A) (0, \[\infty \])
B) (\[-\infty \], 0)
C) \[(-\infty ,\infty )\]
D) None of these
Correct Answer: A
Solution :
\[f(x)=\log (1+x)-\frac{2x}{2+x}\]\[\Rightarrow f'(x)=\frac{1}{1+x}-\frac{(2+x).(2-2x)}{{{(2+x)}^{2}}}\] Þ \[f'(x)=\frac{{{x}^{2}}}{(x+1){{(x+2)}^{2}}}\] Obviously, \[f'(x)>0\]for all \[x>0\] Hence \[f(x)\]is increasing on \[(0,\infty )\].
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