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Q.

Let f(x)=ax2bx+c2,b0 and f(x)0 for all xR. Then

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a

9a3b+c2<0

b

none of these

c

a+c2<b

d

4a+c2>2b

answer is B.

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Detailed Solution

detailed_solution_thumbnail

Here, ax2bx+c2=0 does not have real roots. So,

D<0b24ac2<0a>0

Therefore, f(x) is always positive. So,

f(2)>04a2b+c2>0

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