Question

Let $f\left(x\right)$ be a polynomial of degree three, if the curve $y=f\left(x\right)$ has relative extrema at $x=\pm \frac{2}{\sqrt{3}}=4x^2+y^2=4$ in two parts. Then find the integral part of areas of these two parts.

Answer 1

Since $y=f\left(x\right)$ has relative extrima at $x=\pm \frac{2}{\sqrt{3}}$ and these points are critical points and hence they must be roots of $f^{\prime }\left(x\right)=0;f$ is differentiable everywhere)
& therefore; $f^{\prime }\left(x\right)=a(x-\frac{2}{\sqrt{3}})(x+\frac{2}{\sqrt{3}})=a(x^2-\frac{4}{3})$

$\Rightarrow f\left(x\right)=a(\frac{x^3}{3}-\frac{4x}{3})+b$

This passes through (0, 0) and (1, – 2)

$\therefore b=0\text{and}a(\frac{1}{3}-\frac{4}{3})=-2\Rightarrow a=2$

$\therefore f\left(x\right)=2(\frac{x^3}{3}-\frac{4x}{3})=\frac{2x}{3}(x^2-4)$


Clearly $f\left(x\right)$ meets x-axis at $(0,0),(–2,0),(2,0)$

$f(-x)=-f\left(x\right)$

$\therefore$ The curve is symmetrical about origin
Also $a=2$

$\therefore f^{\prime }\left(x\right)>0\text{for}xϵ(-\infty ,\frac{-2}{\sqrt{3}})\cup (\frac{2}{\sqrt{3}},\infty )\text{and}{f}^{\prime }\left(x\right)<0\text{for}xϵ(\frac{-2}{\sqrt{3}},\frac{2}{\sqrt{3}})$

Area of region $=2\pi -{\int }_{-2}^0|f\left(x\right)|dx+{\int }_0^2|f\left(x\right)|dx$

But ${\int }_{-2}|f\left(x\right)|dx={\int }_0^2|f\left(x\right)|dx$ (due to symmetry)

Area of region $=2\pi$

$\Rightarrow \left[2\pi \right]=6$