A) \[\left( \frac{-1}{2},\,\frac{1}{2} \right)\]
B) \[\left[ \frac{1}{2},\,-\frac{1}{2} \right]\]
C) (? 1, 1)
D) [1, ?1]
Correct Answer: B
Solution :
\[f(x)=4x+\frac{1}{x}\] \[\frac{d}{dx}f(x)=\frac{d}{dx}\left[ 4x+\frac{1}{x} \right]\] = \[4-\frac{1}{{{x}^{2}}}\] For extremum, \[\frac{dy}{dx}=0\] Þ \[4-\frac{1}{{{x}^{2}}}=0\] Þ \[x=\frac{1}{2},\,-\frac{1}{2}\] \[f\ \left( \frac{1}{2} \right)=4.\frac{1}{2}+\frac{1}{1/2}\] = \[2+2=4\] \[f\ \left( -\frac{1}{2} \right)=4.\left( -\frac{1}{2} \right)+\frac{1}{-1/2}=-2-2=-4\] Hence \[f(x)\] is decreasing in interval \[[1/2,\,-1/2]\].
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