The difference of two positive numbers is 6 and the sum of their squares is 90. Find the numbers?

let the numbers be x and y with y > x > 0

y - x = 6 ..1

y^2 + x^ 2 = 90 ..2

we know (y-x)^2 + (y+x)^2 = 2 (y^2 + x^2)

or 6^2 + (y+x)^2 = 2 * 90 = 180

(y + x) ^2 = 180- 36 = 144

or y + x = 12 .. 3 ( as y + x > 0)

from 1 and 3 we get x= 3 and y = 9

W = sinx + icosx and u = cscx + isecx .then | w / u | =

|w| = sqrt( sin ^2 x + cos^2 x) = 1

|u]^2 = csc^2 x + sec^2 x = (1/ sin ^2 x) + (1/ cos^2 x) = ( cos^2 x + sin ^2 x)/ ( sin ^2 x cos ^2 x)
= 1/ ( sin ^2 x cos^2 x)

so |u| = 1/ ( sin x cos x)

so |w/u] = |w|/|u| = sin x cos x

2012/035) Show that if a is any integer, then either 5|a^2 or 5|(a^2-1) or 5|(a^2+1)

a^2 = a^2
a^2 - 1 = (a-1)(a+1)
a^2 + 1 = a^2 - 4 ( note that 1 = - 4 mod 5) = (a+2)(a-2)

5 must divide a or a+ 1 or a-1 or a+ 2 or a- 2 ( 5 consecutive numbers)

if 5 divides a the a^2 is divisible

if 5 divides (a+1) or (a-1) the a^2 - 1 id divisible

else if 5 divides (a+2) or (a-2) then a^2-4 or a^2+ 1 is divisible by 5

so one of them is divisible

2012/034) Given f(1) + f(2) + ... + f(n) = (n^2)*[f(n)] f(1) = 999 Find f(1998)

From the given condition
f(1) + f(2) + ... + f(n) = (n^2)*[f(n)] ...1

f(1) + f(2) + ... + f(n-1) = (n-1)^2)*[f(n-1)] ..2

subtract (2) from (1)

f(n) = n^2f(n) - (n-1)^2 f(n-1)

or f(n) = (n-1)^2/(n^2-1) f(n-1) = (n-1)/(n+1) f(n-1)

f(2) = 1/3 f(1)
f(3) = 2/4 f(2) = 2f(1) / ( 3 * 4)
f(4) = 3/5 f(3)

from the above f(n) = 2f(1)/(n * (n+1))

so f(1998) = 2 * 999/( 1998 *1999) = 1/1999

2012/033) If cosα + cosβ + cosγ = 0 and sinα + sinβ + sinγ = 0 …Then what is cos(β – γ) + cos(γ –α) + cos(α – β)

cosα + cosβ + cosγ = 0
=>cosα + cosβ = - cosγ
=> (cosα + cosβ)^2 = cos^2γ
=> cos^2α + cos^2β + 2 cosα cosβ = cos^2γ
=> 2 cosα cosβ = cos^2γ - (cos^2α + cos^2β) ... 1

similarly 2 cosβ cosγ = cos^2 α - (cos^2β + cos^2γ) ...2

and 2 cosγ cosα =cos^2β- (cos^2γ + cos^2 α) .. 3

adding (1) (2) and (3)

we get 2( cosα cosβ + cosβ cosγ + cosγ cosα) = - (cos^2 α + cos^2β + cos^2γ) ... 4

similarly from sinα + sinβ + sinγ = 0 replacing cos with sin we get

2( sin α sin β + sin β sin γ + sin γ sin α) = - (sin ^2 α + sin ^2β + sin ^2γ) ... 5

adding (4) and (5) and reordering we get

2((cosα cosβ + sin α sin β) + (cosβ cosγ + sin β sin γ) + (cosγ cosα + sin γ sin α)) = - 3
or
2(cos(α-β) + cos(β-γ) + cos(γ - α)) = - 3

or cos(α-β) + cos(β-γ) + cos(γ - α) = - 3/2

2012/032) If a=2π/2013, what is cos2a+cos4a+cos6a+..........+cos 2012a…

First we need to show that

Cos a + cos 2a + cos 3a + … + cos na = 0 ..1

When a = 2π/n

For a proof I propose as below

2 cos ka sin a/2 = sin (k+1/2) a – sin(k-1/2) a

If we take it a sum from k = 1 to n we get telescopic sum

2 ( cos a + … + cos na) sin a/2 = sin (na +1/2) a – sin ½ a = sin (1/2 a) – sin ½ a = 0

So cos a + cos 2a + cos 3a + …. + cos n a = 0 …1

Using this we can prove the desired result

As we know that n = 2013 and an = 2pi

Cos a = cos 2012 a
Cos 3a = cos 2010 a
So for each odd on left there is an even on the right and exception is cos 2013a = 1

Cos a + cos 3a + cos 5a + …+ cos 2011 a = cos 2a + cos 4a + cos 6a + … + cos 2012 a ..2

From (1) group even terms together and odd ones also)

Cos 2a + cos 4a + cos 6a + … + cos 2012 a + ( cos a + cos 3a + cos 5a + … cos 2011 a ) + 1 = 0

Putting from (2) we get
2(cos 2a + cos 4a + … cos 2012a) + 1 = 0
Or (cos 2a + cos 4a + … cos 2012a) = - 1/2

2012/031) Show that for each integer n, n^13 = n mod 2730

first factorize 2730 = 13*7*5*3*2

So we are done if we can show that n^13 == n (mod k), for k=2,3,5,7, and 13.

Use both forms of the Little Theorem...

n^p == n (mod p), for all prime p and a^(p-1) == 1 (mod p) for prime p, when p does not divide n.

Clearly n^13 == n mod 13 ---- (1)

n^13 = n = 0 mod 7 if n is multiple of 7 

if n is co-prime to 7 then 

Also n^7 == n (mod 7) and
n^6 == 1 (mod 7), so
(n^7)*(n^6) = n^13 == n mod 7 ---(2)

n^13 = n = 0 mod 5 if n is multiple of 5

if n is coprime to 5
Again, n^5 == n (mod 5)
and n^4 == 1 (mod 5) => n^8 == 1 (mod 5)
Hence n^13 == n (mod 5) ---(3)

n^13 = n = 0 mod 3 if n is multiple  of 3

if n is coprime to3

n^2 = 1 mod 3
so n^12 = 1 mod 3 Taking power 6

n^13 = n mod 3 . (4)

n^13 = n = 0 mod 2 if n is multiple of 2

if n is coprime to 2

n^2 = n mod 2

so n^12 = n^ 6 mod 2 = n^3 mod 2

so n^13 = n^4 mod 2 = n^2 mod 2 = n mod 2 ...5

so from 1,2,3,4,5 we get

n^ 13 = n mod 2730

Source(s):

http://en.wikipedia.org/wiki/Fermat%27s_…

2012/030) x^2 = at^2 + b. Acceleration is proportional to:?

a) 1/x^3

b) 1/x - 1/x^2

c) - t/x^2

d) None

Answer:
acceleration is d^2x/dt^2

so differentiate both sides to get

xdx/dt = 2at

so differetiate again to get

x d^2x/dt^2 + (dx/dt)^2 = 2a

or x d2x/dt^2 + (2at/x)^2 = 2a
or d^2x/dt^2 + 4at^2/x^3 = 2a/x
or d^x /dt^2 = (2ax^2 - 4a^2t^2)/x^2
= 2a (x^2-2at^2)/x^3
= 2ab/x^3

hence ans is a)

2012/029) find the value of arctan 1 + arctan 2 + arctan 3

tan (x+y + z) = (tan x + tan y + tan z - tan x tan y tan z)/(1- tan x tan y - tan y tan z - tan z tan x)

we see tan (arctan 1 + arctan 2 + arctan3) = 0

so arctan 1 + arctan 2 + tarctan 3 = n π ( we need to calculate n)

as each of arcran 1, arctan 2, artctan 3 are > 0 and < π /2 so sum > 0 and < 3 π /2

so n > 0 and < 3/2 and n is integer so n = 1

hence arctan 1 + arctan 2 + tarctan 3 = π

2012/028) If ab+ bc+ca = 1 Then show (a+b)/(1-ab) +( b+c)/(1-cb) + (c+a)/(1-ca) = 1/(abc)

ab + bc+ ca = 1

so bc + ca = 1- ab

or c(a+b) = (1- ab) or (a+b)/(1-ab) = 1/c

similarly (c+a)/(1-ac) = 1/b
and (b+c)/(1-bc) = 1/a

so (a+b)/(1-ab) + (b+c)/(1-cb) + (c+a)/(1-ca) = 1/c+ 1/a+ 1/b = (ab+bc+ca)/(abc) = 1/(abc)