A) \[-1/4\]
B) \[-1/8\]
C) \[1/12\]
D) \[1/16\]
Correct Answer: B
Solution :
The function \[f(x)=\frac{x}{{{x}^{2}}+16}\] Þ \[{f}'(x)=\frac{({{x}^{2}}+16).1-x.(2x)}{{{({{x}^{2}}+16)}^{2}}}\] =\[\frac{{{x}^{2}}+16-2{{x}^{2}}}{{{({{x}^{2}}+16)}^{2}}}=\frac{16-{{x}^{2}}}{{{({{x}^{2}}+16)}^{2}}}\] ??(i) Put \[{f}'(x)=0\] Þ \[16-{{x}^{2}}=0\] Þ \[x=4,\,-4\] Again, \[f''(x)=\frac{{{({{x}^{2}}+16)}^{2}}(-2x)-(16-{{x}^{2}})2({{x}^{2}}+16)2x}{{{({{x}^{2}}+16)}^{4}}}\] At \[x=4\],\[{f}''(x)>0\]and at \[x=-4\], \[{f}''(x)>0\] \ Least value of \[f(x)=\frac{-4}{16+16}=-\frac{1}{8}\].
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