A) \[\frac{4}{3}\]
B) \[\frac{5}{2}\]
C) \[\frac{13}{4}\]
D) \[\frac{3}{4}\]
Correct Answer: A
Solution :
Expression\[\frac{1}{4{{x}^{2}}+2x+1}\]is maximum when\[4{{x}^{2}}+2x+1\]is minimum. Let \[y=4{{x}^{2}}+2x+1\] ...(i) \[\frac{dy}{dx}=8x+2\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=8\] For maxima and minima, \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[8x+2=0\] \[\Rightarrow \] \[x=-\frac{1}{4}\] At \[x=-\frac{1}{4},\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}>0\]this is minimum. Thus, maximum value of expression \[=\frac{1}{4{{\left( -\frac{1}{4} \right)}^{2}}+2\left( -\frac{1}{4} \right)+1}\] \[=\frac{1}{\frac{1}{4}-\frac{1}{2}+1}\] \[=\frac{1}{3/4}=\frac{4}{3}\] Alternative Method From Eq. (i), \[y=4\left( {{x}^{2}}+\frac{x}{2}+\frac{1}{4} \right)\] \[=4\left[ {{\left( x+\frac{1}{4} \right)}^{2}}-\frac{3}{16} \right]\] y is minimum at\[x=-\frac{1}{4}\] \[\therefore \]Maximum value of given expression is\[\frac{4}{3}\]
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