x - Balliol College - University of Oxford

SJ Roberts - July 2011 Revision 1/ page 12
(ii)
2 1
1 tanh x ; sinh2x
2cosh x sinh x ;
2
cosh x
2cosh x sinh x
.
cosh x
2
Hence tan xsinh2x
2tanh
x
1
2
d
dx
x
x x x
d
. Hence cosh x sinh x and
dx
30. e
e /
2 e
e /
2
similarly
d
dx
sinh x cosh x .
31. (i) 3 5i ; (ii) 4 7i ; (iii) 11 2i
2 2
32. Note y
;
x is the modulus of z (and of z too for that matter).
33. 11/
25
i2/
25
34. Solutions are 1
i
. Yes: for a complex soln. The usual formula
b b
2 4ac
2
gives roots as
. For complex roots, b 4ac
0 ,
2a
giving the imaginary part and the signs always give conjugate
pairs with the same real part. Note though if b 4ac
0 then the
two real solutions are different.
2 2
35. Square to find cos
sin 2i
sin
cos
1
v 37. (i) , 3, 3
6
36. ˆ i j 2k
2
, hence result.
3 , (ii) 1,
2, 3
3
14