1Can someone tell me, from which book this question has been taken?
If a1<a2<a3<a4<a5<a6, then equation (x-a1) (x-a3) (x-a5) + 3(x-a2) (x-a4) (x-a6)=0 has:
A) three real roots
B) a root in (-∞,a1)
C) no real root in (a1,a2)
D) no real root in (a5,a6)
There may be more than one choice correct
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3 Answers
30P(x)=(x-a1) (x-a3) (x-a5) + 3(x-a2) (x-a4) (x-a6)
P(a1)<0
P(a2)>0
implies there's a root in (a1,a2)
again,
P(a3)>0
P(a4)<0
P(a5)<0
P(a6)>0
Therefore, there are 3 real roots in the intervals (a1,a2) , (a3,a4), (a5,a6).
Hence, only correct option (a). [1]