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

Solve by factorization: (m)/(n)x^(2)+(n)/(m)=1-2x

("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))

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))

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

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))

If x^(n)=a^(m)cos^(4)theta and y^(n)=b^(m)sin^(4)theta then (i)(x^((n)/(2)))/((m)/(2))+(y^((n)/(2)))/(b^((m)/(2)))=1(ii)(x^(n))/(a^(m))+(y^(n))/(b^(m))=1( iii) (x^((n)/(2)))/(y^((n)/(2)))+(a^((m)/(2)))/(y^((m)/(2)))=1 (iv) None of these

If lim_(x rarr0)(1-cos(1-cos((x)/(2))))/(2^(m)x^(n)) is equal to the left hand derivative of f(x)=e^(-(|x|)) at x=0, then the value of |n+m| is

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)]