For non-negative integers m and n a function is defined as follows f(m,n) = {n+1 if m=0; f(m-1,1) if m!=0, n=0}` Then the value of f(1,1) is

For non-negative integers m and n a function is defined as follows: f(m,n)={n+1, if m=0 and f(m-1,1) if m!=0,n=0 and f(m-1,f(m,n-1)) if m!=0,n Then the value of f(1,1) is:

For non-negative integers ma n dn a function is defined as follows: f(m ,n)={n+1ifm=0f(m-1,1)ifm!=0,n=0f(m-1,f(m ,n-1))ifm!=0,n!=0 Then the value of f(1,1) is a. 1 b. 12 c. 3 d. 4

For non-negative integers ma n dn a function is defined as follows: f(m ,n)={n+1ifm=0f(m-1,1)ifm!=0,n=0f(m-1,f(m ,n-1))ifm!=0,n!=0 Then the value of f(1,1) is a. 1 b. 12 c. 3 d. 4

For non-negative integers ma n dn a function is defined as follows: f(m ,n)={n+1ifm=0f(m-1,1)ifm!=0,n=0 f(m-1,f(m ,n-1))ifm!=0,n!=0 Then the value of f(1,1) is a. 1 b. 12 c. 3 d. 4

For non-negative integers ma n dn a function is defined as follows: f(m ,n)={n+1ifm=0 f(m-1,1)ifm!=0,n=0 f(m-1,f(m ,n-1))ifm!=0,n!=0} Then the value of f(1,1) is a. 1 b. 12 c. 3 d. 4

For non-negative integers m and n a function is defined as follows: f(m,n)={n+1,ifm=0andf(m−1,1)ifm≠0,n=0andf(m−1,f(m,n−1))ifm≠0,n≠0}Then the value of f(1,1) is:

Let n be a positive integer,and a function f be defined as follows: {(0,n=1),(f([n/2])+1,ngt1):} Then find f(35).

Let f be a function from the set of all non-negative integers into itself such that f(n + 1) = 6f(n) + 1 and f(0) = 1 , then the value of f(4) – f(3)

Let f(x)=x^(2)+ax+b be a quadratic polynomial in which a and b are integers. If for a given integer n, f(n) f(n+1)=f(m) for some integer m, then the value of m is