If `alpha,beta` are the roots of `x^2+p x+q=0a n dx^(2n)+p^n x^n+q^n=0a n di f(alpha//beta),(beta//alpha)` are the roots of `x^n+1+(x+1)^n=0,t h e nn( in N)` a. must be an odd integer b. may be any integer c. must be an even integer d. cannot say anything

Options

A. must be an odd integer
B. may be any integer
C. must be an even integer
D. cannot say anything

Answer

3

Explanation

We have,
`alpha + beta = -p and alphabeta = q" "(1)`
Also, since `alpha, beta` are the roots of `x^(2n) + p^(n)x^(n) + q^(n) = 0`,
we have
`alpha^(2n) + p^(n)alpha^(n) + q^(n) = 0 and beta^(2n) + p^(n)beta^(n) + q^(n) = 0`
Subtracting the above relations, we get
`(alpha^(2n) - beta^(2n)) + p^(n) (alpha^(n) - beta^(n)) = 0`
`therefore alpha^(n) + beta^(n) = -p^(n)" "(2)`
Given, `alpha//beta` or `beta//alpha` is a root of `x^(n) + 1 + (x + 1)^(n) = 0`. So,
`(alpha//beta)^(n) + 1 + [(alpha//beta) + 1]^(n) = 0`
`rArr (alpha^(n) + beta^(n)) + (alpha + beta)^(n) = 0`
`rArr -p^(n) + (-p)^(n) = 0 " "` [Using (1) and (2)]
It is possible only when n is even.