2011/074) If the ratio of GM and HM of 2 numbers is 5: 4 find the ratio of numbers?

without loss of generality let the numbers be a and at

as GM^2 = AM * HM
GM/HM = AM/GM = (a+at)/(2a sqrt(t)) = 5/4

so (1+t)/( 2sqrt(t) = 5/4 and putting sqrt(t) = x

(1+x^2) / (2x)= 5/4

or 4(1+x^2) = 10 x
o (4x^2-10x + 4) = 0 or (2(2x-1)(x-2) = 0 or x = 1/2 or 2 or t = 1/4 or 4

so ratio = 1 : 4 (or 4:1)

2011/073) Solve the system of quadratic equations

5x^2 + 4xy + 5y^2 + 3x + 3y = 74
x^2 + 2xy + y^2 - 6x - 6y = -8

5x^2 + 4xy + 5y^2 + 3x + 3y = 74

the coefficent of x^ and y^2 are same and so is of x and y

so we combine accordingly

5(x+y)^2 - 6xy + 3(x+y) = 74 ... 1


from 2nd equation by same anology

(x+y)^2 - 6(x+y) = -8

putting x + y = t

t^2 - 6t +8 = 0 or (t-4)(t-2) = 0 so t = 4 or 2

5(x+y)^2 - 6xy + 3(x+y) = 74
80-6xy + 12 = 74
or 6xy = 18
or xy - 3

x+ y= 4 .. 2
xy = 3 ... 3

from 2 and 3 we get (x-y)^2 = (x+y)^2 - 4xy = 16-12 = 4 or x-y = 2 or -2

x+y = 4 and x- y = 2 give x = 3 y = 1
x+y = 4 x-y = -2 give x = 1 and y = 3

putting x+ y =2 in 1 we get
5(x+y)^2 - 6xy + 3(x+y) = 74
20 - 6xy + 6 = 74
6xy = 48
xy = -8

so x = 4 y = -2 or x= -2 y = 4 (we can clove by above method)

so we get
(1,3), (3,1)(4,-2),(-2,4) 4 set of solutions

2011/072) Prove that ( 1+ cos π/8)( 1+ cos 3π/8)( 1+ cos 5π/8)( 1+ cos 7π/8) =1/8?

(1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8)
= (1 + cos π/8)(1 + cos 3π/8)(1 - cos 3π/8)(1 - cos π/8), since cos(π - t) = -cos t
= (1 - cos² π/8)(1 - cos² 3π/8)
= (1 - cos² π/8)(1 - sin² π/8), via cos(π/2 - t) = sin t
= (1 - cos² π/8)(1 - sin² π/8)
= sin ^2 π/8 cos ^2 π/8
= (1/4) (2 sin π/8 cos π/8)²
= (1/4) (sin 2π/8)², by double angle formula
= (1/4) (sin π/4)²
= (1/4) (1/√2)²
= (1/4)(1/2)
= 1/8.

2011/071) Show that 2 tan 20˚ + 4 tan 40˚ + 8 tan 80˚ = 9 (cot 10˚ - tan 10˚)

we first prove one relationship
cot x - tan x = cos x/ sin x - sin x/ cos x = (cos^2 x - sin ^2 x) /( sinx cos x) = 2 cot 2x

so

cot x - tan x = 2 cot 2x ...1

also
tan x = cot x - 2 cot 2x ...2

from 2 we generate following 3 relationships

2 tan 20= 2 cot 20 - 4 cot 40 ... 3
4 tan 40 = 4 cot 40 - 8 cot 40... 4
8 tan 80 = 8 cot 80 - 16 cot 160... 5
adding we get
2 tan 20˚ + 4 tan 40˚ + 8 tan 80 = 2 cot 20 - 16 cot 160
= 2 cot 20 + 16 cot(180-160)as cot x = - cot(180-x)
= 2 cot 20 + 16 cot 20
= 18 cot 20
= 9( 2 cot 20)
= 9( cot 10 - tan 10)from 1
proved

20111/070) If a+b+c=1, ab+bc+ca=2, abc=3, then find the value of 1/(a+bc)+1/(b+ca)+1/(c+ab).

we have
a+b+c=1 ...1
ab+bc+ca=2 .. 2
abc=3 ...3

so a b c are roots of equation

x^3-x^2 + 2x - 3 = 0

further

as a+ b+ c = 1

so a = 1- (b+c)
so a + bc = 1-(b+c) + bc = (1-b)(1-c)

similarly b+ ca = (1-c)(1-a)
c + ba = (1-b)(1-a)

we need to find 1/(a+bc)+1/(b+ca)+1/(c+ab) = 1/ ((1-b)(1-c)) + 1/ ((1-c)(1-a)) + 1/ ((1-b)(1-a))

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



so we need to form an equation whose roots are a-1 , b- 1 and c- 1

as a b and c are roots of f(x) = x^3-x^2 + 2x - 3

so a-1 b- 1 and c- 1 are roots of

f(x+1) = (x+1)^3 - (x+1)^2 + 2(x+1) - 3 = x^3 + 2 x^2 + 3x -1

so (a-1) + (b-1) + (c-1) = - 2

(a-1)(b-1)(c-1) = 1

so given expression = ((a-1) + (b-1) + (c-1))/((a-1)(b-1)(c-1)) = - 2

2011/069 ) Factor the following?

(x - 9753)(x -7531) + 234321

this can be factored by multiplying out and solving as quadratic however let is take x –a = t where a = (9753+7531)/2 = 8642

now 9753-8642 = 8642-7531 = 1111

so (x - 9753)(x -7531) + 234321
= (x-8642-1111)(x-8642 + 1111) + 234321
= (t-1111)(t+1111) + 234321 taking t = x- 8642
= t^2 – 1111^2 + 234321
= t^2- 1234321 + 234321
= t^2- 1000000
= t^2- 1000^2
= (t+1000)(t-1000)
= (x-8642+1000)(x-8642-1000)
= (x-7642)(x-9642)

This trick helps in keeping expression simple

2011/068) Find all ordered pairs of integers (x,y) which satisfy x^3 + y^3 - 3x^2 + 6y^2 + 3x + 12y + 6 = 0

Rewrite this using x terms and y terms in groups

(x^3 - 3x^2 + 3x) + (y^3 + 6y^2 + 12y) = -6
==> (x^3 - 3x^2 + 3x - 1) + (y^3 + 6y^2 + 12y + 8) = -6 - 1 + 8
==> (x - 1)^3 + (y + 2)^3 = 1
as difference of 2 cubes cannot be 1 so one of them is zero and another is 1 and there is no other choice

x - 1 = 1 and y + 2 = 0 ==> (x, y) = (2, -2), or
x - 1 = 0 and y + 2 = 1 ==> (x, y) = (1, -1).

so solution sets are (2,-2) and (1,-1)