2011/006) If x > 1, y > 1, z > 1 are in G.P.

If x > 1, y > 1, z > 1 are in G.P., then 1/(1+In x), 1/(1+In y),1/(1+ In z) are in :
(A) A.P.
(B) H.P.
(C) G.P.
(D) none of these

(this is objective and should not take more than a minute)

ans is B

reason
x > 1, y > 1, z > 1 are in G.P. so ln x, ln y, ln z are in AP and >0

so 1 + ln x, and 1+ in y and 1 + ln z in AP and hence the result

2011/005) if cos A + cos B + cos C = 0 = sin A + sin B + sin C

prove that

cos 3A + cos 3B + cos 3C = 3 cos( A+ B+ C)

proof:

we have

cos A + cos B + cos C = 0 ...1

sin A + sin B + sin C = 0 ....2

multiply 2nd by i and add

(cos A + i sin A ) + ( cos B + i sin B) + (cos C + i sin C) = 0

or e^(iA) + e^(iB) + e^(iC)= 0 as e^ix = cos x+ i sin x

and as if (x+y+z) = 0 then x^3+y^3 + z^3 = 3xyz

so e^(i3A) + e^(i3B ) + e^(i3C) = 3 e^i(A+B+C)

so ( cos 3A + i sin 3 A) + (cos 3B + i sin 3B ) + ( cos 3C + i sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)

or (cos 3A + cos 3 B + cos 3C) + i( sin 3A + sin 3B + sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)

equating real and imaginary parts on both sides we get
cos 3A + cos 3 B + cos 3C = 3 cos (A+B+C)
sin 3A + sin 3B + sin 3C = 3 sin (A+B+ C)

2011/004) Using mathematical induction prove that tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(n^2+n+ 1) = tan ^-1(n/(n+2))

in mathematical induction there are 2 steps

number one base step

for n =1 LHS = tan ^-1 (1/3) and RHS = tan ^-1 (1/(1+2)) = tan ^-1 (1/3)

hence true

then the induction step

let it be true for n = k that is

tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(k^2+k+ 1) = tan ^-1(k/(k+2))

for n = k + 1 we have
tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(k^2+k+ 1) + tan ^-1(1/(k+1)^2 + k+ 1+ 1)
= tan ^-1(1/(k+2)) + + tan ^-1(1/(k+1)^2 + k+ 1+ 1)
= tan ^-1(1/(k+2)) + + tan ^-1(1/(k^2+3k + 3)

The RHS = tan ^-1((k+1)/(k+3))


As tan (A+ B) = (tan A + tan B)/(1- tan A tan B)

So A + B = tan ^-1(tan A + tan B)/(1- tan A tan B))

So tan ^-1(k/(k+2)) + tan ^-1(1/(k^2+3k + 3)
= tan ^-1 ( k/(k+2) + 1/(k^2+3k + 3)/(1-k/(k+2)1/(k^3+3k+ 3)
= tan ^-1(( k^3+3k^2+3k + k+ 2)/((k+2)(k^2+3k+3)-k)
= tan ^-1((k^3+3k^2 +4k + 2) /(k^3+ 5k^2 + 8 k + 6))

We have k^3+3k^2+ 4k + 2 = (k^3+ 3k^2 + 3 k + 1) + (k+1) = (k+1)^3 + k+ 1= (k+1)(k^2 + 2k + 2)

(k^3+ 5k^2 + 8 k + 6) = k^3 + 3k^2 + 2k^2 + 8k + 6
= k^2(k+3) + 2( k^2 + 4k + 3) = k^2(k+3) + 2( k+1) (k+ 3) = (k+ 3)( k^2 + 2k + 2)

So tan ^-1((k^3+3k^2 +4k + 2) /(k^3+ 5k^2 + 8 k + 6)) = tan ^-1 ((k+1)/(k+ 3)

= RHS

So induction step is proved

hence proved

2011/003) If 1 , w , w^2 are cube root of unity

If 1 , w , w^2 are cube root of unity show that (a+bw+cw^2)/(b+cw+aw^2) + (a+bw+cw^2)/(c+aw+bw^2) = -1

Proof:

As 1 w and w^2 are cube root of 1 we have

1+w+w^2 = 0 … 1

w^3 = 1…2


now
(a+bw+cw^2) = aw^3 + bw+ cw^2 = bw+cw^2 + aw^3= w( b+ cw + aw^2)

Hence (a+bw+cw^2)/(b+cw^2+aw^2) = w

Further
As (a+bw+cw^2) = a + bw + c/w (as w^2 = 1/w) = 1/w(aw+bw^2+c)

So (a+bw+cw^2)/ (c+ aw+bw^2 ) = 1/w = w^2

Hence (a+bw+cw^2)/(b+cw^2+aw^2) + (a+bw+cw^2)/ (c+ aw+bw^2 ) = w + w^2 = - 1 (from 1)

2011/002) product of sum of squares is sum of squares

Prove that product of two numbers, each of which can be expressed as sum of two squares, can itself be expressed as sum of two squares.

Let 1st number be a^2+b^2 and second be c^2 + d^2
one way

(a^2+b^2)(c^2+d^2)
= (a^2c^2 + b^2 d^2 + a^2d^2 + b^2 c^2)
= (a^2c^2 + b^2 d^2 + 2 abcd + a^2d^2 + b^2 c^2- 2abcd)
= (ac+bd)^2 + ( ad - bc)^2

also
(a^2+b^2)(c^2+d^2)
= (a^2c^2 + b^2 d^2 + a^2d^2 + b^2 c^2)
= (a^2c^2 + b^2 d^2 - 2 abcd + a^2d^2 + b^2 c^2+ 2abcd)
= (ac-bd)^2 + ( ad + bc)^2

so we can do in two different ways

further we can prove using complex numbers

(a^2+b^2)(c^2+d^2) = | a + ib|^2 |c + id|^2
= | (a + ib)( c+ id)| ^ 2
= | (ac - bd) + (ad +bc) i | ^2
= (ac-bd)^2 + (ad + bc)^2

and taking (a^2+b^2)(c^2+d^2) = | a + ib|^2 |c - id|^2

we get (ac+bd)^2 + (ad - bc)^2

2011/001) a problem in inequality

Let a,b,c,d be positive real numbers such that abcd = 1. Show that,?
(1 + a)(1 + b)(1 + c)(1 + d) >= 16

proof:

as a is real

(1+a) = (1-sqrt(a))^2 + 2 sqrt(a) or

1+ a >= 2 sqrt(a) ( we can also show it by AM GM inequality)

similarly

(1+b) > = 2 sqrt(b)

(1+c) > = 2 sqrt(c)

(1+d) > = 2 sqrt(d)

by multiplying

(1+a)(1+b)(1+c)(1+d)>= 16 sqrt(abcd) or > 16 as abcd = 1

2010/065) an example of chinese remainder theorem

What would be the least total no. which give the remainder 1,2 and 3 when divided by 7,9 and 11?

1,2 and 3 when divided by 7,9 and 11?

hence
x= 1 mod 7
x= 2 mod 9
x =3 mod 11

now 7 ,9 and 11 are pair wise coprime

x = 1 a1b1 + 2 a2b2 + 3 a3b3 mod 7*9*11(or 693)


where a1 = 9*11 = 99
a2 = 7 * 11 = 77
a3 = 7 * 9 = 63

and
a1b1 = 1 mod 7
a2b2 = 1 mod 9
a3b3 = 1 mod 11

a1 = 99 so b1 = 99 mod 7 or 1 mod 7 so b1 = 1

77b2 mod 9 = 1 so 5b2 mod 9 = 1 so b2= 2 (as 5*2 = 10 mod 9)

63b3 = 1 mod 11 or 8b3 =1 mod 11 so b3 = 7 as 7 * 8 = 56 = 1 mod 11

So x = 1 * 99 * 1 + 2 * 77 * 2 + 3 * 63 * 7 mod 693 = 1730 mod 693

= 344 mod 693

So x if of the form 693n + 344 or lowest x = 344