2010/007) prove 64 {Cos^8(x) + Sin^8(x)} = cos8 x + 28cos 4x + 35

We know

(a+b)^8 = a^8 + 8 a^7b + 28 a^6b^2 + 56 a^5b^3 + 70a^4b^4 + 56 a^3b^5 + 28 a^2b^6+8ab^7+b^8

And (a-b)^8 = = a^8 - 8 a^7b + 28 a^6b^2 - 56 a^5b^3 + 70a^4b^4 - 56 a^3b^5 + 28 a^2b^6 -8ab^7+b^8

So (a+b)^8 + (a-b)^8 = 2(a^8 + b^8) + 56 (a^6b^2+a^2b^6) + 140(a^4b^4)

Putting a = e^ix and b = e^-ix

LHS = (e^ix+e^-ix)^8 + (e^ix-e^-ix)^8 = 2(e^i8x + e^-i8x)+ 56(e^4ix + e^-4ix) + 140

Or (2 cos x)^8 + (2i sin x)^8 = 2 ( 2 * cos 8x) + 56 * 2 * cos 4x + 140

Or 2^8(cos ^8x + sin ^8x) = 4 * ( cos 8x + 28 cos 4x + 35)

Or 2^6(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

Or 64(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

2010/006) equation sin(w) = (a^2 + b^2 + c^2)/(ab+bc+ca), where a,b,c are fixed nozero real numbers

has a solution for w
A) Whatever be a,b,c;
B) iff a^2 + b^2 + c^2 < 1;
C) iff a,b and c all lie in the interval (-1,1);
D) iff a = b = c;
Explain your answer.

ans

we know by GM AM enaquality a^2+b^2 >= 2ab
a^2+c^2 >= 2ac
and b^2+c^2 >= 2bc

adding all 3 we get 2(a^2+b^2+c^2) >= 2(ab+bc+ca)

or (a^2+b^2+c^2)/(ab+bc+ca)>= 1 and equal only when a= b= c

sin (w) cannot be > 1 so ans is when (a^2+b^2+c^2)/(ab+bc+ca) = 1 or a= b =c

that is d

2010/005) If 0 < x < 1, then the sum of the infinite series (1/2)X^2 + (2/3)X^3 + (3/4)X^4 + ... is ...

let f(x) = (1/2)x^2 + (2/3)x^3 + (3/4)x^4 + ...

df/dx = x + 2 x^2 + 3x^3 + ... ....
= (x+x^2 + x^3...) + (x^2 + x^3 ...) + (x^3+x^4 + ....)
= x/(1-x) + x^2/(1-x) + ...
= x/(1-x)^2
= (1-x-1)/(1-x)^2
= 1/(1-x) - 1/(1-x)^2
integrating you get
f = log(1-x) + 1/(1-x) + constant of integration log(1-x) as 1- x > 0
this can be checked to be zero
so f(x) = 1/(1-x) + log(1-x)

2010/004) Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number.

Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number. Show that n?
is a prime.

Proof:
1 + x + x^2 + x^3 + ... + x^n-1 = (x^n-1)/(x-1)

now in case n is not a prime then say n= pq

now x^n-1 = (x^(pq)-1) = (y^q-1)/(x-1) where y = x^p
= (y-1)/(x-1)(y^(q-1) + y^(q-2) + ...1)
= (x^p-1)/(x-1) (y^(q-1) + y^(q-2) + ...1)
= (x^p-1 + +1)((y^(q-1) + y^(q-2) + ...1)
has got 2 factors and hence not a prime
so n must be prime

2010/003) The equation x^3 + 2x^ + 2x + 1 = 0 and x^200 + x^130 + 1 = 0 have ...?

exactly one common root;
b) no common root;
c) exactly three common roots;
d) exactly two common roots;
Kindly explain...

ans:
x^3 + 2x^2 + 2x + 1 = x(x+1)^2 + (x+1) = (x+1)(x^2 + x + 1) has zeros -1, w and w^2 where w is cube root of 1

-1 is not a zero of x^200 + x^130 + 1 since the expression equals 3 when x = -1.

So we are down to 2 possible roots in common.

let f(x) = x^200 + x^130 + 1

f(w) = w^200 + w^130 + 1= w^2+ w + 1 = 0 so w is a root
f(w^2) = w^400 + w^260 + 1 = w+w^2+1 = 0 so w^2 is a root

so there are exactly 2 common roots and hence d)

2010/002) The numbers 12n + 1 and 30n + 2 are relatively prime for ...?

any positive integer n;
b) infinitely many, but not all, integers n;
c) for infinitely many integers n;
d) none of the above;
Kindly explain...

Ans:
GCD(12n + 1 ,30n + 2)
= GCD(12n + 1 ,2(15n+1))
= GCD(12n + 1 ,15n+1) as 12n+1 is odd
= GCD(12n+1,(15n+1-(12n+1)) as GCD(a,b) = GCD(a,b-a)
= GCD(12n+1,3n)
= GCD(3n, 12n+1- 4*3n) as GCD(a,b) = GCD(b,a- n * b))
= GCD(3n,1)
= 1 for all n
hence ans is a

2010/001) Prove that (2222^5555)+(5555^2222) is divisible by 7

Prove that (2222^5555)+(5555^2222) is divisible by 7

Proof:
we know

2222 mod 7 = 3

and 5555 mod 7 = 4 or -3

so (2222^5555)+(5555^2222) mod 7

= 3^5555 + (-3)^ 2222 mod 7
= 3^5555+ 3^2222 mod 7
= 3^2222(3^3333 + 1) mod 7


as 3^2222 is not divisible by 7 so

we need to show that

3^3333 + 1 mod 7 = 0

now as 7 is prime so as per format's little theorem

3^6 mod 7 = 1

3^3333 mod 7 = 3^(3333 mod 6) mod 7
= 3^ 3 mod 7 = 27 mod 7

so 3^3333 + 1 mod 7 = 28 mod 7 = 0

hence proved