Let J_n,m=ʃ0 to1/2 x^n/(x^m-1) dx,ꓯn greater than m and n ,m ∊N. Consider a matrix A=[a_ij]3ₓ3 where

Let f(x)=x^(m/n) for x in R where m and n are integers , m even and n odd and 0

If I(m,n)=int_0^1x^(m-1)(1-x)^(n-1)dx , then

If x^m y^n=(x+y)^(m+n) , prove that (d^2y)/(dx^2)=0 .

x^(3m) + x^(3n-1) + x^(3r-2) , where, m,n,r in N is divisible by

If mgt0, ngt0 , the definite integral I=int_0^1 x^(m-1)(1-x)^(n-1)dx depends upon the values of m and n is denoted by beta(m,n) , called the beta function.Obviously, beta(n,m)=beta(m,n) .Now answer the question:If int_0^oo x^(m-1)/(1+x)^(m+n)dx=k int_0^oo x^(n-1)/(1+x)^(m+n)dx , then k is equal to (A) m/n (B) 1 (C) n/m (D) none of these

If I_(m,n)= int_(0)^(1) x^(m) (ln x)^(n) dx then I_(m,n) is also equal to

Let m ,n are integers with 0ltnltm . A is the point (m ,n) on the Cartesian plane. B is the reflection of A in the line y=x . C is the reflection of B in the y-axis, D is the reflection of C in the x-axis and E is the reflection of D in the y-axis. The area of the pentagon A B C D E is a. 2m(m+n) b. m(m+3n) c. m(2m+3n) d. 2m(m+3n)

Let m ,n are integers with 0

Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(2","2) is

Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(6","2) is