A) An increasing function
B) A decreasing function
C) Both increasing and decreasing function
D) None of these
Correct Answer: B
Solution :
\[f(x)=\frac{1}{x+1}-\log (1+x)\] Þ \[{f}'(x)=-\frac{1}{{{(x+1)}^{2}}}\,-\,\frac{1}{1+x}\] \[{f}'(x)=-\left[ \frac{1}{x+1}+\frac{1}{{{(x+1)}^{2}}} \right]\] \[{f}'(x)=-ve\], when \[x>0\]or \[{f}'(x)<0\], \[\forall x>0\] \\[f(x)\] is decreasing function.
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