Question

draw wavy curve for the inequality​

Answer 1

Given inequality is,

$\longrightarrow\dfrac{1}{x-2}-\dfrac{1}{x}\leq\dfrac{2}{x+2}$

First make the LHS a simple fraction.

$\longrightarrow\dfrac{x-x+2}{x(x-2)}\leq\dfrac{2}{x+2}$

$\longrightarrow\dfrac{2}{x(x-2)}\leq\dfrac{2}{x+2}$

Divide both sides by 2.

$\longrightarrow\dfrac{1}{x(x-2)}\leq\dfrac{1}{x+2}$

Make RHS zero by bringing the fraction to left.

$\longrightarrow\dfrac{1}{x(x-2)}-\dfrac{1}{x+2}\leq0$

Now make LHS a unique fraction.

$\longrightarrow\dfrac{x+2-x(x-2)}{x(x-2)(x+2)}\leq0$

$\longrightarrow\dfrac{x+2-x^2+2x}{x(x-2)(x+2)}\leq0$

$\longrightarrow\dfrac{2+3x-x^2}{x(x-2)(x+2)}\leq0$

Multiply both sides by -1 (Note the sign change).

$\longrightarrow\dfrac{x^2-3x-2}{x(x-2)(x+2)}\geq0$

Let us factorise the numerator.

$\longrightarrow\dfrac{x^2+\left(\frac{\sqrt{17}-3}{2}-\frac{\sqrt{17}+3}{2}\right)x-\left(\frac{\sqrt{17}-3}{2}\right)\left(\frac{\sqrt{17}+3}{2}\right)}{x(x-2)(x+2)}\geq0$

$\longrightarrow\dfrac{x^2+\left(\frac{\sqrt{17}-3}{2}\right)x-\left(\frac{\sqrt{17}+3}{2}\right)x-\left(\frac{\sqrt{17}-3}{2}\right)\left(\frac{\sqrt{17}+3}{2}\right)}{x(x-2)(x+2)}\geq0$

$\longrightarrow\dfrac{\left(x+\frac{\sqrt{17}-3}{2}\right)\left(x-\frac{\sqrt{17}+3}{2}\right)}{x(x-2)(x+2)}\geq0$

Now let us draw wavy curve.

Let $f(x)=\dfrac{\left(x+\frac{\sqrt{17}-3}{2}\right)\left(x-\frac{\sqrt{17}+3}{2}\right)}{x(x-2)(x+2)}.$

Conditions for the equality are (RHS in decreasing order),

• $x=\dfrac{3+\sqrt{17}}{2}$

• $x=2$

• $x=0$

• $x=\dfrac{3-\sqrt{17}}{2}$

• $x=-2$

The inequality holds true, i.e., $f(x)$ is non - negative for $x\geq\dfrac{3+\sqrt{17}}{2}.$

Here each condition for equality is occurred only one (odd) time. Hence $f(x)$ has alternate signs at left of each condition.

Means,

• $f(x)$ is non - negative for $x\geq\dfrac{3+\sqrt{17}}{2}.$

• $f(x)$ is non - positive for $\dfrac{3+\sqrt{17}}{2}\geq x\geq 2.$

• $f(x)$ is non - negative for $2\geq x\geq0.$

• $f(x)$ is non - positive for $0\geq x\geq\dfrac{3-\sqrt{17}}{2}.$

• $f(x)$ is non - negative for $\dfrac{3-\sqrt{17}}{2}\geq x\geq -2.$

• $f(x)$ is non - positive for $-2\geq x.$

Hence wavy curve for the inequality is,

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){105}}\multiput(15,-0.1)(15,0){5}{\circle*{1}}\put(12,-5){ -2 }\put(21,-6){ \frac{3-\sqrt{17}}{2} }\put(44,-5){0}\put(59,-5){2}\put(74,-6){ \frac{3+\sqrt{17}}{2} }\qbezier(0,-10)(7.5,-7.5)(15,0)\multiput(0,0)(30,0){2}{\qbezier(15,0)(22.5,10)(30,0)\qbezier(30,0)(37.5,-10)(45,0)}\qbezier(90,10)(82.5,7.5)(75,0)\multiput(6,-3)(30,0){3}{ - }\multiput(21,1)(30,0){3}{ + }\end{picture}$

We can conclude that the solution to our inequality is,

$\longrightarrow\underline{\underline{x\in\left[-2,\ \dfrac{3-\sqrt{17}}{2}\right]\cup\Big[\,0,\ 2\,\Big]\cup\left[\dfrac{3+\sqrt{17}}{2},\ \infty\right)}}$