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 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]

These type of equations can generally be solved by grouping the terms well.

= (12x -1)(3x-1)*(6x-1)(4x-1)=5

implies

(36x2 -15x +1)*(24x2  - 10x +1)=5

take 12x2 - 5x =t

implies

(3t+1)(2t+1)=5

Now you can easily solve for t.

That solves the problem!

Edit:I didnt see konichiwa2x posted already.
 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
The Obvious solution is f(x) = -g(x).
But consider,
let g be any real valued,continuous  function that satisfies f + g =0 for at least one value of x (1.4)

Let H(x) =  ( f + g)dx

By LMVT for H over (1,4),

H'(c) = [H(4) - H(1)]/3

But H'(c) = f + g

if f + g =0 for at least one c  (1.4),

Then H'(c) = 0 for at least one c  (1.4),

implies , a solution for g can be got if ,<Though this is not the only solution>

H(4) - H(1) =0

implies [1][4]( f + g)dx = 0

implies

[1][4](g)dx = -3

Hence any function that satisfies [1][4](g)dx = -3

is one class of  solution.

example g(x) = (-2x/5)

That solves the problem!
 Discussion Forums -> This Post 45 points    (9    in 9 votes )   [?]

The Problem isnt very difficult unless I am very much mistaken,

Let  a=(1+k) 1/3  and b =(k)1/3

Then the denominator reduces to  a2 + b2 + ab =( a3 - b3 )/(a-b)

But  a3 - b3     =   1

therefore substituting and by telescopic cancellation,

ans is 6 - 1 =5

That solves the problem!
 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]
This is the basic definition of Definite Integration.

If F(x) = f(x)dx

[a][b] f(x)dx = F(b) - F(a)

Hence

[0][2]  2xf(x2)dx = F(2) - F(0)

As I have assumed F(x)=
2xf(x2)dx

Actually I have a better proof:

Consider       ,

Let t=x

The Integral becomes [0][2]  2xf(x2)dx   =
[0][1]  2xf(x2)dx  + [1][2]  2xf(x2)dx

Consider F(x) = 2xf(x2)dx

By LMVT for F over (0,1),

for at least one a (0,1) ,

F'(a) = (F(1) - F(0))/(1-0) =
2af(a2) = F(1) - F(0)

similarly
By LMVT for F over (1,2),

for at least one b (1,2) ,

2bf(b2) = F(2) - F(1)

Adding these two results,we get
F(2) - F(0) = [0][2]  2xf(x2)dx = 2af(a2) + 2bf(b2)

That Solves the problem!

 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
Consider       ,

Let t=x

The Integral becomes [0][2]  2xf(x2)dx

Consider F(x) = 2xf(x2)dx

By LMVT for F over (0,2),

for at least one c (0,2) ,

F'(c) = (F(2) - F(0))/(2-0)

But (F(2) - F(0)) = [0][2]  2xf(x2)dx = I (say)

Hence I = 2F'(c) = 2 *2cf(c2) =4cf(c2) = 2cf(c2) + 2cf(c2)

Actually I dont see a necessity for 2 variables a and b.

But one possible explanation is:

if there exists more than one 'c' (a and b) that satisfies LMVT,then

I/2 =F'(a) =F'(b)

Therefore F'(a) + F'(b) =I

else

a=b

That solves the problem!
 Discussion Forums -> This Post 25 points    (5    in 5 votes )   [?]
What is given is a non recurring,non terminating decimal .
Hence not a rational number.

Note that sum of infinite rational terms need not be rational.

ex = 1 +x +x2 /2! ....

is obviously not rational even for rational values of x.

That solves the problem!
 Discussion Forums -> This Post 4 points    (0    in 2 votes )   [?]
See,

if
 f2008(x) =f2007(x)
abs(1 - f2007) = f2007

implies
f2007 =1/2,
But,
abs(1 - f2006) = f2007
implies
abs(1 - f2006) = 1/2

implies    f2006   = 1/2 or 3/2

similarly,
considering the two cases for  f2006   = 1/2 or 3/2,

we get f2005   = 1/2 or 3/2 or  5/2  <not -1/2 as abs(x) >0>
similarly
f2004   = 1/2 or 3/2 or 5/2 or  7/2

thereby,we can generalize and expect   each lower  fn   to have  1 more solution than previous one.
therefore there will be 2007 solutions for f1.

finally we will have 2007*2 solutions for x.

Therefore no of solutions is 4014.

That solves the problem!

//<proper solution posted below by Aviator>
 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
1) Just use properties of integrals

[b-1 ][b ] e-tdt/(t-b-1), replace t = (-m) + b

implies

[1 ][0 ] (em-b)dm/(1 + m)  = (-e-b) [0 ][1 ] (em)dm/(1 + m) = -a(e-b)

That solves the problem.

2) let In =
[0 ][pi ] {sin(n+1/2)x} dx/(sin(x/2))}

consider

In - In-1  = 2
[0 ][pi ]  (cos(nx))sin(x/2))(dx)/(sin(x/2)) ,implies

In - In-1  = 2  [0 ][pi ]  (cos(nx))dx

implies,

In - In-1  = 0

implies In = In-1 = . . . =I1 = I0 = pi

That solves the problem!
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
Proofs for such problems can generally be given by Calculus

consider f(x)=x^(1/x)

f'(x) = (1/x)(x)^(1/x -1) - x^(1/x -2)(log x)

=x^(1/x -2)[ 1- (log x)] <=0     for      x>=e

therefore f decreases in[e,infinity)
implies,

(pi)^(1/pi) < e^(1/e), as pi>e
implies,

(pi)^(e) < (e)^(pi)

That solves the problem!

 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]
As only negative solutions are reqd,

The question can easily be rewritten as

x2.2x+1 + 25-x  =  x2.27-x  + 2x-1     implies

2x-1 (4x2 -1) = 25-x (4x2 -1)    implies

x=3 <not possible >

or x=+-1/2

but -1/2 is not an integer.

no solution exists.

That solves the problem!

 This Post 0 points    (0    in 0 votes )   [?]
1)0
2)100^101
3)always perfect square
4)dont know it means
5)2/pi
 Discussion Forums -> This Post 20 points    (4    in 4 votes )   [?]
300! > 100300

This is a direct consequence of Stirling's  approximation of factorial values.

Stirling's approximation for higher factorials states that

$n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$
There is also another relation

(n/3)n  < n! < (n/2)n

That solves the Problem!
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
To prove:
tanxsec4x+tan4x=tanx+tan4xsec2
x          (ie.)

tanx  + sin4x  =   tanx cos4x  +  2sin2x     (ie)

tanx (1- cos4x) = 2sin2x(1  - cos2x)           (ie)

<(1- cos4x) = 2sin2 2x>

tanxsin2x =  1-cos2x                                (ie)

2sin2x  =     2sin2 x

Proved.

That solves the problem!

 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
consider,

1/(x+1) >= 3

implies

(3x - 4)/(x-1) <=0

or x lies in (1,4/3]   (upper bound(4/3) is the root of numerator and lower bound(1) is root of denominator)

similarly for

1/(x-1)  +  2/(x-2) >=3    implies

(3x2 - 11x +9)/(x-1)(x-2)  <=0

or x lies in (1, (11 - 13)/6 )(2, (11 + 13)/6 )

notice that again lower bounds and upper bounds form roots of numerator and denominator respectively.

you will find a similar case for

1/(x-1)  +  2/(x-2) + 3/(x-3) >=3

where lower and upper bounds form the roots.

therefore , for    1/(x-1)  +  2/(x-2) + 3/(x-3) ....... 120/(x-120) >=3

the roots of numerator and denominator of the simplified expression  will give upper and lower bounds.

What is asked is the sum of the roots of the numerator - sum of roots of denominator.

which you can easily find to be 2420.

That solves the problem!
the roots of the denominator will o
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
Is the answer 2420?
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
Given,
angle C =pi/3

We have to prove,
1/(a+c) + 1/(b+c) = 3/(a+b+c)     (ie)

1 +  b/(a+c)  + 1 + a/(b+c) =3     (ie)

b2 + bc = (a+c)(b+c-a)               (ie)

a2 + b2 -c2 = ab                         (ie)

(
a2 + b2 -c2 )/2ab = 1/2  = cos(pi/3)  which is cosine formula for angle C.

therefore proved.

That soles the problem!
 Discussion Forums -> This Post 7 points    (1    in 2 votes )   [?]
I think the contradictions arise because we are assigning a finite value (s) for an infinite/indeterminate  qty.

for example, in 1)

we assign s= 1-1+1-1+1-1+----------------------------------upto infinity
which is basically (infinity -infinity) which is not defined.
therefore we cant write it as
s=1-

2) here also s = infinity therefore we cant substitute.

But consider,

s= 1 + 1/2  + 1/4 + 1/8 ...
2s = 2+s,
s=2

here it works as 's' is known to have a  convergent,finite value.

I think That solves the problem!
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
Let

[x][0] {(1^x+2^x+3^x.........+n^x)/n}^1/x  = L

implies<taking log>

[x][0] {1/x}Log{(1^x+2^x+3^x.........+n^x)/n} = Log(L)

=

[x][0] {1/x}(Log{(1^x+2^x+3^x.........+n^x)} - Log(n)) = Log L

this is 0/0 form,therefore L'Hopital's rule is applicable

[x][0] 1/{(1^x+2^x+3^x.........+n^x)} * {log(1) + log(2) + log(3) ....log(n)} = LogL

implies

Log(n!)/n  = LogL

implies L = (n!)^(1/n)            (ie) C)

That solves the problem!
 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]
Hi ramyani!

see, cos(cos(cos(cosx))) will always be >0 as cos x is an even func.

therefore I am considering only values of sin x>0 because for values of sin x<0

sin(sin(sin x))<0 which cant be a solution for sin(sin(sinx)) + cos(cos(cosx)) = pi/2
as
cos(cos(cosx)) then would have to be >pi/2

therefore

0 <sinx <1

as sine is an increasing function in (0,1) ,Applying sine will preserve inequality.

therefore

sin0 <sin(sin x) <sin 1

implies

0<sin(sin x)<sin(sin1)

sorry if  my explanation was inadequate.

That solves the problem!
 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
I have another method:

sin(sin(sin(sinx)))=cos(cos(cos(cosx)))

implies

sin(sin(sin(sinx)))=sin(pi/2 - cos(cos(cosx)))

hence,

sin(sin(sinx)) + cos(cos(cosx)) = pi/2
<here we can equate angles as angles will obviously lie between 0 and ip/2>

but consider,
obviously,

0 < sin(sinx) < sin1    and   cos1<cos(cosx)<1

implies,

applying sin                           and             applying cos

0 < sin(sin(sinx)) < sin(sin1)    and      cos1 < cos(cos(cosx)) < cos(cos1)

<inequality reverses as cos is a decreasing func. in (0,pi/2)>

implies

sin(sin(sinx)) + cos(cos(cosx)) < sin(sin1) + cos(cos1)

but
sin(sin1) + cos(cos1) < pi/2  as root(2)<pi/2

therefore no solution exists for the given eqn.

That solves the problem!
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