If `f(x)=alpha x+beta` and `f(0)=f^(1)(0)=1` then `f(2)`=

If f(x)= alpha x+beta and f(0)=f^(')(0)=1 then f(2) =

If f(x)=alpha x+beta and f ={(1, 1), (2, 3), (3, 5), (4, 7)}, then the values of alpha,beta are :

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'(x)=x^2+5 and f(0)=-1 then f(x)=

If f'(x)>0,f(x)>0AA x in(0,1) and f(0)=0,f(1)=1, then prove that 'f(x)f^(^^)-1(x)

If a function 'f' satisfies the relation f(x)f^('')(x)-f(x)f^(')(x) -f^(')(x)^(2)=0 AA x in R and f(0)=1=f^(')(0) . Then find f(x) .

If y=f(x) satisfies f(x+1)+f(z-1)=f(x+z) AA x, z in R and f(0)=0 and f'(0)=4 then f(2) is equal to