If `x,y` is positive real numbers then maximum value of `(x^my^m)/((1+x^(2m))(1+y^(2n)))` is (A) `1/4` (B) `1/2` (C) `(m+n)/(6mn)` (D) `1`

Let x, y be positive real numbers and m, n be positive integers, The maximum value of the expression (x^(m)y^(n))/((1+x^(2m))(1+y^(2n))) is

If x and y are positive real numbers and m,n are any positive integers,then (x^(n)y^(m))/((1+x^(2n))(1+y^(2m)))<(1)/(4)

If x,y are positive real numbers and m, n are positive integers, then prove that (x^(n) y^(m))/((1 + x^(2n))(1 + y^(2m))) le (1)/(4)

If a,m,n are positive integers,then {anm}^(mn) is equal to a^(mn)( b) a(c)a^((m)/(n))(d)1

For the natural number m, n, if ( 1- y ) ^(m) (1 + y) ^( n) =1 + a _(1) y + a _(2) y ^(2) + ...+ a _( m + n) y ^( m +n) and a _(1) = a _(2) = 10, then the value of (m+n) is equal to :

If 1<=m<=n,m in n, then the value of lim_(x rarr oo)(x_(1),x_(2)......x_(n)) is

If tan x=n tan y,n in R^(+), then the maximum value of sec^(2)(x-y) is equal to (a) ((n+1)^(2))/(2n) (b) ((n+1)^(2))/(n)( c) ((n+1)^(2))/(2) (d) ((n+1)^(2))/(4n)

y-1=m_1(x-3) and y - 3 = m_2(x - 1) are two family of straight lines, at right angled to each other. The locus of their point of intersection is: (A) x^2 + y^2 - 2x - 6y + 10 = 0 (B) x^2 + y^2 - 4x - 4y +6 = 0 (C) x^2 + y^2 - 2x - 6y + 6 = 0 (D) x^2 + y^2 - 4x - by - 6 = 0

For natural numbers m, n if (1-y)^(m)(1+y)^(n) = 1+a_(1)y+a_(2)y^(2) + "……." and a_(1) = a_(2) = 10 , then (m,n) is :