Suppose that `f(x)f(f(x))=1` and `f(1000)=999` then which of the following is trueA. `f(500)=(1)/(500)`B. `f(199)=(1)/(199)`C. `f(x)=(1)/(x)AA x in R-{0}`D. `f(1999)=(1)/(1999)`

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

1.	Suppose that `f(x)f(f(x))=1` and `f(1000)=999` then which of the following is trueA. `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» Correct Answer - A::B `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.