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Q.
Maximum value of the expression $ \frac{1}{4{{x}^{2}}+2x+1} $ is
Rajasthan PETRajasthan PET 2002
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} $