If x ,1,`(1)/(x)` are in G.P then x can't have a value

If f(x) = (1)/(1 -x) x ne 1 and g(x) = (x-1)/(x) , x ne0 , then the value of g[f(x)] is :

If 5x-y,2x+y,x+2y are in "A.P" and (x-1)^(2),xy+1,(y+1)^(2) are in G.P (x!=0) then x+y=

If x,1,z are in A.P. and x,2,z are in G.P., then x,4,z are in

If f(x)=x^2+(1)/(x^2), g(x)=x^4+(1)/(x^4) and a+(1)/(a)=3 , then the respectively values of f(a) and g(a) are

P(x) be a polynomial of degree 3 satisfying P(-1) =10 , P(1) =-6 and p(x) has maxima at x = -1 and p(x) has minima at x=1 then The value of P(1) is