2012/017) Z1, z2 & z3 are 3 complex no.'s that lie on a circle w/ centre (0,0). show that the reciprocals are in a circle

z1 = r cis t1
z2 = r cis t2
z3 = r cis t3

as they are in a circle with center (0,0)

1/z1 = 1/r cis (- t1)
1/z2 = 1/r cis (- t2)
1/z3 = 1/r cis (- t3)

they are also in a circle or radius 1/r

2012/016) In a triangle ABC , 3sin A + 4 cos B = 6 and 3 cos A + 4 sin B = 1 , then find angle C

square both and add

(9 sin^2 A + 16 cos^2 B + 24 sin A cos B) + (9 cos^2 A + 16 cos^2 B + 24 cos A sin B) = 37

or 9+ 16 + 24( sin A cos B + cos A sin B) = 37

so sin (A+B) = 1/2

sin C = sin (A+B) as A+B+C = 180

so sin C = 1/2

C = 30 or 150

cos A < =1/3 so A > 60 degrees and C cannot be 120 so C = 30

2012/015) find Nth term of this sequence? 14, 29, 54, 89, 134

The 1st order differences are 15,25,35,45

The second order differences are 10, 10,10,

So the equation for the 1st order differences are 5 + 10n

Now the original terms are equation is 2nd order that is an^2+bn+c

difference is

a(n+1)^2 + b(n+1) + c – an^2 –bn – c
= a(2n+1) + b = 2an + (a+b) = 10n + 5
So a = 5 and b= 0

So we have nth term = 5n^2+ c

for n=1 gives 5n^2 + c = 14 or c= 9

so nth term = 5n^2+9

2012/014) The no. of integers that are both multiples of 2002^2004 and factors of 2002^2006 is

we know 2002^2006/2002^2004 = 2002^2

the no. of integers that are both multiples of 2002^2004 and factors of 2002^2006 is number of inetgers factor of 2002^2

now 2002 =2 * 7 * 11 * 13

so 2002^2 = 2^2 * 7^2 * 11^2 * 13 ^2
number of factors = (2+1)(2+1)(2+1)(2+1) = 3^ 4 = 81

( rationale:
Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.

Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.

So the number of factors for your number would be (a+1)*(b+1)*...(k+1).)

2012/013) solve √x+1 + √x-1 =1

(√x+1)^2 =(1 - √x-1)^2
so square both sides
x+1 = 1 - 2 √x-1 + x-1 = x - 2 √x-1

or 1 = - 2 √x-1

LHS is positive and RHS = -ve so contradiction thus no solution

2012/012) simplify ³√[(√980/27) + 6] - ³√[(√980/27) - 6]?

let x = ³√(√980/27) + 6

and y = ³√- ([(√980/27) - 6)


x^3 + y^3 = 12
xy = - ³√(980/27-36) = - ³√(980/27-36) = - ³√8/27 = - 2/3
we need to find x + y

(x^3+y^3) = (x+y)^3 - 3xy(x+y)
or 12 = (x+y)^3 + 2(x+y)
or (x+y)^3 + 2(x+y) - 12 = 0

if x + y = t

t^3 + 2t - 12 = 0

t = 2 is a real root and and other roots are complex

hence we have ³√[(√980/27) + 6] - ³√[(√980/27) - 6] = 2

2012/011) The sum of 1+1/3+1/6+1/10+1/15......2005 terms is

the nth term = 2/(n(n+1) = 2(1/n- 1/n+1)

we can check on that from 1,3,6,10 the common difference is 2,3,4,5...

1st tern = 2/1 - 2/2
2nd term = 2/2- 2/3
2005th term = 2/2005 - 2/2006

sum up to 2005 terms is 2- 2/2006 = (2006*2-2)/2006 = 2005/1003

note:
sum up to n terms = 2- 2/(n+1) = 2(1- 1/(n+1)) = 2n/(n+1)