If `(5 + 2 sqrt(6))^(n) = I + f `, where `I in N, n in ` N and
` 0 le f le 1, ` then I equals

Options

A. a natural number
B. a negateive integer
C. a prime number
D. an irrational number

Answer

b

Explanation

We have ,
` (5 + 2sqrt(6))^(n) = (5 _ sqrt(24))^(n)`
Now,let
` I + f = (5 + sqrt(24))^(n)` …(i)
` 0 le f lt 1` …(ii)
and ` f' = (5 - sqrt(24))^(n)` …(iii)
` 0 lt f' lt 1` …(iv)
On adding Eqs.(i) and (iii) , we get
` I + f + f' = 2k ` (even integer)
` rArr f' + f= 1`
` rArr f'=1 - f `
` therefore f^(2) - f + If - I = f (f-1) + (f-1)`
`= (f-1)(I+f)`
` =-(1-f)(I +f)= - f'(I +f)`
`= - (5 - sqrt(24))^(n) (5 + sqrt(24))^(n)`
` = - (25 - 24)^(n) =-1`
= a negative integer .