Question

# If a1 < a2< a3 < a4 < a5 < a6, then the equation (x−a1)(x−a3)(x−a5)+2(x−a2)(x−a4)(x−a6) = 0 has

A

Three real roots

B

One real root

C

One real root in each interval ( ,), ( , ) and ( , )

D

None of these

Solution

## The correct option is D One real root in each interval ( ,), ( , ) and ( , ) f(x)=(x−a1)(x−a3)(x−a5)+2(x−a2)(x−a4)(x−a6) = 0  a1 < a2< a3 < a4 < a5 < a6 f( a1 ) = 2 (a1−a2)(a1−a4)(a1−a6)<0 f( a2 ) = 2 (a2−a1)(a2−a3)(a2−a5)>0 ∴   at least one real root lies in (a1 ,a2 ) Similarly, at least one real root lies in each interval (a3 , a4 )  and (a5 , a6 ) But f(x) is cubic, therefore there are only three roots. Hence, the equation f(x) = 0 has one real root in each interval (a1  ,a2 ), (a3  , a4 ) and (a5  , a6 ) Mathematics

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