A) 4/3
B) 2/3
C) 1
D) ¾
Correct Answer: A
Solution :
\[f(x)=\frac{1}{4{{x}^{2}}+2x+1}\]Þ \[{f}'(x)=\frac{-(8x+2)}{{{(4{{x}^{2}}+2x+1)}^{2}}}\] Put \[{f}'(x)=0\] Þ \[8x+2=0\] Þ \[x=-1/4\]. \[{f}''\,(x)=\frac{-[{{(4{{x}^{2}}+2x+1)}^{2}}8-(8x+2)\,2\,(4{{x}^{2}}+2x+1)(8x+2)]}{{{(4{{x}^{2}}+2x+1)}^{4}}}\] \[{f}''(-1/4)=-ve\] (point of maxima) \ \[f{{(-1/4)}_{\text{max}\text{.}}}\]= \[\frac{1}{4\times \frac{1}{16}-2\times \frac{1}{4}+1}=\frac{4}{3}\].
You need to login to perform this action.
You will be redirected in
3 sec
