

[6]
Question 15
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=excos⁡xf(x)=e^x\cos x.

A

the Maclaurin series for f(x)f(x) up to and including the term in x4x^4.

[5]
Question 16
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let

f(x)=ln⁡(1+x)1−xf(x)=\frac{\ln(1+x)}{1-x}
A

the Maclaurin series for f(x)f(x) up to and including the term in x3x^3.

[4]
B

lim⁡x→0f(x)−x−x22x3\lim_{x\to 0}\frac{f(x)-x-\dfrac{x^2}{2}}{x^3}
[2]

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Question 17
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The Maclaurin series for tan⁡x\tan x may be found using the series for sin⁡x\sin x and cos⁡x\cos x.

A

tan⁡x=x+x33+2x515+⋯\tan x=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\cdots.

[4]
B

Using this series, the smallest positive value of xx for which tan⁡(2x)−2x=0.100\tan(2x)-2x=0.100.

[2]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A function yy satisfies the differential equation

y′′+xy′+y=0y''+xy'+y=0

with y(0)=1y(0)=1 and y′(0)=0y'(0)=0.

A

y′′(0)y''(0) and y′′′(0)y'''(0).

[2]
B

the Maclaurin series for yy up to and including the term in x6x^6.

[4]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let L(x)=ln⁡(cos⁡x)L(x)=\ln(\cos x).

A

the Maclaurin series for L(x)L(x) up to and including the term in x6x^6.

[4]
B

lim⁡x→0ln⁡(cos⁡x)+x22+x412x6\displaystyle \lim_{x\to 0}\frac{\ln(\cos x)+\frac{x^2}{2}+\frac{x^4}{12}}{x^6}.

[2]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let p(x)=arctan⁡(ex−1)p(x)=\arctan(e^x-1).

A

the Maclaurin series for p(x)p(x) up to and including the term in x4x^4.

[4]
B

Using this series, the smaller positive solution (the solution close to x=0x=0) of arctan⁡(ex−1)=0.500\arctan(e^x-1)=0.500.

[2]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the third-degree Maclaurin polynomial for ln⁡(1−2x)\ln(1-2x).

A

this polynomial and the interval of convergence of the corresponding infinite series.

[3]
B

Use a GDC to the largest positive value of xx for which ∣ln⁡(1−2x)−P3(x)∣=0.005|\ln(1-2x)-P_3(x)|=0.005.

[2]
Question 22
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=(1+x)−12ln⁡(1+2x)f(x)=(1+x)^{-\frac12}\ln(1+2x), where xx is close to 00.

A
I.

the Maclaurin series for (1+x)−12(1+x)^{-\frac12} up to and including the term in x3x^3.

[2]
II.

the Maclaurin series for ln⁡(1+2x) \ln(1+2x) up to and including the term in x3x^3.

[2]
B

the Maclaurin series for f(x)f(x) up to and including the term in x3x^3.

[3]
C

lim⁡x→0f(x)−2x+3x2x3\lim_{x\to 0}\frac{f(x)-2x+3x^2}{x^3}

[2]
Question 23
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function GG is defined by G(x)=∫0xarctan⁡tt dtG(x)=\int_0^x \frac{\arctan t}{t}\,\mathrm{d}t where the integrand is defined by its limiting value at t=0t=0.

A
I.

Starting from 11+t2\dfrac{1}{1+t^2}, obtain the Maclaurin series for arctan⁡t\arctan t up to and including the term in t7t^7.

[3]
II.

the interval of convergence of the series for arctan⁡t\arctan t.

[1]
B

the Maclaurin series for G(x)G(x) up to and including the term in x7x^7.

[3]
C

Using the first four non-zero terms of this series, an approximate value for G(1)G(1) as a single fraction.

[2]
Question 24
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

For x≠0x\ne0, let f(x)=sin⁡xxf(x)=\dfrac{\sin x}{x}, and define f(0)=1f(0)=1.

A
I.

the Maclaurin series for f(x)f(x) up to and including the term in x6x^6.

[2]
II.

the Maclaurin series for 1f(x)\dfrac{1}{f(x)} up to and including the term in x4x^4. If you did not obtain the series in part (a)(i), use f(x)=1−x26+x4120+⋯f(x)=1-\dfrac{x^2}{6}+\dfrac{x^4}{120}+\cdots.

[3]
B

lim⁡x→0xsin⁡x−1−x26x4\lim_{x\to0}\frac{\dfrac{x}{\sin x}-1-\dfrac{x^2}{6}}{x^4}

[3]

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Question 25
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function FF is defined by F(x)=∫0x(1+t2)−32 dtF(x)=\int_0^x (1+t^2)^{-\frac32}\,\mathrm{d}t

A
I.

the Maclaurin series for (1+t2)−32(1+t^2)^{-\frac32} up to and including the term in t6t^6.

[3]
II.

the Maclaurin series for F(x)F(x) up to and including the term in x7x^7.

[2]
B
I.

F(x)=x1+x2F(x)=\dfrac{x}{\sqrt{1+x^2}}.

[2]
II.

lim⁡x→0F(x)−x+x32−3x58x7\lim_{x\to0}\frac{F(x)-x+\frac{x^3}{2}-\frac{3x^5}{8}}{x^7}

[2]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let f(x)=ex2cos⁡xf(x)=e^{x^2}\cos x

A
I.

the Maclaurin series for ex2e^{x^2} and for cos⁡x\cos x up to and including the term in x6x^6.

[2]
II.

the Maclaurin series for f(x)f(x) up to and including the term in x6x^6.

[2]
B

Using the series from part (a), approximate ∫01f(x)−1x2 dx\int_0^1 \frac{f(x)-1}{x^2}\,\mathrm{d}x

[3]
C

why every odd derivative of ff at 00 is zero.

[2]
Question 27
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A function yy satisfies

y′′+y=xy''+y=x

with y(0)=1y(0)=1 and y′(0)=0y'(0)=0.

A

By writing y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots, the Maclaurin series for yy up to and including the term in x5x^5.

[6]
Question 28
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The polynomial P5(x)P_5(x) is the Maclaurin polynomial for sin⁡x\sin x up to and including the term in x5x^5.

A
I.

P5(x)P_5(x).

[1]
II.

why the error function E(x)=sin⁡x−P5(x)E(x)=\sin x-P_5(x) is an odd function.

[2]
B

Use a GDC to the largest positive value of xx for which

∣sin⁡x−P5(x)∣=0.001|\sin x-P_5(x)|=0.001
[2]
C

the interval, symmetric about zero, on which P5(x)P_5(x) approximates sin⁡x\sin x with absolute error less than 0.0010.001.

[2]
D

why the Maclaurin series for sin⁡x\sin x itself converges for all real xx, but the polynomial P5(x)P_5(x) does not give a uniformly accurate approximation for all real xx.

[2]

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Question 29
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A function yy satisfies

y′=x+y2y'=x+y^2

with y(0)=1y(0)=1.

A

the Maclaurin series for yy up to and including the term in x4x^4.

[5]
B

Use this series to y(0.2)y(0.2).

[1]
Question 30
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For x≠0x\ne 0, define F(x)=(1+x)1/xF(x)=(1+x)^{1/x}, and define F(0)=eF(0)=e by its removable continuous extension. the Maclaurin series for F(x)F(x), up to and including the term in x3x^3, is e(1−x2+11x224−7x316+⋯ )e\left(1-\dfrac{x}{2}+\dfrac{11x^2}{24}-\dfrac{7x^3}{16}+\cdots\right).

A

the Maclaurin series for F(x)F(x), up to and including the term in x3x^3, is e(1−x2+11x224−7x316+⋯ )e\left(1-\dfrac{x}{2}+\dfrac{11x^2}{24}-\dfrac{7x^3}{16}+\cdots\right).

[5]
B

lim⁡x→0(1+x)1/x−ex\displaystyle \lim_{x\to 0}\frac{(1+x)^{1/x}-e}{x}.

[1]
Question 31
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

The value of π\pi can be approximated using π=6arcsin⁡(12)\pi=6\arcsin\left(\frac12\right). In this question, a Maclaurin series for arcsin⁡x\arcsin x is obtained by integration.

A
I.

Use the binomial expansion to 11−t2=1+t22+3t48+5t616+⋯\frac{1}{\sqrt{1-t^2}}=1+\frac{t^2}{2}+\frac{3t^4}{8}+\frac{5t^6}{16}+\cdots

[3]
II.

the Maclaurin series for arcsin⁡x\arcsin x up to and including the term in x7x^7.

[2]
B

Use the series from part (a) to π\pi.

[1]
C

The next term in the series for arcsin⁡x\arcsin x is 35x91152\frac{35x^9}{1152}. You may use that, for the terms of this series evaluated at x=12x=\frac12, the ratio of each term to the preceding term is less than 14\frac14. Use this to an upper bound for the error in your in part (b), and whether the is an underestimate or an overestimate.

[2]
Question 32
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A function yy satisfies the differential equation y′′+2xy′+2y=0y''+2xy'+2y=0 with y(0)=1y(0)=1 and y′(0)=0y'(0)=0.

A
I.

y′′(0)y''(0) and y′′′(0)y'''(0).

[3]
II.

y(4)(0)y^{(4)}(0) and y(5)(0)y^{(5)}(0). If you did not obtain y′′(0)=−2y''(0)=-2 and y′′′(0)=0y'''(0)=0, use these values in this part.

[3]
B

the Maclaurin series for yy up to and including the term in x5x^5.

[3]
C

Using this series, approximate ∫01y dx\displaystyle\int_0^1 y\,\mathrm{d}x.

[2]

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Question 33
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For a real constant aa, define fa(x)=eaxcos⁡xf_a(x)=e^{ax}\cos x

A
I.

the Maclaurin series for fa(x)f_a(x) up to and including the term in x4x^4.

[4]
II.

Given that the coefficient of x3x^3 is zero and a>0a>0, aa.

[2]
B

lim⁡x→0e3xcos⁡x−1−3x−x2x4\lim_{x\to0}\frac{e^{\sqrt3 x}\cos x-1-\sqrt3 x-x^2}{x^4}

[3]
Question 34
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A function yy satisfies the differential equation y′′=xyy''=xy with y(0)=0y(0)=0 and y′(0)=1y'(0)=1. Suppose y=a0+a1x+a2x2+a3x3+⋯y=a_0+a_1x+a_2x^2+a_3x^3+\cdots

A
I.

a0a_0, a1a_1 and a2a_2.

[2]
II.

By substituting the series into the differential equation, (n+2)(n+1)an+2=an−1for n≥1(n+2)(n+1)a_{n+2}=a_{n-1}\quad \text{for } n\ge1

[2]
B

the Maclaurin series for yy up to and including the term in x7x^7.

[3]
C

Using this series, an approximate value for y(1)y(1) as a single fraction.

[2]
Question 35
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let

f(x)=ln⁡(1+sin⁡x)f(x)=\ln(1+ \sin x)
A
I.

the Maclaurin series for sin⁡x\sin x up to and including the term in x5x^5.

[1]
II.

Using the Maclaurin series for ln⁡(1+u)\ln(1+u), the Maclaurin series for f(x)f(x) up to and including the term in x5x^5.

[5]
B

lim⁡x→0ln⁡(1+sin⁡x)−x+x22−x36+x412x5\lim_{x\to0}\frac{\ln(1+\sin x)-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}}{x^5}

[2]
Question 36
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let h(x)=arctan⁡(sin⁡x)h(x)=\arctan(\sin x)

A
I.

the Maclaurin series for arctan⁡u\arctan u up to and including the term in u5u^5.

[1]
II.

the Maclaurin series for h(x)h(x) up to and including the term in x5x^5.

[4]
B
I.

h′(0)h'(0), h′′′(0)h'''(0) and h(5)(0)h^{(5)}(0).

[3]
II.

lim⁡x→0h(x)−x+x32x5\lim_{x\to0}\frac{h(x)-x+\frac{x^3}{2}}{x^5}

[2]

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Question 37
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A function yy satisfies y′=(1+x)y+x2y'=(1+x)y+x^2 with y(0)=1y(0)=1.

A
I.

y′(0)y'(0), y′′(0)y''(0), y′′′(0)y'''(0) and y(4)(0)y^{(4)}(0).

[5]
II.

the Maclaurin series for yy up to and including the term in x4x^4. If you did not obtain the derivative values in part (a)(i), use y′(0)=1y'(0)=1, y′′(0)=2y''(0)=2, y′′′(0)=6y'''(0)=6 and y(4)(0)=12y^{(4)}(0)=12.

[2]
B

Using this series, approximate ∫01y dx\displaystyle\int_0^1 y\,\mathrm{d}x.

[2]
Question 38
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let L(x)=ln⁡(1+x1−x),−1<x<1L(x)=\ln\left(\frac{1+x}{1-x}\right),\qquad -1<x<1

A
I.

Using the Maclaurin series for ln⁡(1+x)\ln(1+x) and ln⁡(1−x)\ln(1-x), the Maclaurin series for L(x)L(x) up to and including the term in x5x^5.

[3]
II.

the interval of convergence of the series for L(x)L(x).

[1]
B

Let M(x)=1−x2 L(x)M(x)=\sqrt{1-x^2}\,L(x). the Maclaurin series for M(x)M(x) up to and including the term in x5x^5.

[4]
C

lim⁡x→0M(x)−2x+x33x5\lim_{x\to0}\frac{M(x)-2x+\frac{x^3}{3}}{x^5}

[2]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function ff is defined by

f(x)=e−x2cos(3x),x∈Rf(x)=e^{-x^2}\\cos(3x), \quad x\in\mathbb{R}

The polynomial P6(x)P_6(x) is the Maclaurin polynomial for ff up to and including the term in x6x^6.

A
I.

the Maclaurin series for e−x2e^{-x^2} and for cos⁡(3x)\cos(3x) up to and including terms in x6x^6.

[2]
II.

P6(x)P_6(x).

[3]
B

Use P6(x)P_6(x) to a polynomial approximation for

I(a)=∫0ae−x2cos⁡(3x) dxI(a)=\int_0^a e^{-x^2}\cos(3x)\,\mathrm{d}x
[2]
C

Using the approximation from part (b), the two positive values of aa for which I(a)=0.240I(a)=0.240.

[3]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

g(x)=ln⁡(1+sin⁡x)g(x)=\ln(1+\sin x)
A
I.

the Maclaurin series for sin⁡x\sin x up to and including the term in x5x^5.

[1]
II.

the Maclaurin series for g(x)g(x) up to and including the term in x5x^5.

[4]
B

Using your series, the smallest positive solution of ln⁡(1+sin⁡x)=0.250\ln(1+\sin x)=0.250.

[2]
C

the exact value of

lim⁡x→0g(x)−x+x22−x36+x412x5\lim_{x\to 0}\frac{g(x)-x+\dfrac{x^2}{2}-\dfrac{x^3}{6}+\dfrac{x^4}{12}}{x^5}
[3]

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Question 41
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A function yy satisfies the differential equation

y′′+xy′+2y=0y''+xy'+2y=0

with y(0)=1y(0)=1 and y′(0)=0y'(0)=0.

A
I.

y′′(0)y''(0) and y′′′(0)y'''(0).

[2]
II.

By writing y=a0+a1x+a2x2+⋯y=a_0+a_1x+a_2x^2+ \cdots, a recurrence relation for the coefficients.

[3]
B

the Maclaurin series for yy up to and including the term in x6x^6.

[3]
C

Using the polynomial found in part (b), the first positive value of xx for which y=0y=0.

[2]
Question 42
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Consider

F(x)=11+x+x2F(x)=\frac{1}{1+x+x^2}

and

J(x)=∫0xF(t) dtJ(x)=\int_0^x F(t)\,\mathrm{d}t
A
I.

By writing F(x)=a0+a1x+a2x2+⋯F(x)=a_0+a_1x+a_2x^2+ \cdots, a recurrence relation for the coefficients.

[3]
II.

write the Maclaurin series for F(x)F(x) up to and including the term in x6x^6.

[2]
B

Use the series in part (a) to the Maclaurin series for J(x)J(x) up to and including the term in x7x^7.

[2]
C

J(x)=23(arctan⁡(2x+13)−π6)J(x)=\frac{2}{\sqrt{3}}\left(\arctan\left(\frac{2x+1}{\sqrt{3}}\right)-\frac{\pi}{6}\right)
[2]
D

, using a GDC, the positive solution of J(x)=0.500J(x)=0.500. Then the value given by solving the polynomial approximation from part (b).

[2]
Question 43
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The inverse sine function may be written as

arcsin⁡x=∫0x11−t2 dt,∣x∣<1\arcsin x=\int_0^x \frac{1}{\sqrt{1-t^2}}\,\mathrm{d}t, \quad |x|<1
A
I.

Use the binomial expansion to the Maclaurin series for (1−x2)−12(1-x^2)^{-\frac12} up to and including the term in x6x^6.

[3]
II.

the Maclaurin series for arcsin⁡x\arcsin x up to and including the term in x7x^7.

[2]
B

Use the series in part (a) with x=12x=\dfrac12 to obtain an approximation for π\pi.

[2]
C

the absolute error in the approximation from part (b).

[1]
D

Using a GDC and the series generated by integrating the binomial expansion, the least odd power x2n+1x^{2n+1} that must be included so that the resulting approximation to π\pi has absolute error less than 10−610^{-6}.

[2]
Question 44
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

H(x)=ex2ln⁡(1+2x)H(x)=e^{x^2}\ln(1+2x)
A
I.

the Maclaurin series for ex2e^{x^2} and ln⁡(1+2x)\ln(1+2x) up to the terms needed to H(x)H(x) as far as x5x^5.

[2]
II.

the Maclaurin series for H(x)H(x) up to and including the term in x5x^5.

[3]
B

Use your series to

∫00.25ex2ln⁡(1+2x) dx\int_0^{0.25} e^{x^2}\ln(1+2x)\,\mathrm{d}x
[2]
C

Use a GDC to evaluate the integral in part (b) directly and the absolute error in the .

[2]

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Question 45
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A function yy satisfies

y′=1+xyy'=1+xy

with y(0)=1y(0)=1.

A
I.

Let y=a0+a1x+a2x2+⋯y=a_0+a_1x+a_2x^2+ \cdots. a recurrence relation for the coefficients.

[3]
II.

the Maclaurin series for yy up to and including the term in x6x^6.

[2]
B

Using the polynomial found in part (a), the positive value of xx for which y=2y=2.

[2]
C

The exact solution may be written in the form

y=ex2/2(1+∫0xe−t2/2 dt)y=e^{x^2/2}\left(1+\int_0^x e^{-t^2/2}\,\mathrm{d}t\right)

this expression and use a GDC to the positive value of xx for which the exact solution has value 22.

[3]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A damped oscillation is modelled by f(x)=e−xcos⁡(2x)f(x)=e^{-x}\cos(2x) for 0≤x≤0.50\le x\le 0.5. A fourth-degree Maclaurin polynomial P4(x)P_4(x) is to be used as a simple approximation to f(x)f(x) near x=0x=0.

A
I.

the Maclaurin series for e−xe^{-x} and cos⁡(2x)\cos(2x) up to and including the term in x4x^4.

[2]
II.

P4(x)P_4(x) for f(x)f(x).

[3]
B
I.

Use P4P_4 to f(0.30)f(0.30).

[2]
II.

Using a GDC, the largest value of kk in 0≤k≤0.50\le k\le 0.5 such that ∣f(x)−P4(x)∣≤0.005|f(x)-P_4(x)|\le 0.005 for all 0≤x≤k0\le x\le k. If you did not obtain P4(x)P_4(x), use P4(x)=1−x−3x22+11x36−7x424P_4(x)=1-x-\frac{3x^2}{2}+\frac{11x^3}{6}-\frac{7x^4}{24}.

[3]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For 0≤x≤0.50\le x\le 0.5, define A(x)=2arctan⁡x−arctan⁡(2x)A(x)=2\arctan x-\arctan(2x). This function represents the small difference between two angle measurements.

A
I.

the first four non-zero terms of the Maclaurin series for \rctanx\rctan x and for \rctan(2x)\rctan(2x).

[2]
II.

the first three non-zero terms in the Maclaurin series for A(x)A(x).

[3]
B

Use the series from part (a) to A(0.4)A(0.4).

[1]
C

that A(x)>0A(x)>0 for 0<x<0.50<x<0.5.

[2]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A function yy is defined by the differential equation y′=1+xyy'=1+xy with y(0)=0y(0)=0.

A
I.

By writing y=∑n=0∞anxny=\sum_{n=0}^{\infty}a_nx^n, a recurrence relation for the coefficients.

[3]
II.

the Maclaurin series for yy up to and including the term in x7x^7.

[2]
B

Use your series to y(0.6)y(0.6).

[1]
C

a general expression for the non-zero coefficients of the series.

[2]

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Question 49
HL • Paper 3
Hard
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HL • Paper 3
Hard
Calculator Permitted

The curve y=sec⁡xy=\sec x is to be approximated near x=0x=0 by an even polynomial S6(x)=1+ax2+bx4+cx6S_6(x)=1+ax^2+bx^4+cx^6.

Expression

Role

y=sec⁡xy=\sec x

Curve near x=0x=0

S6(x)=1+ax2+bx4+cx6S_6(x)=1+ax^2+bx^4+cx^6

Even polynomial approximation

A
I.

Use the identity cos⁡xsec⁡x=1\cos x\sec x=1 to aa, bb and cc.

[4]
B

Use S6S_6 to ∫00.4sec⁡x dx\int_0^{0.4}\sec x\,\mathrm{d}x.

[2]
C

Given that the next non-zero term in the Maclaurin series for sec⁡x\sec x is positive, whether your in part (b) is an underestimate or an overestimate. your answer by considering the sign of the entire omitted tail, not only the next term.

[2]
Question 50
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a real parameter aa, let Ha(x)=eaxln⁡(1+2x)H_a(x)=e^{ax}\ln(1+2x) A value of aa is chosen so that the quadratic term in the Maclaurin series of HaH_a is zero.

A
I.

the coefficient of x2x^2 in the Maclaurin series for Ha(x)H_a(x).

[3]
II.

the required value of aa, and the Maclaurin series for Ha(x)H_a(x) up to and including the term in x4x^4.

[2]
B

lim⁡x→0exln⁡(1+2x)−2xx3\lim_{x\to0}\frac{e^x\ln(1+2x)-2x}{x^3}

[1]
C

Using the polynomial found in part (a), the small positive solution (the solution closest to x=0x=0) of exln⁡(1+2x)=0.25e^x\ln(1+2x)=0.25. If you did not obtain a polynomial, use 2x+5x33−2x42x+\frac{5x^3}{3}-2x^4.

[2]
Question 51
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For small positive xx, the difference D(x)=sin⁡x−arctan⁡xD(x)=\sin x-\arctan x may be approximated by a polynomial.

A
I.

the Maclaurin series for D(x)D(x) up to and including the term in x5x^5.

[3]
B

the value of aa for which lim⁡x→0sin⁡x−arctan⁡x−ax3x5\lim_{x\to0}\frac{\sin x-\arctan x-ax^3}{x^5} is finite and non-zero, and the value of the limit.

[3]
C

Using the polynomial from part (a), the smallest positive solution of sin⁡x−arctan⁡x=0.020\sin x-\arctan x=0.020.

[2]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^{+}, define

Sn=4∑r=0n(−1)r((1/2)2r+12r+1+(1/3)2r+12r+1)S_n=4\sum_{r=0}^{n}(-1)^r\left(\frac{(1/2)^{2r+1}}{2r+1}+\frac{(1/3)^{2r+1}}{2r+1}\right)
A
I.

arctan⁡12+arctan⁡13=π4\arctan\dfrac{1}{2}+\arctan\dfrac{1}{3}=\dfrac{\pi}{4}.

[2]
II.

why SnS_n is an approximation to π\pi.

[2]
B

S2S_2 and the absolute error ∣S2−π∣|S_2-\pi|.

[3]
C

Using the alternating series error ,

∣Sn−π∣≤42n+3(122n+3+132n+3)|S_n-\pi|\le \frac{4}{2n+3}\left(\frac{1}{2^{2n+3}}+\frac{1}{3^{2n+3}}\right)
[2]
D

the least value of nn for which this error bound is less than 10−610^{-6}.

[2]

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Question 53
HL • Paper 2
Hard
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HL • Paper 2
Hard
Calculator Permitted

For sufficiently small xx, let ww be defined implicitly by

wew=xw e^w=x

Assume that

w=a1x+a2x2+a3x3+a4x4+⋯w=a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots
A
I.

Write the Maclaurin series for ewe^w up to and including the term in w4w^4.

[1]
II.

By substituting the assumed series into wew=xwe^w=x,

w=x−x2+3x32−8x43+⋯w=x-x^2+\frac{3x^3}{2}-\frac{8x^4}{3}+\cdots
[5]
B

Use this series to the solution of wew=0.300we^w=0.300.

[2]
C

Use a GDC to the actual solution of wew=0.300we^w=0.300 and the percentage error in the from part (b).

[2]
Question 54
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A rational approximation to exe^x is given by

R(x)=1+ax1+bx+cx2R(x)=\frac{1+ax}{1+bx+cx^2}

where aa, bb and cc are constants. The coefficients of xx, x2x^2 and x3x^3 in the Maclaurin series for R(x)R(x) are to match those of exe^x.

A
I.

the Maclaurin series for (1+bx+cx2)−1(1+bx+cx^2)^{-1} up to and including the term in x3x^3.

[3]
II.

a=13a=\dfrac13, b=−23b=-\dfrac23 and c=16c=\dfrac16.

[3]
B

Use R(1)R(1) to approximate ee.

[2]
C

the percentage error of this approximation with that of the third-degree Maclaurin polynomial for exe^x at x=1x=1.

[2]
Question 55
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A function yy is represented by a Maclaurin series and satisfies

xy′′+y′+xy=0xy''+y'+xy=0

with y(0)=1y(0)=1. Assume that y′(0)=0y'(0)=0.

A
I.

Let y=a0+a1x+a2x2+⋯y=a_0+a_1x+a_2x^2+ \cdots. a1=0a_1=0 and that, for n≥1n\ge 1,

(n+1)2an+1+an−1=0(n+1)^2a_{n+1}+a_{n-1}=0
[3]
II.

the Maclaurin series for yy up to and including the term in x8x^8.

[3]
B

Using this polynomial approximation, the first positive zero of yy.

[2]
C

why the approximation in part (b) should be treated with caution.

[2]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A function yy satisfies the differential equation y′′+xy′−3y=0y''+xy'-3y=0 with y(0)=1y(0)=1 and y′(0)=2y'(0)=2. Let y=∑n=0∞anxny=\sum_{n=0}^{\infty}a_nx^n

A
I.

an+2=3−n(n+2)(n+1)ana_{n+2}=\frac{3-n}{(n+2)(n+1)}a_n for n≥0n\ge0.

[3]
II.

the Maclaurin series for yy up to and including the term in x6x^6.

[2]
B

Use your series to y(0.4)y(0.4).

[1]
C

that all odd-power coefficients after a3a_3 are zero.

[3]

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Question 57
HL • Paper 3
Hard
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HL • Paper 3
Hard
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For x≠0x\ne0, define F(x)=ln⁡(1+x2)x2F(x)=\frac{\ln(1+x^2)}{x^2} and define F(0)F(0) by continuity. The integral I=∫00.8F(x) dxI=\int_0^{0.8}F(x)\,\mathrm{d}x is to be estimated using a Maclaurin series.

A
I.

the Maclaurin series for F(x)F(x).

[2]
II.

Use the first three non-zero terms to II.

[2]
B

The integrated series is alternating for 0≤x≤0.80\le x\le0.8. the minimum number of terms of the series for F(x)F(x) that must be integrated so that the alternating-series error bound for II is less than 10−410^{-4}.

[4]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A function yy satisfies y′′=xyy''=xy with y(0)=1y(0)=1 and y′(0)=−1y'(0)=-1. Let y=∑n=0∞anxny=\sum_{n=0}^{\infty}a_nx^n

A
I.

a recurrence relation for the coefficients ana_n.

[3]
II.

the Maclaurin series for yy up to and including the term in x6x^6.

[2]
B

Use your series to y(1)y(1).

[1]
C

that a3k+2=0a_{3k+2}=0 for all integers k≥0k\ge0.

[2]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The function D(x)=e−x2∫0xet2 dtD(x)=e^{-x^2}\int_0^x e^{t^2}\,\mathrm{d}t occurs in a model of diffusion. Its Maclaurin series may be found using products of series.

A
I.

the Maclaurin series for ∫0xet2 dt\int_0^x e^{t^2}\,\mathrm{d}t up to and including the term in x7x^7.

[2]
II.

the Maclaurin series for D(x)D(x) up to and including the term in x7x^7.

[3]
B

Use the series from part (a) to D(0.5)D(0.5).

[1]
C

DD satisfies the differential equation D′=1−2xDD'=1-2xD, and use this equation to the coefficient of x7x^7 in your series.

[3]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For x≠0x\ne0, define L(x)=ln⁡(sin⁡xx)L(x)=\ln\left(\frac{\sin x}{x}\right) with L(0)L(0) defined by continuity. This function is used to study the central peak of a wave pattern.

A
I.

sin⁡xx=1−x26+x4120−x65040+⋯\frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{5040}+\cdots

[2]
II.

the Maclaurin series for L(x)L(x) up to and including the term in x6x^6.

[3]
B

Use your answer to part (a) to the positive value of xx for which sin⁡xx=0.95\frac{\sin x}{x}=0.95.

[2]
C

why the Maclaurin series for L(x)L(x) contains only even powers of xx.

[1]

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Kinematics