(m)/(n)x^(2)+(n)/(m)=1-2x

lim_(x rarr0)((2^(m)+x)^((1)/(m))-(2^(n)+x)^((1)/(n)))/(x) is equal to (1)/(m2^(m))-(1)/(n2^(n)) (b) (1)/(m2^(m))+(1)/(n2^(n))(1)/(m2^(-m))-(1)/(n2^(-n))( d) (1)/(m2^(-m))+(1)/(n2^(-n))

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|))

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. (a) (|m+n|)/((m-n)^2) (b) 2/(|m+n|) 1/((|m+n|)) (d) 1/((|m-n|))

("lim")_(xvec 0)((2^m+x)^(1/m)-(2^n+x)^(1/n))/xi se q u a lto 1/(m2^m)-1/(n2^n) (b) 1/(m2^m)+1/(n2^n) 1/(m2^(-m))-1/(n2^(-n)) (d) 1/(m2^(-m))+1/(n2^(-n))

("lim")_(xto0)((2^m+x)^(1/m)-(2^n+x)^(1/n))/x is equal t o (a) 2 (1/(m2^m)-1/(n2^n))' (b) (1/(m2^m)+1/(n2^n)) (c) 1/(m2^(-m))-1/(n2^(-n)) (d) 1/(m2^(-m))+1/(n2^(-n))

lim_(x->0) [(2^m +x)^(1/m)-(2^m+x)^(1/n)]/x a) 1/(m2^m)-1/(n2^n) b) 1/(m2^(m-1))-1/(n2^(n-1)) c) m/2^(m-1)-n/2^(n-1) d) none of these

The mth term of an arithmetic progression is x and nth term is y.Then the sum of the first (m+n) terms is: a.(m+n)/(2)[x-y+(x+y)/(m+n)] b.(1)/(2)[(x+y)/(m+n)+(x-y)/(m-n)]c(1)/(2)[(x+y)/(m+n)-(x-y)/(m-n)]d(m+n)/(2)[x+y+(x-y)/(m-n)]

If 2 cos alpha = x + (1)/(x) and 2 cos beta = y + (1)/(y) , show that x^(m) y^(n) + (1)/(x^(m)y^(n)) = 2 cos (m alpha + n beta)