2011/102) Prove z/ (1 + z^2) is real if and only if z is real or |z| =1.

if z is zero it is true so let us take for the case z not zero

so z/(1+z^2) = 1/(z + 1/z) is real

let z = r cos t + ir sin t

so 1/z = 1/r cos t - 1/r i sin t

z + 1/z imaginary part is zero

r sin t = - 1/r sin t
sin t = 0 or r = -1/r or |r| = 1 and hence |z| = 1


sin t = 0 means real

hence z is real or |z| = 1

2011/101) Prove this inequality? 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x (x is a positive number)

we shall use the AM GM enaquality to prove it

we know (a+b)/2 > = sqrt(ab) or (a+b) > = 2 sqrt(ab)

a = 4x^10, b = x^8 gives 4x^10 + x^ 8 > = 4 x^9
a = 4x^10, b = 1 gives 4x^10 + 1 > = 4 x^5
a = x^8, b = 4x^2 gives x^ 8 +4x^2 > = 4x^5
a = 4x^2, b = 1 gives 4x^2 + 1 > = 4 x

adding we get

8x^10 + 2x^2+ 8x + 2 >=4x^9 + 8x^5 + 2

dividing by 2 we get 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x

2011/100) Solve for x: (2x + 14) / (5x - 1) ≤ x + 3.

we need to get rid of denominator so multiplying by (5x-1) is problematic as we do not know whether it is positive or negative

so multiply by (5x-1)^2 which is positive to get

(2x+ 14) (5x -1 ) <= (5x-1)^2(x+3)

or (2x+14) (5x-1) - (5x-1)^2 (x+ 3) <= 0

or (5x-1) ((2x+14 - (5x- 1 )(x+3)) < = 0

or (5x-1)(2x+14 - (5x^2 + 14x - 3)) <= 0
or (5x-1)(-5x^2 - 12x + 17) < = 0
or (5x-1)(5x^2+ 12x - 17) >= 0
or (5x-1) (5x+17)(x-1) > = 0

there are 4 ranges - infinite to -17/5
- 17/5 to 1/5 (in this it is positive but x cannot be 1/5 as 5x-1 is in denominator)
1/5 to 1
and 1 to infinite (in this it is positive

x belongs to [- 17/5, 1/5) U [1, ∞).

2011/099) Find cube root of 9 sqrt 3+11 sqrt 2

because there is sqrt(3) and sqrt(2) in the number the cube root is a combination of them
let it be x sqrt (3) + y sqrt (2)
so (x sqrt (3) + y sqrt (2))^3 = 9 sqrt 3+11 sqrt 2
or 3 x^3 sqrt (3) + 9x^2y sqrt 2 + 6xy^2 sqrt (2) + 2y^3 sqrt 2 = 9 sqrt 3+11 sqrt 2
comparing coefficient of sqrt (3) and sqrt(2) we get
3x^3 + 6x y^2 = 9 ..1
9x^2 y + 2y^3 = 11 .. 2
multiply (1) by 11 and 2nd one by 9 to get
33 x^3 + 66 xy^2 = 81 x^2y + 18y^3
divide both sides by 311 x^3 + 22xy^2 = 27x^2y + 6y^3
or 11x^3 - 27 x^2y + 22 xy^2 - 6y^3 = 0
putting x/y = t we get
11t^3-27 t^2 + 22t- 6 = 0
f(1) = 0 gives t = 1 is a sqrt ( as 11 + 22 = 27+ 6)
so t = 1 or 11t^2-16t + 6 = 0
or x = y is a solution
and
from (1) 9x^3 = 1 or x = 1 and y = 1
so result = sqrt(3) + sqrt (2)

2011/098) if a=0.5(5-root 21) then find a^3 + a^-3- 5a^2- 5a^-2+ a+ a^-1.?

a = .5(5-sqrt(21)

so 1/a = 2/(5- sqrt(21) = 2(5+ sqrt(21)/(5^2-21) = .5(5 + sqrt(21))

so a + 1/a = 5

a^3+ 1/a^3- 5 ( a^2+ 1/a^2) + (a+ 1/a)
= (a+1/a)^3 - 3 ( a + 1/a) - 5((a+1/a2^2) - 2) + (a+ 1/a)
= 5^3 - 3 * 5 - 5(5^2 - 2) + 5
= 125 - 15 - 125 + 10 + 5 = 0

2011/097) Prove the identity: (1+cosx+sinx)/(1+cosx-sinx) = secx+tanx?

regular solution
(1+ cos x + sin x)/(1+ cos x - sin x)
= (1+ cos x + sin x)^2/ (( 1+ cos x + sin x)(1+ cosx - sin x))
= (1+ cos x + sin x)^2/ ((1+ cos x) ^2 - sin ^2 x)
= (1+ cos x + sin x)^2/ ((1+ 2cos x + cos ^2 x - sin ^2 x)
= (1+ cos x + sin x)^2/ ((cos ^2 x+ 2cos x + cos ^2 x)
= (1+ cos x + sin x)^2/ (2(cos^2 x + 2 cos x))
= (1+ cos x + sin x)^2/ (2 cos x( 1+ cos x))
= ( 1 + 2 cos x + 2 sin x + 2 sin x cos x + cos ^2 x + sin ^2 x)/ (2 cos x( 1+ cos x))
= ( 2 + 2 cos x + 2 sin x + 2 cos x sin x)/(2 cos x( 1+ cos x))
= 2(1+ cos x)(1+ sin x)/ (2 cos x (1+ cos x))
= (1+ sin x)/cos x
= sec x + tan x


alternatively ( simpler solution)

1+ cos x/ sin x = ( 2 cos ^2 x/2)/(2 cos x/2 sin x/2)
= (cos x/2)/ sin x/2

using compnondo dividendo
we get
(1+cosx+sinx)/(1+cosx-sinx) = (cos x/2 + sin x/2)/( cos x/2 - sin x/2)
= ( cos x/2 + sin x/2)^2 / (( cos x/2 + sin x/2)(cos x/2 - sin x/2)
= ( cos ^2 x/ 2 + sin ^2 x/2 + 2 cos x/2 sin x/2)/(cos^2 x/2 - sin ^2 x/2)
= (1 + sin x)/ cos x
= sec x + tan x

2011/096) The equation P(x) = x^4 – 16x^3 + 94x^2 +px + q = 0 has two double roots. Solve

Because it has 2 double roots so it is square of a quadratic polynomial

Let it be (x^2+ax+b)^2

= x^4 + 2ax^3 + x^2(2b+a^2) + 2abx+ b^2

Comparing coefficients

2x = -16 or a = - 8

2b+a^2 = 94 or b = 15

p = 2ab = - 240

q = b^2 = 225

so equation

= x^4-16x^2 + 94x^2 – 240x + 225

It is (x^2 -8 x + 15) ^2 = (x-3)^2(x-5)^2 and roots are x = 3,3,5,5