Question

If $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\left[4{\mathrm{x}}^2+2\mathrm{x}+1\right]}$, then its maximum value is

• A. $\frac{4}{3}$
• B. $\frac{2}{3}$
• C. 1
• D. $\frac{3}{4}$

Answer 1

The correct option is A

$\frac{4}{3}$

Step 1: Apply quotient rule differentiation

Given function

$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\left[4{\mathrm{x}}^2+2\mathrm{x}+1\right]}$

$\therefore \text{If}f\left(x\right)=\frac{s\left(x\right)}{t\left(x\right)},\text{then}f\text{'}\left(x\right)=\frac{t\left(x\right)\times s\text{'}\left(x\right)-s\left(x\right)\times t\left(x\right)}{{\left\{t\left(x\right)\right\}}^2}$

$\begin{array}{rcl}\therefore \mathrm{f}\text{'}\left(\mathrm{x}\right)& =& \frac{\left(4{\mathrm{x}}^2+2\mathrm{x}+1\right)*{\displaystyle \frac{d}{dx}}\left(1\right)-1*{\displaystyle \frac{d}{dx}}\left(4{\mathrm{x}}^2+2\mathrm{x}+1\right)}{{\left(4{\mathrm{x}}^2+2\mathrm{x}+1\right)}^2}\left[\because \frac{d}{dx}\left(\mathrm{constant}\right)=0\mathrm{and}\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\right]\\ & =& \frac{-\left(8x+2\right)}{{\left(4{\mathrm{x}}^2+2\mathrm{x}+1\right)}^2}\end{array}$

For maxima or minima put$f\text{'}\left(x\right)=0,$

$\begin{array}{rcl}0& =& 8\mathrm{x}+2\\ 8\mathrm{x}& =& -2\\ \mathrm{x}& =& \frac{-2}{8}\\ \mathrm{x}& =& \frac{-1}{4}\end{array}$

Step 2: Again differentiating, we get

$\mathrm{f}"\left(\mathrm{x}\right)=\frac{-\left[{\left(4{\mathrm{x}}^2+2\mathrm{x}+1\right)}^{2}8-\left(8\mathrm{x}+2\right)2(4{\mathrm{x}}^2+2\mathrm{x}+1)(8\mathrm{x}+2\right]}{{\left[4{\mathrm{x}}^2+2\mathrm{x}+1\right]}^4}$

At $\mathrm{x}=\frac{-1}{4}$, $f"\left(x\right)$ is negative

The function $f\left(x\right)$ is maximum at $\mathrm{x}=\frac{-1}{4}$

Therefore, the maximum value is

$\begin{array}{rcl}\therefore \mathrm{f}{\left(\frac{-1}{4}\right)}_{\max }& =& \frac{1}{4\times {\displaystyle \frac{1}{16}}-2\times {\displaystyle \frac{1}{4}}+1}\\ \mathrm{f}{\left(\frac{-1}{4}\right)}_{\max }& =& \frac{1}{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2}}+2}\\ & =& \frac{4}{1-2+4}\\ & =& \frac{4}{3}\end{array}$

Hence, the correct option is an option (A).