If -0- -1- -2- - -n-1 are nth roots of unity- then- k -cos2 k- n - i sin2 k- n - where0 - k - n -1- Also x n -1- x -0 x -1 - x - n -1Value of 1-01-1 -1-n-1 isA- 1-1- n -1B- 3 C- -1 nD- 0

The correct option is A 1+( 1)^ n+1Put x = 1, so that ( 1)^n 1 = ( 1 α _0) ( 1 α _1)… ( 1 α _n 1) ⇒ ( 1)^n [1+ ( 1)^ n +1) =( 1)^n (1 + α _0) (1 + ...

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Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

If α0, α1, α2...

Question

If $α_{0}, α_{1}, α_{2}, \dots, α_{n - 1}$ are $n^{t h}$ roots of unity, then

$α_{k} = c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})$ ; where

$0 \leq k \leq n - 1$ , Also

$(x^{n} - 1) = (x - α_{0}) (x - α_{1}) \dots (x - α_{n - 1})$

Value of $(1 + α_{0}) (1 + α_{1}) \dots (1 + α_{n - 1})$ is

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(- 1)^{n}

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0.0

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1 + (- 1)^{- n + 1}

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Solution

The correct option is D $1 + (- 1)^{- n + 1}$
Put x = -1, so that

$(- 1)^{n} - 1 = (- 1 - α_{0}) (- 1 - α_{1}) \dots (- 1 - α_{n - 1})$

$\Rightarrow (- 1)^{n} [1 + (- 1)^{- n + 1}) = (- 1)^{n} (1 + α_{0}) (1 + α_{1}) \dots (1 + α_{n - 1})$

$\Rightarrow (1 + α_{0}) (1 + α_{1}) \dots (1 + α_{n - 1}) = 1 + (- 1)^{n - 1}$

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Similar questions

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Q. If

1, α_{1}, α_{2}, α_{3}, . . . . ., α_{n - 1}

are the nth roots of unity then show that

n = (1 - α_{1}) (1 - α_{2}) (1 - α_{3}) . . . . . (1 - α_{n - 1})

Q. If n is an odd positive integer and

I, α_{1}, α_{2}, . . . . α_{n - 1}

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Q. If

1, α_{1}, α_{2}, α_{3}, . . ., α_{n - 1}

are the nth roots of unity, then the value of

1. α_{1} . α_{2} . α_{3} . . . α_{n - 1}

Q. If

α_{1}

α_{2}

, ....

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If α0,α1,α2,…,αn−1 are nth roots of unity, then αk=cos(2kπn)+i sin(2kπn); where 0≤k≤n−1, Also (xn−1)=(x−α0)(x−α1)…(x−αn−1) Value of (1+α0)(1+α1)…(1+αn−1) is

The correct option is D 1+(−1)−n+1Put x = -1, so that (−1)n−1=(−1−α0)(−1−α1)…(−1−αn−1) ⇒(−1)n[1+(−1)−n+1)=(−1)n(1+α0)(1+α1)…(1+αn−1) ⇒(1+α0)(1+α1)…(1+αn−1)=1+(−1)n−1