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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.
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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!
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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!
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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=x2
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!
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Consider ,
Let t=x2
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!
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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!
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See,
if
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>
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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!
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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!
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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!
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1)0
2)100^101
3)always perfect square
4)dont know it means
5)2/pi
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300! > 100300
This is a direct consequence of Stirling's approximation of factorial values.
Stirling's approximation for higher factorials states that
There is also another relation
(n/3)n < n! < (n/2)n
That solves the Problem!
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To prove:
tanxsec4x+tan4x=tanx+tan4xsec2x (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!
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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
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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!
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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-s
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!
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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!
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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!
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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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Vriti Edustore
