`I fy=f(x)` satisfies the condition `f(x+1/x)=x^2+1/(x^2)(x!=0)` then `f(x)=`

Ify=f(x) satisfies the condition f(x+(1)/(x))=x^(2)+(1)/(x^(2))(x!=0) then f(x)=

Consider the function y=f(x) satisfying the condition f(x+(1)/(x))=x^(2)+1/x^(2)(x!=0) Then the domain of f(x) is R domain of f(x) is R-(-2,2) range of f(x) is [-2,oo] range of f(x) is (2,oo)

Consider the function y=f(x) satisfying the condition f(x+(1)/(x))=x^(2)+(1)/(x^(2))(x!=0)

If a function satisfies the condition f(x+1/x)=x^(2)+1/(x^(2)),xne0 , then domain of f(x) is

The function y=f(x) satisfying the condition f(x+(1)/(x))=x^(3)+(1)/(x^(3)) is

A polynomial function f(x) satisfies the condition f(x)f((1)/(x))=f(x)+f((1)/(x)) . If f(10)=1001, then f(20)=

A function f:R rarr R satisfies the condition x^(2)f(x)+f(1-x)=2x-x^(4). Then f(x) is

A polynomial function f(x) satisfies the condition f(x)f((1)/(x))=f(x)+f((1)/(x)) for all x in R,x!=0. If f(3)=-26, then f(4)=