A) \[{{2}^{0.4}}\]
B) \[{{2}^{-0.4}}\]
C) \[1\]
D) \[{{2}^{0.6}}\]
Correct Answer: C
Solution :
\[f(x)=\frac{0.6{{(1+x)}^{-0.4}}(1+{{x}^{0.6}})-0.6{{x}^{-0.4}}{{(1+x)}^{0.6}}}{{{(1+{{x}^{0.6}})}^{2}}}\] \[=0.6.\frac{(1+{{x}^{0.6}})-{{x}^{-0.4}}.{{(1+x)}^{1}}}{{{(1+{{x}^{0.6}})}^{2}}{{(1+x)}^{0.4}}}\] \[=0.6.\frac{(1+{{x}^{0.6}}){{x}^{0.4}}-(1+x)}{{{(1+{{x}^{0.6}})}^{2}}{{(1+x)}^{0.4}}.{{x}^{0.4}}}\] \[=0.6.\frac{{{x}^{0.4}}-1}{{{(1+{{x}^{0.6}})}^{2}}{{(1+x)}^{0.4}}.{{x}^{0.4}}}<0\] Hence,\[f(x)\]is decreasing. \[\therefore \] \[f{{(x)}_{\max }}=f(0)=1\]
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