Question

The area between the curve $y=2x^4-x^2$ , the $x$ -axis and the ordinates of the two minima of the curve is

• A. $\frac{11}{60}sq.$ units
• B. $\frac{7}{120}sq.$ units
• C. $\frac{1}{30}sq.$ units
• D. $\frac{7}{90}sq.$ units

Answer 1

Answer: (B)

The equation of the curve is $y=2x^4-x^2=\left(2x^2-1\right)x^2$
The curve is symmetrical about the $y$ -axis.
Also, it is a polynomial of degree four having roots $0,0,\pm \frac{1}{\sqrt{2}}.$
$x=0$ is repeated root. Hence, the graph touches $x$ -axis at $\left(0,0\right)$ and intersects the $x$ -axis at
$A\left(-\frac{1}{\sqrt{2}},0\right)\&B\left(\frac{1}{\sqrt{2}},0\right)$
$\Rightarrow \frac{dy}{dx}=8x^3-2x$
$=2x\left(4x^2-1\right)=0$
$x=0,x=\pm \frac{1}{2}$
$\left(\frac{d^{2}y}{d{x}^2}\right)>0$ at $x=\frac{1}{2}$ and at $x=\frac{-1}{2}$
So, $x=\pm \frac{1}{2}$ are the points of local minima
Thus, the graph of the curve is shown in the diagram.

Here, $y\le 0$ , as $x$ varies from $x=-\frac{1}{2}$ to $x=\frac{1}{2}$
$\therefore$ The required area
$=2$ Area $OCDO$
$=2\left|{\int }_0^{\frac{1}{2}}ydx\right|$
$=2\left|{\int }_0^{\frac{1}{2}}\left(2x^4-x^2\right)dx\right|$
$=\frac{7}{120}sq.$ units