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If roots of an equation x^n-1=0 are 1,a1...

If roots of an equation `x^n-1=0` are `1,a_1,a_2,........,a_(n-1)` then the value of `(1-a_1)(1-a_2)(1-a_3)........(1-a_(n-1))` will be (a) `n` (b) `n^2` (c) `n^n` (d) `0`

Text Solution

Verified by Experts

We can write,
`x^n - 1 = (x-1)(x-a_1)(x-a_2)...(x-a_n)`
`=>(x^n - 1)/(x-1) = (x-a_1)(x-a_2)...(x-a_n)`
Applying limits `Lim_(x->1)` on both sides,
`Lim_(x->1) (x-a_1)(x-a_2)...(x-a_n) = Lim_(x->1) (x^n - 1)/(x-1)`
`=>(1-a_1)(1-a_2)...(1-a_n) = Lim_(x->1) (x^n - 1)/(x-1)`
`Lim_(x->1) (x^n - 1)/(x-1)` is a `0/0` form. So, we will apply L`'` hospital rule.
`Lim_(x->1) (x^n - 1)/(x-1) = Lim_(x->1) (nx^(n-1))/1 = n`
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