For natural numbers `m ,n ,if(1-y)^m(1+y)^n=1+a_1y+a_2y^2+... , and a_1=a_2=10 , then` a. `m < n` b. `m > n` c. `m+n=80` d. `m-n=20`

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 :

If m\ a n d\ n are the order the degree of the differential equation (y_2)^5+(4(y_2)^3)/(y_3)+y_3=x^2-1, then a. m=3,\ n=3 b. m=3,\ n=2 c. m=3,\ n=5 d. m=3,\ n=1

Explain , giving reason , which of the following sets of quantum number are not possible {:(a,n= 0 ,l = 0, m_(1) = 0, m_(s) = +1//2),(b,n= 1 ,l = 0, m_(1) = 0, m_(s) = -1//2),(c,n= 1 ,l = 1, m_(1) = 0, m_(s) = +1//2),(d,n= 2 ,l = 1, m_(1) = 0, m_(s) = -1//2),(e,n= 3 ,l = 3, m_(1) = -3, m_(s) = +1//2),(f,n= 3 ,l = 1, m_(1) = 0, m_(s) = +1//2):}

If \ ^m C_1=\ \ ^n C_2 then which is correct a. 2m=n b. 2m=n(n+1) c. 2m=(n-1) d. 2n=m(m-1)

If m =! n and (m + n)^(-1) (m^(-1) + n^(-1)) = m^(x) n^(y) , show that : x + y + 2 = 0.

If 1/ (m+i n) - (x-iy)/(x+iy) =0, where x,y,m,n are real and x+iy!=0 and m+i n!=0 , prove that m^2+n^2=1 .

Simplify m - n+2 [ m - {m - (m+n-2) -1} +1] Hence , find its value when m = n approx 1

If f and g are two functions defined on N , such that f(n)= {{:(2n-1if n is even), (2n+2 if n is odd):} and g(n)=f(n)+f(n+1) Then range of g is (A) {m in N : m= multiple of 4 } (B) { set of even natural numbers } (C) {m in N : m=4k+3,k is a natural number (D) {m in N : m= multiple of 3 or multiple of 4 }

The area of the parallelogram formed by the lines y=m x ,y=x m+1,y=n x ,a n dy=n x+1 equals. (|m+n|)/((m-n)^2) (b) 2/(|m+n|) 1/((|m+n|)) (d) 1/((|m-n|))

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