Question

The function $f\left(x\right)=\frac{1}{4x^2+2x+1}$, 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 $\frac{4}{3}$

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

On differentiating, we get

$\Rightarrow f^{\prime }\left(x\right)=\frac{-(8x+2)}{{(4x^2+2x+1)}^2}$

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

$\Rightarrow 8x+2x=-\frac{1}{4}$

Again differentiating, we get

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

At $x=-\frac{1}{4},f^{\prime \prime }(-\frac{1}{4})=-ve$

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

$\therefore$ Maximum value is $f{(-\frac{1}{4})}_{max}=\frac{1}{4\times \frac{1}{16}-2\times \frac{1}{4}+1}=\frac{1}{\frac{1}{4}-\frac{2}{4}+1}$

$=\frac{4}{1-2+4}=\frac{4}{3}$