If ` x + (1)/(x) = 1 " and" p = x^(4000) + (1)/(x^(4000)) ` and q is the digit at
unit place in the number ` 2^(2^(n)) + 1, n in N abd n gt 1` , then p + q is .

Options

A. 8
B. 6
C. 7
D. None of these

Answer

b

Explanation

Given , `(x + (1)/(x) = 1 rArr x^(2) - x + 1 = `
` rArr (x + omega ) (x + omega^(2)) = 0 `
` rArr = x - omega - omega^(2)`
` therefore p= (-omega)^(4000) + (1)/((-omega)^(4000)) = omega ^(4000) + (1)/(omega ^(4000))`
` = omega + (1)/(omega) = (omega^(2) + 1)/(omega)= - (omega)/(omega)= -1`
Also , for `x = omega^(2) , p = - 1`
For ` n gt 1,2^(2) = 4k, k in N`
` therefore 2^(2^(n)) = 2^(4k) = (16)^(k)` = last digit number is 6
Now , q = unit digit at unit place in the number ` (2^(2^(n)) + 1)`
` = 6 + 1 7`
` therefore p + q = - 1 + 7 = 6` .