Suppose that `f(x)f(f(x))=1` and `f(1000)=999` then which of the following is true

Options

A. `f(500)=(1)/(500)`
B. `f(199)=(1)/(199)`
C. `f(x)=(1)/(x)AA x in R-{0}`
D. `f(1999)=(1)/(1999)`

Answer

A, B

Explanation

`f(1000)f(f(1000))=1`
`rArr" "f(1000)f(999)=1`
`rArr" "999f(999)=1`
`therefore" "f(999)=(1)/(999)`
The numbers 999 and `(1)/(999)` are in the range of f.
Hence, by intermediate value property of continuous function, function takes all values between 999 and `(1)/(999)`, then there exists
`alpha in ((1)/(999),999)` such that `f(alpha)=500`
Then `f(alpha)f(f(alpha))=1 rArr f(500)=(1)/(500)`
Similarly, `199 in ((1)/(199),999),` thus `f(199)=(1)/(199)`
But there is nothing to show that 1999 lies in the range of f.
Thus (d) is not correct and (c) is alos incorrect.