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Question

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

A
a+c2 < b
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B
4a+c2>2b
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C
9a+c2<3b
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D
None of the above.
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Solution

The correct option is B 4a+c2>2b
f(x)0forallxRdiscriminantislessthanzeroax2bx+c2=0b24ac2<0a>0sof(x)>0xRf(2)=4a2b+c2>04a+c2>2b

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