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Date November 2020 Marks available 2 Reference code 20N.3.AHL.TZ0.Hca_4
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Write down Question number Hca_4 Adapted from N/A

Question

The function f is defined by f(x)=ln(1+x2) where -1<x<1.

The seventh derivative of f is given by f(7)(x)=1440x(x6-21x4+35x2-7)(1+x2)7.

Use the Maclaurin series for ln(1+x) to write down the first three non-zero terms of the Maclaurin series for f(x).

[2]
a.i.

Hence find the first three non-zero terms of the Maclaurin series for x1+x2.

[4]
a.ii.

Use your answer to part (a)(i) to write down an estimate for f(0.4).

[1]
b.

Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating f(0.4), using the first three non-zero terms of the Maclaurin series for f(x).

[6]
c.i.

With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for f(0.4).

[2]
c.ii.

Markscheme

substitution of x2 in ln(1+x)=x-x22+x33-        (M1)

x2-x42+x63       A1


[2 marks]

a.i.

ddx(ln(1+x2))=2x1+x2        (M1)


Note: Award (M1) if this is seen in part (a)(i).


attempt to differentiate their answer in part (a)        (M1)


2x1+x2=2x-4x32+6x53       M1


Note: Award M1 for equating their derivatives.


x1+x2=x-x3+x5       A1


[4 marks]

a.ii.

f(0.4)0.149         A1


Note: Accept an answer that rounds correct to 2 s.f. or better.


[1 mark]

b.

attempt to find the maximum of |f(7)(c)| for c[0,0.4]       (M1)

maximum of |f(7)(c)| occurs at c=0.199       (A1)

|f(7)(c)|<1232.97  (for all c]0,0.4[)       (A1)

use of x=0.4       (M1)

substitution of n=6 and a=0 and their value of x and their value of f(7)(c) into Lagrange error term       (M1)


Note: Award (M1) for substitution of n=3 and a=0 and their value of x and their value of f(4)(c) into Lagrange error term.


|R6(0.4)|<1232.97(0.4)77!

upper bound =0.000401         A1


Note: Accept an answer that rounds correct to 1 s.f or better.


[6 marks]

c.i.

f(7)(c)<0  (for all c]0,0.4[)       R1


Note: Accept R6(c)<0 or “the error term is negative”.


the answer in (b) is an overestimate       A1


Note: The A1 is dependent on the R1.


[2 marks]

c.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Topic 5 —Calculus » AHL 5.19—Maclaurin series
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Topic 5 —Calculus

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