Cambridge International AS & A Level

*7325464498*

Cambridge International AS & A Level

FURTHER MATHEMATICS Paper 2 Further Pure Mathematics 2

You must answer on the question paper. You will need: List of formulae (MF19)

9231/22 October/November 2021

2 hours

INSTRUCTIONS Answer all questions. Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. Write your name, centre number and candidate number in the boxes at the top of the page. Write your answer to each question in the space provided. Do not use an erasable pen or correction fluid. Do not write on any bar codes. If additional space is needed, you should use the lined page at the end of this booklet; the question

number or numbers must be clearly shown. You should use a calculator where appropriate. You must show all necessary working clearly; no marks will be given for unsupported answers from a

calculator. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in

degrees, unless a different level of accuracy is specified in the question.

INFORMATION The total mark for this paper is 75. The number of marks for each question or part question is shown in brackets [ ].

DC (CE/CGW) 199014/2 ? UCLES 2021

This document has 16 pages.

[Turn over

2

1 It is given that y = sinh (x2) + cosh (x2).

(a) Use standard results from the list of formulae (MF19) to find the Maclaurin's series for y in terms

of x up to and including the term in x4.

[2]

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

(b)

Deduce

the

value

of

d4y dx4

when

x

=

0.

[1]

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

y1 2

(c) Use your answer to part (a) to find an approximation to y dx, giving your answer as a rational

fraction in its lowest terms.

0

[2]

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

? UCLES 2021

9231/22/O/N/21

3 2 Find the solution of the differential equation

dy 4x3y dx + x4 + 5 = 6x

for which y = 1 when x = 1. Give your answer in the form y = f (x).

[7]

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

....................................................................................................................................................................

? UCLES 2021

9231/22/O/N/21

[Turn over

4

3 y 1

0

1

2

n

n

n-1 1 x n

The diagram shows the curve with equation y = 1 - x2 for 0 G x G 1, together with a set of n rectangles

of

width

1 n

.

(a) By considering the sum of the areas of the rectangles, show that

y1 (1

0

-

x2) dx

1

4n2

+ 3n 6n2

-

1

.

[4]

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

? UCLES 2021

9231/22/O/N/21

5

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

y (b) Use a similar method to find, in terms of n, a lower bound for 1 (1 - x2) dx.

[4]

0

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

............................................................................................................................................................

? UCLES 2021

9231/22/O/N/21

[Turn over

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download