Lf n is a multiple of 6 and - - are the roots of x2-x-1-0- then 1-n-1-n-

The correct option is A 2 nbsp;Given x ^ 2 +x+1=0 ⇒ x= 1 ι√( 3 ) nbsp;/ 2 or 1+ι√( 3 ) nbsp;/ 2 ∴ α =ω and β =ω nbsp;^ 2 wh ...

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Standard XII

Mathematics

Higher Order Derivatives

lf n is a m...

Question

lf $n$ is a multiple of $6$ and $α, β$ are the roots of $x^{2} + x + 1 = 0$ , then $(1 + α)^{- n} + (1 + β)^{- n} =$

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Solution

The correct option is A $2$
$G i v e n x^{2} + x + 1 = 0 \Rightarrow x = \frac{- 1 - ι \sqrt{3}}{2} o r \frac{- 1 + ι \sqrt{3}}{2} ∴ α = ω a n d β = ω^{2} w h e n ω = c u b e r o o t o f u n i t y . A l s o l e t n = 6 p w h e n p i s a n a t u r a l n u m b e r . ∴ (1 + α)^{- n} + (1 + β)^{- n} = (1 + ω)^{- 6 p} + (1 + ω^{2})^{- 6 p} = (ω^{2})^{- 6 p} + (ω)^{- 6 p} = (ω^{3})^{- 4 p} + (w^{3})^{- 3 p} = 1 + 1 = 2 A n s w e r - O p t i o n A$

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lf n is a multiple of 6 and α, β are the roots of x2+x+1=0, then (1+α)−n+(1+β)−n=

The correct option is A 2Givenx2+x+1=0⇒x=−1−ι√32or−1+ι√32∴α=ωandβ=ω2whenω=cuberootofunity.Alsoletn=6pwhenpisanaturalnumber.∴(1+α)−n+(1+β)−n=(1+ω)−6p+(1+ω2)−6p=(ω2)−6p+(ω)−6p=(ω3)−4p+(w3)−3p=1+1=2Answer−OptionA

lf $n$ is a multiple of $6$ and $α, β$ are the roots of $x^{2} + x + 1 = 0$ , then $(1 + α)^{- n} + (1 + β)^{- n} =$