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Q.

The maximum value of f(x)=2bx2−x4−3b is g(b), where b>0. If b varies, then the minimum value of g(b) is

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answer is -2.25.

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Detailed Solution

f(x)=−x4−2bx2−3b=−x2−b2+b2−3b∴ g(b)=b2−3b=b2−3b+94−94=b−322−94⇒gmin=−94

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