If α1<α2<α3<α4<α5<α6, then the equa...
Question
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If α1<α2<α3<α4<α5<α6, then the equation  

(x-α1)(x-α3)(x-α5)+3(x-α2)(x-α4)(x-α6)=0

has  

(A) three

(B) no real root in (-∞,α1)

(C) one real root in (α1,α2)

(D) no real root in (α5,α6)

JEE/Engineering Exams
Maths
Solution
127
Rating
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If α1<α2<α3<α4<α5<α6, then the equation    (x-α1)(x-α3)(x-α5)+3(x-α2)(x-α4)(x-α6)=0
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= (x-α1) (x-α3) (x-α5)              +3 (x-α2) (x-α4) (x-α6)    f (α1)=3 (α1-α2) (α1-α4) (α1-α6)      α1<α2  = α1-α2<0=-ve similary α1-α4=-ve                 α1-α6=-ve = f(α1)=-ve f (α2) = (α2+ve-α1) (α2+ve-α3)  (α2+ve-α5) = +ve  f(α5)=-ve f(α6)=+ve  Ans. (c) 

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