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

Options

A. domain of `f(x)` is R
B. domain of `f " is " R-(-2,2)`
C. range of `f(x)" is " [-2,oo)`
D. range of `f(x) " is " [2,oo)`

Answer

B, D

Explanation

`f(x+(1)/(x))=x^(2)+(1)/(x^(2))`
`or f(x+(1)/(x))=x^(2)+(1)/(x^(2))=(x+(1)/(x))^(2)-2`
` or f(y)=y^(2)-2`
Now, `y=x+(1)/(x) ge 2 or le -2`
Hence, the domain of the function is `(-oo, -2] cup [2, oo).`
Also, for these values of `y, y^(2) ge 4 or y^(2) -2 ge 2.`
Hence, the range of the function is `[2,oo).`