A) -10
B) -6
C) -7
D) -8
Correct Answer: A
Solution :
\[\operatorname{f}(x)=\,\,{{x}^{3/2}}+{{x}^{-3/2}}-4\,\left( x+\frac{1}{x} \right)\] \[\operatorname{f}(x)=\,\,{{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{3}}-3\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)-4\left[ {{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}}-2 \right]\]Let \[\sqrt{x}+\frac{1}{\sqrt{x}},\,\,t\,\left( x>0 \right)\] Let \[\operatorname{g}(t)={{t}^{3}}-3t\,-4{{t}^{2}}+8\] \[\operatorname{g}\left( t \right)={{t}^{3}}-4{{t}^{2}}-3t+8\] \[\operatorname{g}\left( t \right)=3{{t}^{2}}-8t\,-3=\left( t\,-3 \right)\left( 3t+1 \right)\] \[\operatorname{g}(t)=\,\,0\,\,\Rightarrow \,\,t=3\,(t\ne 1/3)\] \[\operatorname{g}'(t)=6t\,-8\] \[\operatorname{g}''(3)=\,\,10>\,\,0\,\,\Rightarrow \,\,g\,(3)isminimum\] \[\operatorname{g}(3)=27-9-36+8=-10\]You need to login to perform this action.
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