    Question

# The equation xn=1,n>1,n∈N has roots 1,a1,a2,...,an−1. Then which of the following are correct?

A
n1r=1(1ar)=n
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B
n1r=1(1ar)=n2
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C
n1r=111ar=n+12
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D
n1r=111ar=n12
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Solution

## The correct options are A n−1∏r=1(1−ar)=n D n−1∑r=111−ar=n−121,a1,a2,a−3,...,an−1 are the roots of xn−1=0 ⇒xn−1=(x−1)(x−a1)(x−a2)...(x−an−1) So we can write limx→1xn−1x−1 =limx→1[(x−a1)(x−a2)(x−a3)...(x−an−1)] ⇒n=(1−a1)(1−a2)(1−a3)...(1−an−1) Hence, n−1∏r=1(1−ar)=n Now, log(xn−1)=log(x−1)+log(x−a1) +log(x−a2)+⋯+log(x−an−1) Differentiate with respect to x, we get nxn−1xn−1=1x−1+1x−a1+1x−a2+...+1x−an−1 ⇒nxn−1xn−1−1x−1 =1x−a1+1x−a2+...+1x−an−1 ⇒nxn−1−1(1+x+x2+...+xn−1)xn−1 =1x−a1+1x−a2+...+1x−an−1 ⇒limx→1nxn−1−1(1+x+x2+...+xn−1)xn−1 =limx→1(1x−a1+1x−a2+...+1x−an−1) ⇒n(n−1)−(1+2+...+(n−1))n =11−a1+11−a2+11−a3+...+11−an−1 ⇒n(n−1)−n(n−1)/2n =11−a1+11−a2+11−a3+...+11−an−1 Hence, n−1∑r=1(11−ar)=n−12  Suggest Corrections  0      Similar questions
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