If `A=[(a,b),(0,a)]` is nth root of `I_(2)`, then choose the correct statements :
(i) if n is odd, a=1, b=0
(ii) if n is odd, a=-1, b=0
(iii) if n is even, a=1, b=0
(iv) if n is even, a=-1, b=0

If A=[a b0a] is nth root of I_2, then choose the correct statements: If n is odd, a=1,b=0 If n is odd, a=-1,b=0 If n is even, a=1,b=0 If n is even, a=-1,b=0 a. i, ii, iii, iv b. ii, iii, iv c. i, ii, iii, iv d. i, iii, iv

If f(x)=(sin^2x-1)^("n"),"" then x=pi/2 is a point of local maximum, if n is odd local minimum, if n is odd local maximum, if n is even local minimum, if n is even

Let A=[(0,1),(0,0)] , show that (aI+bA)^(n)=a^(n)I+na^(n-1)bA , where I is the identity matrix of order 2 and n in N .

Find the sequence of the numbers defined by a_(n)={{:(1/n,"when n is odd"),(-1/n,"when n is even"):}

The function f(x)=(4sin^2x−1)^n (x^2−x+1),n in N , has a local minimum at x=pi/6 . Then (a) n is any even number (b) n is an odd number (c) n is odd prime number (d) n is any natural number

If (a^(n + 1) + b^(n + 1))/(a^(n) + b^(n)) is the arithmetic mean between a and b , then n =

If a > b >0, with the aid of Lagranges mean value theorem, prove that n b^(n-1)(a-b) < a^n -b^n < n a^(n-1)(a-b) , if n >1. n b^(n-1)(a-b) > a^n-b^n > n a^(n-1)(a-b) , if 0 < n < 1.

The nth derivative of the function f(x)=1/(1-x^2) [where x in(-1,1) at the point x=0 where n is even is (a) 0 (b) n! (c) n^nC_2 (d) 2^nC_2

Roots of the equation |x m n1a x n1a b x1a b c1|=0 are independent of m ,a n dn independent of a ,b ,a n dc depend on m ,n ,a n da ,b ,c independent of m , na n da ,b ,c

Let A =({:(1,0,1,0),(0,1,0,1),(1,0,0,1):}) , B=({:(0,1,0,1),(1,0,1,0),(1,0,0,1):}) C=({:(1,1,0,1),(0,1,1,0),(1,1,1,1):}) by any three boolean matrices of the same type. Find (i) A vee B , (ii) A wedge B , (iii) (A vee A) wedge C , (iv) (A wedge B) vee C .