MCQ
If $\text{f}(\text{x})=\frac{1}{4\text{x}^{2}+2\text{x}+1}$, then its maximum value is :
  • $\frac{4}{3}$
  • B
    $\frac{2}{3}$
  • C
    $1$
  • D
    $\frac{3}{4}$

Answer

Correct option: A.
$\frac{4}{3}$
Maximum value of $\frac{1}{4\text{x}^{2}+2\text{x}+1} =$ Minimum value of $4\text{x}^{2}+2\text{x}+1$
Now, $\text{f}(\text{x})=\text4\text{x}^{2}+2\text{x}+1$
lmplies that $\text{f}'(\text{x})=8\text{x}+2$
For a local maxima or a local minima, We must have $f\ '(x) = 0$
lmplies that $8\text{x}+2 =0$
lmplies that $8\text{x}=-2$
lmplies that $\text{x}=-14$
Now, $\text{f}\ ''(\text{x})=8$
lmplies that $\text{f}\ ''(\text{1})=8 > 0$
Therefore, $\text{x}=\frac{-1}{4}$ is a local minima.
Thus, $\frac{1}{4\text{x}^{2}+2\text{x}+1}$ is maximum at $\text{x}=\frac{-1}{4}$
lmplies that maximum value of $\frac{1}{4\text{x}^{2}+2\text{x}+1}=\frac{1}{4\Big(\frac{-1}{4}\Big)^{2}+2\Big(\frac{-1}{4}\Big)+1}$
$=\frac{16}{12}=\frac{4}{3}$

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