Question

solve
I < 2
OL
SC+2
Please solve the above question...it is a question of inequality

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm :\longmapsto\:\dfrac{1}{x - 2} - \dfrac{1}{x} \leqslant \dfrac{2}{x + 2}$

For inequality to be exist,

$\rm :\longmapsto\:x \: \ne \: 0, \: - 2, \: 2$

$\rm :\longmapsto\:\dfrac{x - (x - 2)}{x(x - 2)} \leqslant \dfrac{2}{x + 2}$

$\rm :\longmapsto\:\dfrac{x - x + 2}{x(x - 2)} \leqslant \dfrac{2}{x + 2}$

$\rm :\longmapsto\:\dfrac{2}{x(x - 2)} \leqslant \dfrac{2}{x + 2}$

$\rm :\longmapsto\:\dfrac{1}{x(x - 2)} \leqslant \dfrac{1}{x + 2}$

$\rm :\longmapsto\:\dfrac{1}{x(x - 2)} - \dfrac{1}{x + 2} \leqslant 0$

$\rm :\longmapsto\:\dfrac{(x + 2) - x(x - 2)}{x(x - 2)(x + 2)}\leqslant 0$

$\rm :\longmapsto\:\dfrac{x + 2 - {x}^{2} + 2x}{x(x - 2)(x + 2)}\leqslant 0$

$\rm :\longmapsto\:\dfrac{3x + 2 - {x}^{2}}{x(x - 2)(x + 2)}\leqslant 0$

$\rm :\longmapsto\:\dfrac{ - ({x}^{2} - 3x - 2)}{x(x - 2)(x + 2)}\leqslant 0$

$\rm :\longmapsto\:\dfrac{{x}^{2} - 3x - 2}{x(x - 2)(x + 2)}\geqslant 0$

$\green{ \boxed{ \because \: \bf \: x > 0 \: then \: - x < 0}}$

Consider,

$\red{\rm :\longmapsto\: {x}^{2} - 3x - 2 = 0}$

Its a quadratic equation,

Solution is given by

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

$\rm :\longmapsto\:x = \dfrac{ 3 \: \pm \: \sqrt{ {( - 3)}^{2} - 4(1)( - 2)} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ 3 \: \pm \: \sqrt{{{9 + 8}}}}{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ 3 \: \pm \: \sqrt{{{17}}}}{2}$

So, critical points are

$\rm :\longmapsto\:x = - 2, \: 0, \: 2, \: \dfrac{3 + \sqrt{17} }{2}, \: \dfrac{3 - \sqrt{17} }{2}$

Let assume that

$\rm :\longmapsto\:\dfrac{3 + \sqrt{17} }{2} = a$

and

$\rm :\longmapsto\:\dfrac{3 - \sqrt{17} }{2} = b$

Now, order of values are

$\rm :\longmapsto\: - 2 < b < 0 < 2 < a$

Let check the sign in the intervals.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf interval & \bf sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x < - 2 & \sf - ve \\ \\ \sf - 2 < x \leqslant b & \sf + ve \\ \\ \sf b \leqslant x < 0 & \sf - ve\\ \\ \sf 0 < x < 2 & \sf + ve\\ \\ \sf 2 < x \leqslant a & \sf - ve\\ \\ \sf x \geqslant a & \sf + ve \end{array}} \\ \end{gathered}$

So, required solution is

$\rm :\longmapsto\:x \: \in \: ( - 2,b] \: \cup \: (0,2) \: \cup \: [a, \infty )$

where,

$\rm :\longmapsto\:\dfrac{3 + \sqrt{17} }{2} = a$

and

$\rm :\longmapsto\:\dfrac{3 - \sqrt{17} }{2} = b$