Question

Which of the following is /are true for the curve $y=f\left(x\right)=2x^4-x^2$?

• A. Curve is Symmetric about $x$ -axis
• B. Curve touches the $x$ axis at origin
• C. $y=f\left(x\right)$ has two points of local minima
• D. Graph of $y=f\left(x\right)$ when $x\in [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$ is

Answer 1

Answer: (B), (C), (D)

Graph of $y=f\left(x\right)$ when $x\in [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$ is

$y=2x^4-x^2=f\left(x\right)$

$x=0\Rightarrow y=0$

$y=0\Rightarrow x^2(2x^2-1)=0\Rightarrow x=0,\pm \frac{1}{\sqrt{2}}$

$f(-x)=f\left(x\right)\Rightarrow$ curve is Symmetric about $y$ axis

$y^{\prime }=8x^3-2x$

$y^{\prime }=0\Rightarrow x=0,\pm \frac{1}{2}$

$y^{\prime \prime }=24x^2-2$

$y^{\prime \prime }\left(0\right)<0\Rightarrow x=0$ is a point of maxima

$y^{\prime \prime }(\pm \frac{1}{2})>0\Rightarrow x=\pm \frac{1}{2}$ are points of minima

$y^{\prime \prime }=0\Rightarrow x=\pm \frac{1}{\sqrt{12}}$ are points of inflection

Since, $x=0$ is a repeated root of $y=0$ so, curve touches the $x$ axis at origin

$y(\pm \frac{1}{2})=\frac{2}{16}-\frac{1}{4}=-\frac{1}{8}\to \text{local min. value}$

$y\left(0\right)=0\to$ local max. Value

Curvature of $y=f\left(x\right)$ changes at $x=\pm \frac{1}{\sqrt{12}}$