InterviewSolution
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InterviewSolution
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If `alpha,beta`are the roots of `x^2+p x+q=0a d nx^(2n)+p^n x^n+q^n=0a n di lf(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 integerb. may be any integerc. must be an even integer d. cannot say anythingA. must be an odd integerB. may be any integerC. must be an even integerD. cannot say anything |
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Answer» Correct Answer - 3 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. |
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