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Date November 2017 Marks available 2 Reference code 17N.3ca.hl.TZ0.5
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term Hence or otherwise and Find Question number 5 Adapted from N/A

Question

Consider the function f(x)=sin(parcsinx), 1<x<1 and pR.

The function f and its derivatives satisfy

(1x2)f(n+2)(x)(2n+1)xf(n+1)(x)+(p2n2)f(n)(x)=0, nN

where f(n)(x) denotes the n th derivative of f(x) and f(0)(x) is f(x).

Show that f(0)=p.

[2]
a.

Show that f(n+2)(0)=(n2p2)f(n)(0).

[1]
b.

For pR{±1, ±3}, show that the Maclaurin series for f(x), up to and including the x5 term, is

px+p(1p2)3!x3+p(9p2)(1p2)5!x5.

[4]
c.

Hence or otherwise, find limx0sin(parcsinx)x.

[2]
d.

If p is an odd integer, prove that the Maclaurin series for f(x) is a polynomial of degree p.

[4]
e.

Markscheme

f(x)=pcos(parcsinx)1x2     (M1)A1

 

Note: Award M1 for attempting to use the chain rule.

 

f(0)=p     AG

[2 marks]

a.

EITHER

f(n+2)(0)+(p2n2)f(n)(0)=0     A1

OR

for eg, (1x2)f(n+2)(x)=(2n+1)xf(n+1)(x)(p2n2)f(n)(x)     A1

 

Note: Award A1 for eg, (1x2)f(n+2)(x)(2n+1)xf(n+1)(x)=(p2n2)f(n)(x).

 

THEN

f(n+2)(0)=(n2p2)f(n)(0)     AG

[1 mark]

b.

considering f and its derivatives at x=0     (M1)

f(0)=0 and f(0)=p from (a)     A1

f     A1

{f^{(3)}}(0) = (1 - {p^2}){f^{(1)}}(0) = (1 - {p^2})p,

{f^{(5)}}(0) = (9 - {p^2}){f^{(3)}}(0) = (9 - {p^2})(1 - {p^2})p     A1

 

Note:     Only award the last A1 if either {f^{(3)}}(0) = (1 - {p^2}){f^{(1)}}(0) and {f^{(5)}}(0) = (9 - {p^2}){f^{(3)}}(0) have been stated or the general Maclaurin series has been stated and used.

 

px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} + \frac{{p(9 - {p^2})(1 - {p^2})}}{{5!}}{x^5}     AG

[4 marks]

c.

METHOD 1

\mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} +  \ldots }}{3}     M1

= p     A1

METHOD 2

by l’Hôpital’s rule \mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{p\cos (p\arcsin x)}}{{\sqrt {1 - {x^2}} }}     M1

= p     A1

[2 marks]

d.

the coefficients of all even powers of x are zero     A1

the coefficient of {x^p} for (p odd) is non-zero (or equivalent eg,

the coefficients of all odd powers of x up to p are non-zero)     A1

{f^{(p + 2)}}(0) = ({p^2} - {p^2}){f^{(p)}}(0) = 0 and so the coefficient of {x^{p + 2}} is zero     A1

the coefficients of all odd powers of x greater than p + 2 are zero (or equivalent)     A1

so the Maclaurin series for f(x) is a polynomial of degree p     AG

[4 marks]

e.

Examiners report

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Syllabus sections

Topic 9 - Option: Calculus » 9.6
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