Question

The minimum value of the function $y=x^4-2x^2+1$ in the interval $\left[\frac{1}{2},2\right]$ is

• A. 0
• B. 2
• C. 8
• D. 9

Answer 1

Answer: (A)

$\frac{dy}{dx}=\frac{d}{dx}\left(x^4-2x^2+1\right)$
$=4x^3-4x=4x\left(x^2-1\right)$
For max. or min, $\frac{dy}{dx}=0$
$\Rightarrow 4x\left(x^2-1\right)=0$
either $x=0$ or $x=\pm 1$
$x=0$ and $x=-1$
does not belong to $\left[\frac{1}{2},2\right]$
$\frac{d^{2}y}{d{x}^2}=12x^2-4$
$\Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=1}=12(1{)}^2-4=8>0$
$\therefore$ there is minimum value of function at $x=1$
$\therefore$ minimum value is
$y(1)=1^4-2(1{)}^2+1=1-2+1=0$