2009/028) Rationalize the denominator of expression? [1 / {1 + x^1/5}]

Rationalize the denominator of expression?
[1 / {1 + x^1/5}]


ANS:
we know that (x^(1/5))^5 = x

say x^(1/5) = a

1+ x^ 1/5 = (1+ a)

as (1+a^5)/(1+a) = (1-a+a^2-a^3+a^4) and 1+a^5 in terms of x is rationalised

so
[1 / {1 + x^1/5}] = (1-x^(1/5) + x^(2/5) - x(^3/5) + x^(4/5))/(1+x)

2009/027) If α = 2 arctan [(1 + x)/(1 - x)] and β = arcsin [(1 - x^2)/(1 + x^2)], then what is α + β

we know tan(pi/4+y) = tan (pi/4 + tan y)/(1 tan pi/4-tan y)

so tan (pi/4+y) = (1+ tan y)/(1-tan y))

so pi/4 + y = tan ^-1(1+tan y)/(1-tan y)

put x = tan y

so pi/4 + y = arctan ((1+x)/(1-x))

α = 2 arctan ((1+x)/(1-x)) = pi/2 + 2 tan ^- 1 x .1

again cos 2y = cos ^2 y - sin ^2 y

= cos^2 y(1- tan ^2 y)
= (1- tan ^2 y)/sec^2 y
= (1- tan ^2 y)/(1+ tan ^2y)

so put tan y = x

cos 2y = (1-x^2)/(1+x^2)
2y = arccos ((1-x^2)/(1+x^2))

so arcsin ((1-x^2)/(1+x^2)) = pi/2 - 2y = pi/2 - 2 tan ^- x .2

addimg 1 and 2 we get
α + β = π

2009/026) If a , b , c are in Arithmetic Progression prove: 1/(b1/2+c1/2),1/(c1/2+a1/2),1/… also in AP.

If a , b , c are in Arithmetic Progression prove : 1 / (b1/2+c1/2) , 1 / (c1/2+a1/2) , 1 / (a1/2+b1/2) in AP?

to get rid of ^1/2

Let a = x^2 , b= y^2 and c = z^2

We need to prove

1/(y+z) , 1/(x+z) and 1/(x+y) are in AP


As x^2 y^2 and z^2 are in AP

So x^2+xy + yz + xz , y^2 + +xy + yz + xz , z^2+xy + yz + xz (adding same to all terms) are in AP

So (x+y)(x+z), (y+x)(y+z), (z+x)(z+y) are in AP by factoring

So 1/(y+z) , 1/(x+z) and 1/(x+y) are in AP by deviding each term by (x+y)(x+z)(y+z)

2009/025) Let t_n denote the nth triangular number. For what values of n does t_n divide the sum t_1+t_2+...+t_n?

Well we know t_m = (m²+m)/2

We want S_n = Σ_n=1_m [ t_m ] = Σ_n=1_m [ (m²+m)/2 ]
Applying Faulhaber's Formulas,
S_n = [ (2n³+3n²+n)/6 + (n²+n)/2 ] /2
= (2n³+3n²+n + 3n²+3n) /12
= (2n³+6n²+4n) /12
= n(n+1)(n+2) /6

So the question is for what values of n does
n(n+1)/2 | n(n+1)(n+2)/6

RHS/LHS = (n+2)/3 and LHS is a factor if and only if this is an integer

so n + 2 = 0 mod 3 or n = 1 mod 3

so for n = 1 mod 3 the value t_n devides the sum

2009/024) Resolve into factors X^4-3x+20

if this can be factored with real coefficients then it can be as product of 2 quadratic
we shall get
(x^2+ax+b)(x^2+cx+d)

now coeffcient of x^3 = 0 so a +c = 0 or c = -a

so we get(x^2+ax+b)(x^2-ax+d)
= x^4 + (d + b - a^2) x^2 + x(ad-ab) + bd

comparing coefficients

a^2 = b + d
bd = 20
so ad -ab = - 3
by trial and error we see that a = 3 , b= 5 and d= 4 (this is using 20 = 1 * 20 = 2(*10 - 4 * 5 outof which only 4 +5 = 9 we get perfect square) satisfies
this can be factored with rational coefficient as
(x^2+3x+5)(x^2-3x+4)
and not further this can be checked by quadaric factorisation

hence X^4-3x+20 = (x^2+3x+5)(x^2-3x+4)

2009/023) Find values of x: √x + √(x+2) + √[x(x+2)] = 3-x

as product of 1st term and 2nd term is 3rd I would take the 3rd term to the right and get

√x + √(x+2) = (3-x)- √[x(x+2)]

square both sides to get
x + (x+2) + 2√[x(x+2)] = (3-x)^2 + x(x+2)- 2(3-x) √[x(x+2)]

or
2x + 2 + 2√[x(x+2)] = 9-6x+ x^2 + x^2 + 2x - 2(3-x) √[x(x+2)]
= 9-4x+ 2x^2 +(2x-6) √[x(x+2)]

now keeping redicals on one side we get
2x + 2 - (2x^2-4x+9) = (2x-8) √[x(x+2)]
(2x-8) √[x(x+2)] = -2x^2 +6x -7
now by squaring again you can get rid of redicals and proceed

No one has solved so far so I continue

(2x-8)^2x(x+2) = (-2x^2+6x-7)^2

or expanding we get after simplfication

64x^2-212x + 49 = 0

or(4x-1)(16x-49) = 0

x = 1/4 or 49/16
x has to be less than 3 as LHS > 0

so x = 1/4 need to be checked and this is indeed a solution

so x = 1/4

2009/022) Prove that c(r,r)+c(r+1,r)+.....+c(n,r)=c(n+1,r+1)

The right hand side is choosing r+1 objects from n+1 objects,

let us count another way

as we need to chose r+1 objects if we order the numbers from 1 to n+1 and chose r+1 objects the last object shall be in position r+1 to n+1

let last object be in position m ( r+1 <= m <=n + 1)

then we chose r objects from m-1 objects

this can be done in c(m-1,r)

m varies from 1+1 to n+1

so total number of ways = sum c(m-1.,r) m from r+1 to n+ 1

or sum c(m,r) (m from r to n)
= c(r,r) +c(r+1,r)+ c(r+2,r)+ … c(n,r)

As by 2 methods there are 2 results(it has been counted correctly) so both are same and hence

c(r,r) +c(r+1,r)+ c(r+2,r)+ … c(n,r) = c(n+1,r+1)

hence proved