If the equation x n -1-0- n -1- n - N- has roots 1- a 1- a 2- - a n -1- thenA- 1-a11-a2 -1-an-1-nB- - r -1 n -11-2- a r -2 n -1 n -2-1-2 n -1C- -r-1n-11-2-ar-2n-1n-1-1-2n-1D- -r-1n-11-1-ar-n-1-2

The correct options are A (1 a_1)(1 a_2)⋯(1 a_n 1)=n B ∑_r=1^n 112 a_r=2^n 1(n 2)+12^n 1 D ∑_r=1^n 111 a_r=n 12Since, nbsp;1,a_1,a_2, …,a_n 1 are roots of x^n ...

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Byju's Answer

Standard XII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

If the equati...

Question

If the equation $x^{n} - 1 = 0, n > 1, n \in N,$ has roots $1, a_{1}, a_{2}, \dots, a_{n - 1},$ then

(1 - a_{1}) (1 - a_{2}) \dots (1 - a_{n - 1}) = n

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n - 1 \sum r = 1 \frac{1}{2 - a_{r}} = \frac{2^{n - 1} (n - 2) + 1}{2^{n} - 1}

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n - 1 \sum r = 1 \frac{1}{2 - a_{r}} = \frac{2^{n - 1} (n - 1) - 1}{2^{n} - 1}

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n - 1 \sum r = 1 \frac{1}{1 - a_{r}} = \frac{n - 1}{2}

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Solution

The correct options are
A $(1 - a_{1}) (1 - a_{2}) \dots (1 - a_{n - 1}) = n$
B $n - 1 \sum r = 1 \frac{1}{2 - a_{r}} = \frac{2^{n - 1} (n - 2) + 1}{2^{n} - 1}$
D $n - 1 \sum r = 1 \frac{1}{1 - a_{r}} = \frac{n - 1}{2}$
Since, $1, a_{1}, a_{2}, \dots, a_{n - 1}$ are roots of $x^{n} - 1 = 0,$ then
$x^{n} - 1 = (x - 1) (x - a_{1}) (x - a_{2}) \dots (x - a_{n - 1}) \dots (1)$
$\Rightarrow \frac{x^{n} - 1}{x - 1} = (x - a_{1}) (x - a_{2}) \dots (x - a_{n - 1})$
$\Rightarrow lim x \to 1 \frac{x^{n} - 1}{x - 1} = lim x \to 1 [(x - a_{1}) (x - a_{2}) \dots (x - a_{n - 1})]$
$\Rightarrow (1 - a_{1}) (1 - a_{2}) \dots (1 - a_{n - 1}) = n$

From equation $(1)$ ,
$log (x^{n} - 1) = log (x - 1) + log (x - a_{1}) + \dots + log (x - a_{n - 1})$
Differentiating w.r.t. $x$ , we get
$\frac{n x^{n - 1}}{x^{n} - 1} = \frac{1}{x - 1} + \frac{1}{x - a_{1}} + \frac{1}{x - a_{2}} + \dots + \frac{1}{x - a_{n - 1}} \dots (2)$
Putting $x = 2$ in $(2)$ , we get
$\frac{n 2^{n - 1}}{2^{n} - 1} = 1 + \frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + \dots + \frac{1}{2 - a_{n - 1}}$
$\Rightarrow \frac{1}{2 - a_{1}} + \frac{1}{2 - a_{2}} + . . . + \frac{1}{2 - a_{n - 1}} = \frac{n 2^{n - 1}}{2^{n} - 1} - 1$
$= \frac{n 2^{n - 1} - 2^{n} + 1}{2^{n} - 1}$
$= \frac{2^{n - 1} (n - 2) + 1}{2^{n} - 1}$

From equation $(2)$
$\frac{n x^{n - 1}}{x^{n} - 1} - \frac{1}{x - 1} = \frac{1}{x - a_{1}} + \frac{1}{x - a_{2}} + \dots + \frac{1}{x - a_{n - 1}}$
$\Rightarrow \frac{n x^{n - 1} - 1 (1 + x + x^{2} + \dots + x^{n - 1})}{x^{n} - 1} = \frac{1}{x - a_{1}} + \frac{1}{x - a_{2}} + \dots + \frac{1}{x - a_{n - 1}}$
$\Rightarrow lim x \to 1 \frac{n x^{n - 1} - 1 (1 + x + x^{2} + \dots + x^{n - 1})}{x^{n} - 1} = lim x \to 1 (\frac{1}{x - a_{1}} + \frac{1}{x - a_{2}} + \dots + \frac{1}{x - a_{n - 1}})$
$\Rightarrow lim x \to 1 \frac{n (n - 1) x^{n - 2} - (1 + 2 x + \dots + (n - 1) x^{n - 2})}{n x^{n - 1}} = \frac{1}{1 - a_{1}} + \frac{1}{1 - a_{2}} + \dots + \frac{1}{1 - a_{n - 1}}$
$\Rightarrow \frac{n (n - 1) - (1 + 2 + \dots + (n - 1))}{n} = \frac{1}{1 - a_{1}} + \frac{1}{1 - a_{2}} + \dots + \frac{1}{1 - a_{n - 1}}$
$\Rightarrow \frac{1}{1 - a_{1}} + \frac{1}{1 - a_{2}} + \dots + \frac{1}{1 - a_{n - 1}} = \frac{n - 1}{2}$

Suggest Corrections

If the equation xn−1=0,n>1,n∈N, has roots 1,a1,a2,…,an−1, then

If the equation $x^{n} - 1 = 0, n > 1, n \in N,$ has roots $1, a_{1}, a_{2}, \dots, a_{n - 1},$ then