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)

2012/010) solve for x given X^4 + (2-x)^4 = 34

This is quartic and we can convert to sum of quartic of the form (a+x)^4 + (a-x)^4 by choosing x and 2-x equdistance from mean that is 1
let x = 1- t
then 2 - x = t+ 1
x^4 + (2-x)^2 = (1-t)^4 + (1+t)^4 = 2(t^4 + 6t^2+ 1)
hence 2(t^4 + 6t^2+ 1) = 34 or t^ 4 + 6t^2 - 16 =0

or (t^2-2)(t^2-8) = 0

t = sqrt(2) or - sqrt(2) or 2 sqrt(2)i or -2(sqrt(2) i

hence x= 1 + sqrt(2) or 1- sqrt(2) or 1 + 2 sqrt(2) i or 1- 2sqrt(2) i

2012/009) The product of four distinct positive integers a,b,c,d is 8!.The no. also satisfy ab+a+b=322 and bc+b+c=398 find d

add 1 to 1st relation
ab + a + b + 1= 323 or (a+1)(b+1) = 17 * 19 ..1

add 1 to 2nd relation

bc+b+ c = 398

or bc + b+ c + 1 = 399 or (b+1)(c+1) = 19 * 21 ...2

from 1 and 2 we have b+1 = 19 or b= 18 (as b + 1 is common factor and is not 1 as b is not zero)

a + 1 = 17 or a = 16
c= 20
so abc = 16 * 18 * 20

abcd = 8!

so d = 8!/(16 * 18 * 20) = 7

2012/008) Find all complex roots of p(z) = z^4− 10z^3+ 38z^2− 74z + 85 given that 4 + i is a root.

4 + i is a root

so 4 - i is a root

so a factor = z - 4 = i

(z-4)^2 = - 1 or z^2-8z + 17 = 0 or z^2 - 8z + 17 is a factor

so we have z^4− 10z^3+ 38z^2− 74z + 85 = (z^2 + az + b)(z^2-8z+ 17)
= z^4 + z^3(a-8) + z^2(17 + b - 8a) + z (17a - 8b) + 17b


which gives a - 8 = - 10 => a = - 2(coefficient of z^3)
and 17 b = 85 or b = 5(constant)

17+b- 8a = 17 + 5 + 16 = 38( coefficient of z^2)
17 a - 8b = -34 - 40 = 74(coeffificent of z)

so equtions are consistant

so we get z^2-2z + 5 = 0 => (z-1)^2 + 4 = 0 or z = 1 + 2i or 1- 2i
so all roots are 1 + 2i, 1- 2i, 4+i, 4- i

2012/007) find n when ( cosx + isinx )^n . ( sinx + icosx )^n = -1

( cosx + isinx )^n . ( sinx + icosx )^n
= (( cosx + isinx ) ( sinx + icosx ))^n
= ( cos x + i sin x) i ( - i sin x+ cos x ))^n
= (i ( cos x + i sin x)(cos x - i sin x))^n
= ( i ( cos ^2 x + sin ^2 x))^n
= i^n = - 1

as i^2 = 1 and i^4 = i

Hence n = 4m+2, for m∈Z

2012/006) prove that the derivative of an odd function is even and vice versa

1)

because the function is odd

f(-x) = - f(x)

differentiate both sides wrt x using chain rule

- f'(-x) = - f'(x) or f'(-x) = f'(x) and hence derivative is even function

2)
because the function is even

f(-x) = f(x)
differentiate both sides wrt x using chain rule

- f'(-x) = f'(x) or f'(-x) = - f'(x) and hence derivative is odd function

2012/005) If ab + bc + cd +da = 16, what will at least a^2 + b^2 + c^2 + d^2

we have (a-b)^2 = a^2 + b^2 - 2ab
(b-c)^2 = b^2 + c^2 - 2bc
(c-d)^2 = c^2 + d^2 - 2cd
(d-a) ^2 = d^2+a^2 - 2ad

adding (a-b)^2 + (b-c)^2 + (c-d)^2 - (d-a)^2 = 2(a^2 +b^2 +c^2 + d^2) - 2(ab+bc+cd+da)

or 2(a^2 +b^2 +c^2 + d^2) = (a-b)^2 + (b-c)^2 + (c-d)^2 - (d-a)^2 + 2(ab+bc+cd+da)

or (a^2 +b^2 +c^2 + d^2) = ((a-b)^2 + (b-c)^2 + (c-d)^2 - (d-a)^2) /2+ (ab+bc+cd+da)
= ((a-b)^2 + (b-c)^2 + (c-d)^2 - (d-a)^2) /2+ 16
clearly is lowest when a= b=c=d and value= 16

2012/004) P(x) = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5

we have
(x+2)^5 = x^5 + 2 * 5x^4 + 2^ 2 * 10x^3 + 2^ 3 * 10 x^2+ 2^4 * 5x + 32
= x^5 + 10x^4 + 40x^3 + 80 x^2+ 80 x + 32
= P(x)

so P(x) = (x+2)^5

so p(-2 + 2i) = (-2 + 2i + 2)^5 = (2i)^5 = 32 i

2012/003) If 3a^2+2a-4=0 & 3b^2+2b-4=0 then, (1/a)+(1/b)=

a and b are root of equation

3x^2 + 2x - 4 = 0

so x^ 2+ (2/3) x - 4/3 = 0

sum of roots a + b = - 2/3
product of roots ab = - 4/3

so 1/a + 1/b = (b+a)/ab = (-2/3) / (-4/3) = 1/2

another method

a and b are root of equation

3x^2 + 2x - 4 = 0

so 1/a and 1/b are roots of equation

3(1/x)^2 + 2(1/x) - 4 = 0
or 4x^2-2x- 3 = 0

so 1/a + 1/b = 2/4 = 1/2(-coefficient of x/ coefficient of x^2)

2012/002) Why must you flip the inequality symbol when you divide by a negative number?

this follows from the analysis below

say x < y

now subtract x + y from both sides

x - (x+y) < y - (x+y)

or -y < - x

hence dividing or multiplying by -1 changes the sign

divide by a positive number does not change the sign
so divide by - number changes sign

(this being the last entry of the year this was a small but useful analysis)

2012/001) Arithmetic sequence only 1 perfect square?

Can an infinite arithmetic sequence of positive integers contain exactly one perfect square? If so, which one?

Ans : NO

reason :
if it has one square without loss of generality we can assume that it is the 1st term ( deleting all terms to the left and let it be x^2 and common difference is d


then x^2 + 2dx + d^2 = x^2 + d ( 2x + d)

so the term (2x +d+1)th term is is perfect square

we can generalize if as

x^2 + 2ndx + (nd)^2 = x^2 + d(2nx + n^2d) and putting n = 1 on wards it shall have infinite perfect squares

so if it has at least one perfect square then it shall have infinite perfect squares