Question

Maximum value of the expression $\frac{1}{4x^2+2x+1}$ is

• A. $\frac{4}{3}$
• B. $\frac{5}{2}$
• C. $\frac{13}{4}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

Expression $\frac{1}{4x^2+2x+1}$ is maximum when
$4x^2+2x+1$ is minimum.
Let $y=4x^2+2x+1$ ...(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 $\frac{4}{3}$