A) \[-\frac{1}{3}\]
B) \[-\frac{1}{4}\]
C) \[\frac{1}{5}\]
D) \[\frac{1}{6}\]
Correct Answer: D
Solution :
Given,\[f(x)=\frac{x}{4+x+{{x}^{2}}}\] Let\[f(x)=\frac{1}{u}\] \[\therefore \] \[u=\frac{4+x+{{x}^{2}}}{x}\] \[=\frac{4}{x}+1+x\] \[\frac{du}{dx}=-\frac{4}{{{x}^{2}}}+1,\,\,\frac{{{d}^{2}}u}{d{{x}^{2}}}=\frac{8}{{{x}^{3}}}\] For maximum or minimum, put\[\frac{du}{dx}=0\] \[\Rightarrow \] \[1-\frac{4}{{{x}^{2}}}=0\] \[\Rightarrow \] \[{{x}^{2}}-4=0\] \[\Rightarrow \] \[x=\pm 2\] \[\therefore \]At\[x=\pm 2,\,\,\frac{{{d}^{2}}u}{d{{x}^{2}}}=-\frac{8}{{{(\pm 2)}^{2}}}<0\], maximum It is a decreasing function. \[\Rightarrow \]\[f(x)\]is an increasing function\[[\because \,\,f(x)=\frac{1}{u}]\] \[\Rightarrow \]\[f(x)\]is an increasing function in\[[-1,\,\,1]\]. \[\therefore \]The maximum value at \[x=1\] is \[f(x)=\frac{1}{4+1+1}=\frac{1}{6}\] Note If \[u\] is increasing, then reciprocal of \[u\] is decreasing.
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