Your Practice Set – Applications and Interpretation for IBDP Mathematics
Solution
( a ) The area of the triangle OPQ 1
� ( OQ )( PQ ) ( M1 ) for valid approach
2
( b )
|
1 � ( 2cos � )( 2sin
� )
2
|
( M1 ) for substitution |
|
�
2sin�cos�
|
A1 |
1 ( 2 )
2 2sin cos S � � � � �
( M1 ) for valid approach
4 S �� � 2sin� cos� A1 dS
( c ) ( i ) � 0 � 2 (( cos �)( cos �) � ( sin �)( � sin �))
( M1 ) for product rule d� dS
2 2
2cos � 2sin � d� � � � A1
[ 3 ]
[ 2 ]
( ii ) dS
0 d� � 2 2
��2cos �� 2sin � � 0 By considering the graph of
2 2 y 2cos � 2sin �
� � � , � � 0.7853981634 . ( M1 ) for valid approach �� � 0.785 rad
A1
( iii )
2 d S
2 d�
� �0.7853981634
� 8 ( sin 0.7853981634 )( cos 0.7853981634 ) �4� 0
R1
Thus , S attains its minimum at
� � 0.785 rad . A1
( iv ) The minimum value of S
�sin 0.7853981634 � ��
�2 � �
( M1 ) for substitution �� cos0 . 7853981634 � � 2.141592654 � 2.14
A1
�
( d ) � � 0,
� � A2 2
[ 8 ]
[ 2 ]
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