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).
Hence find the first three non-zero terms of the Maclaurin series for x1+x2.
Use your answer to part (a)(i) to write down an estimate for f(0.4).
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).
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).
Markscheme
substitution of x2 in ln (1+x)=x-x22+x33-… (M1)
x2-x42+x63 A1
[2 marks]
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]
f(0.4)≈0.149 A1
Note: Accept an answer that rounds correct to 2 s.f. or better.
[1 mark]
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]
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]