118138 If α is a non-real root of x6=1, then α5+α3+α+1α2+1 is equal to
C Given,⇒α3=1α5+α3+α+1α2+1=α2+1+α+11+α2=(0+1)−α=−1α=−1⋅α2α3α5+α3+α+1α2+1=−α2∵(1+α+α2=−1)α is a non real root of x6=1⇒α6=1⇒1+α+α2=0,α3=1 i.e. Cube roots of units
118139 If α,β,γ are the roots of x3+2x2−3x−1=0then α−2+β−2+γ−2=
B Given α,β,γ are the roots of the equation, x3+2x2−3x−1=0∴α+β+γ=−2αβ+βγ+αγ=−3αβγ=+1α−2+β−2+γ−2=1α2+1β2+1γ2=β2γ2+α2γ2+α2β2(αβγ)2=(βγ)2+(αγ)2(αβ)2(+1)2We know,(αβ)2+(αγ)2+(βα)2=(αβ+βγ+αγ)2−2(αβ2γ+αβγ2+α2βγ)=[(αβ)+(βγ+αγ)]2−2αβγ(α+β+γ)=(−3)2−2×(+1)(−2)=9+4=13α−2+β−2+γ−2=13
118180 If the roots of the equation 5x2−7x+k=0 are reciprocal of each other, then value of k is
A Given the equation -5x2−7x+k=0, Let the roots areα,β Given α=1βHere, α+β=75α⋅β=k5Given,αβ=1⇒k=5
118140 If the roots of the equation 4x3−12x2+11x+k= 0 are in arithmetic progression, then k is equal to
A Given the roots of the equation4x3−12x2+11x+k=0 is in P. ⇒x3−3x2+114x+k4=0 Let the roots are α,β,γ, We have 2β=α+γα+β+γ=3αβ+βγ+αγ=11/4αβγ=−k4∵α+β+γ=32β+β=3∴β=1,α+γ=2β(α+γ)+γα=11/41×2+αγ=11/4αγ=114−2=34∴34×1=−k4∴k=−3∴ Option (a) is correct. Let the roots are α,β,γ, We have∴ Option (a) is correct.
