If the results of 6 9 i a bi 7 2 i is a real number find the value of ba 4

# If the results of 6 9 i a bi 7 2 i is a real number

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be non-zero real numbers. If the results of (6 + 9 i )( a + bi )(7 – 2 i ) is a real number, find the value of b a . (4 marks) Solution: (6 + 9 i )( a + bi )(7 – 2 i ) = (6 a + 6 bi + 9 ai – 9 b )(7 – 2 i ) 1M = 42 a – 12 ai + 42 bi + 12 b + 63 ai + 18 a – 63 b + 18 bi = (42 a + 12 b + 18 a – 63 b ) + (–12 a + 42 b + 63 a + 18 b ) i = (60 a – 51 b ) + (51 a + 60 b ) i 1A The result of (6 + 9 i )( a + 6 i )(7 – 2 i ) is a real number. 51 a + 60 b = 0 1M 51 a = –60 b b a = 51 60 1A <bk=4A><ch=1><type=L2><mark=6> [10167225] (a) Simplify ) ( ) ( 2 2 2 2 y x y x y x y x , where x y 0. (b) Hence, simplify 1 1 1 1 2 2 y y y y , where 0 y 1. (6 marks) Solution: (a) ) ( ) ( 2 2 2 2 y x y x y x y x i y x y x i y x y x 2 2 2 2 1M ] ] [ ] [ 2 2 2 2 2 2 2 2 ) ( ) ( ) ( ) ( [ ] [ i y x y x i y x y x i y x y x i y x y x 1M ) )( ( ) ( 2 ) )( ( )]} ( [ ) ){( ( ) ( ) ( ) )( ( 2 ) ( 2 2 2 2 2 2 2 2 2 2 ] [ y x y x y x i y x y x y x y x y x y x y x y x y x y x y x i y x y x y x i x y x x y x y x 2 2 2 ) ( 2 2 1M OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.108
i x y x x y 2 2 1A (b) Substitute x = 1 into the result of (a). i y y y y y y 1 1 1 1 1 1 1 2 2 2 2 1M i y y 2 1 1A <bk=4A><ch=1><type=L2><mark=7> [10167296] In the table below, identify the results of each of the following as an integer, a rational number, an  irrational number or a non-real number by putting a ‘ ’ in the appropriate box. 4 5 6 . 2 2 1 3 2 3 24 26 ) 5 ( 2 o 60 cos 1 Integer Rational number Irrational number Non-real  number (7 marks) Solution: 333 884 4 5 6 . 2 6 7 6 3 6 4 2 1 3 2 24 3  =  13 824 1 26 25 26 ) 5 ( 2 2 2 1 1 60 cos 1 4 5 6 . 2 2 1 3 2 3 24 26 ) 5 ( 2 o 60 cos 1 Integer Rational number OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.109
Irrational number Non-real  number 1A for each correct answer <bk=4A><ch=1><type=L2><mark=7> [10167401] (a) Convert each of the following recurring decimals into a fraction. (i) 3 0 0 . (ii) 0 6 0 . (b) Using the results of (a), solve x x 0 6 . 0 3 0 . 0 3 0 . 1 . (7 marks) Solution: (a) (i) Let x = 3 0 0 . . x = 0.033 33 ……………………(i) 10 x = 0.333 33 ……………………(ii) 3 . 0 9 (i) (ii) x 1M 90 3 x 30 1 30 1 3 0 . 0 1A (ii) Let y = 0 6 0 . . y = 0.606 060 …………………..…(i) 100 y = 60.606 060 ……………………(ii) 60 99 (i) (ii) y 1M 33 20 99 60 y 33 20 0 6 . 0 1A (b) x x 0 6 . 0 3 0 . 0 3 0 . 1 3 0 . 0 0 6 . 0 3 0 . 1 x x 30 1 33 20 30 1 1 x x 1M 30 1 33 20 30 31 x x 30 1 110 47 x 1M OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.110
141 11 x 1A <bk=4A><ch=1><type=L2><mark=7> [10168169] (a) Simplify n n ) 1 ( 1 , where n 0. (b) Using the result of (a), simplify 4 5 1 3 4 1 2 3 1 1 2 1 . (c) Hence, or otherwise, simplify 4 5 1 3 4 1 2 3 1 1 2 1 24 25 1 23 24 1 . (7 marks) Solution: (a) n n ) 1 ( 1 i n i n 1 1 1M ) 1 ( 1 n n i ) 1 ( ) 1 ( 2 n n i n n i 1M i n n ) 1 ( 1A (b) 4 5 1 3 4 1 2 3 1 1 2 1 i i i i ) 4 5 ( ) 3 4 ( ) 2 3 ( ) 1 2 ( 1M i ) 5 1 ( i ) 5 1 ( 1A (c) 24 25 1 3 4 1 2 3 1 1 2 1 OXFORD UNIVERSITY PRESS 2014 4A Chapter 1 P.111
i ) 25 1 ( 1M i 4 1A <bk=4A><ch=1><type=L2><mark=8> [10168196] (a) Given that 2 is a solution of the equation x 2 + bx + c = 0, find the values of the real numbers b and c .