Analysis and Approaches for IBDP Maths Ebook 1 | Page 83

Front Page
28
Paper 1 Section A – Simple identities
Example
(a) Show that 5 � 1 � 1 .
36 4 9
2m
�1 1 1
(b) Prove that � �
2 2 2 2
m ( m �1) ( m �1)
m
for m � 1.
[2]
[3]
Solution
(a)
(b)
R.H.S.
1 1
� �
4 9
1�9 1�4
� �
4�9 9�4
M1
9 4
� �
36 36
A1
5
�
36
� L.H.S.
5 1 1
� � �
36 4 9
AG N0
R.H.S.
1 1
� �
( m�1)
m
2 2
2 2
m ( m�1)
� �
( m �1) m m ( m �1)
2 2 2 2
2 2
m �( m�1)
�
2 2
m ( m�1)
( m � ( m �1))( m � ( m �1))
�
2 2
m ( m�1)
(2m
�1)(1)
�
2 2
m ( m�1)
2m
�1
� �L.H.S.
2 2
m ( m�1)
2m
�1 1 1
� � �
m ( m �1) ( m �1)
m
2 2 2 2
M1
M1
A1
for m � 1
AG N0
[2]
[3]
8
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