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Question

Let f(x)=3ax24bx+c(a,b,cR,a0) where a,b,c are in A.P. Then the equation f(x)=0 has

A
No real solution.
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B
Two unequal real roots.
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C
Sum of roots always negative.
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D
Product of roots always positive.
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Solution

The correct option is C Two unequal real roots.

Since a,b,c are in A.P., so,

2b=a+c

4b2=(a+c)2

The discriminant of the given functionf(x)=3ax24bx+c is,

D=16b212ac

=4(a+c)212ac

=4[(a2+c2+2ac)3ac]

=4(a2+c2ac)

=4(a2+c22ac+ac)

=4((ac)2+ac)

Case 1: If a andc are of opposite signs, then, D=(+)ve.

Case 2: If a andc are of same signs, then, D=(+)ve.

This shows that f(x)=0 has two unequal real roots.


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