If R = `(sqrt(2) + 1)^(2n+1) and f = R - [R]`, where [ ]
denote the greatest integer function, then [R] equal (a) `f+1/f` (b) `f-1/f` (c) `1/f-f` (d) None of these

Options

A. `f+(1)/(f)`
B. `f-(1)/(f)`
C. `(1)/(f) -f`
D. none of these

Answer

c

Explanation

Let G ` = (sqrt(2) - 1)^(2n +1!)`. Then
`R - G = (sqrt(2) +1)^(2n +1) - (sqrt(2) -1)^(2n+1)`
`rArr R - G = {""^(2n +1)C_(1) (sqrt(2))^(2n) + ""^(2n+1)C_(3) (sqrt(2))^(2n=1)+...}`
`rArr R-G` = An even integer
`rArr [R] + f - G` = An even integer = 1 (say)
`rArr f- G = 0 rArr f - G`
Now,
f = G
`rArr (1)/(f) = (1)/(G) = (1)/(sqrt(2) -1)^(2n-1) = (sqrt(2) +1)^(2n +1)= R`
`rArr (1)/(f) = R rArr (1)/(f) = f | [R] rArr [R] rArr [R] = (1)/(f) f` .