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 p∈R.
The function f and its derivatives satisfy
(1−x2)f(n+2)(x)−(2n+1)xf(n+1)(x)+(p2−n2)f(n)(x)=0, n∈N
where f(n)(x) denotes the n th derivative of f(x) and f(0)(x) is f(x).
Show that f′(0)=p.
Show that f(n+2)(0)=(n2−p2)f(n)(0).
For p∈R∖{±1, ±3}, show that the Maclaurin series for f(x), up to and including the x5 term, is
px+p(1−p2)3!x3+p(9−p2)(1−p2)5!x5.
Hence or otherwise, find limx→0sin(parcsinx)x.
If p is an odd integer, prove that the Maclaurin series for f(x) is a polynomial of degree p.
Markscheme
f′(x)=pcos(parcsinx)√1−x2 (M1)A1
Note: Award M1 for attempting to use the chain rule.
f′(0)=p AG
[2 marks]
EITHER
f(n+2)(0)+(p2−n2)f(n)(0)=0 A1
OR
for eg, (1−x2)f(n+2)(x)=(2n+1)xf(n+1)(x)−(p2−n2)f(n)(x) A1
Note: Award A1 for eg, (1−x2)f(n+2)(x)−(2n+1)xf(n+1)(x)=−(p2−n2)f(n)(x).
THEN
f(n+2)(0)=(n2−p2)f(n)(0) AG
[1 mark]
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]
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]
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]