Let `f(n)=a^n+b^n and |(3, 1+f(1), 1f(2)),(1+f(1), 1+f(2), 1+f(3)),(1+f(2), 1+f(3), 1+f(4))|=k(1-a)^2(1-b)^2(a-b)^2`, then k= (A) 0 (B) 1 (C) -1 (D) 4

If alpha,beta!=0 , and f(n)""=alpha^n+beta^n and |3 1+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)|=K(1-alpha)^2(1-beta)^2(alpha-beta)^2 , then K is equal to (1) alphabeta (2) 1/(alphabeta) (3) 1 (4) -1

If f(1) = 1 and f(n + 1) = 2 f(n) + 1, if n ge 1, then f(n) is.

Let f(n)=2cos nx AA n in N, then f(1)f(n+1)-f(n) is equal to (a)f(n+3) (b) f(n+2)(c)f(n+1)f(2)(d)f(n+2)f(2)

Let f(x+y)=f(x).f(y) for all x and y and f(1)=2. If in as triangle ABC, a=f(3), b=f(1)+f(3), c=f(2)+f(3), then 2A= (A) C (B) 2C (C) 3C (D) 4C

If f(x)=|(log)_(2)x|, then f(1^(+))=1 (b) f(1^(-))=-1f(1)=1( c) f'(1)=-1

f (1) = 1, n> = 1f (n + 1) = 2f (n) +1 then f (n) =

If f(n)=|(n!, (n+1)!, (n+2)!),((n+1)!, (n+2)!, (n+3)!), ((n+2)!, (n+3)!, (n+4)!)| Then the value of 1/1020[(f(100))/(f(99))] is