Question

Q1.please answer with proper proof!​

Answer 1

Step-by-step explanation:

Consider the inequality,

$\small\displaystyle\longrightarrow\dfrac{(x-12)(x-4)}{(x^2-16)} > 0$

$\small\displaystyle\longrightarrow\dfrac{(x-12)(x-4)}{(x+4)(x-4)} > 0$

Performing wavy curve method,

$\setlength{\unitlength}{1mm}\begin{picture}(0,0)\thicklines\put(10,0){\line(1,0){60}}\multiput(20,0)(20,0){3}{\circle{2}}\qbezier(60,0)(65,5)(70,10)\multiput(20,0)(20,0){2}{\qbezier(0,0)(10,-10)(20,0)}\qbezier(10,10)(15,5)(20,0)\put(15,-5){ -4 }\put(39,-5){ 4 }\put(60,-5){ 12 }\end{picture}$

Therefore,

$\small\displaystyle\longrightarrow x\in(-\infty,\ -4)\cup(12,\ \infty)\quad\quad\dots(1)$

Consider the inequality,

$\small\displaystyle\longrightarrow\dfrac{(x^2+16)(x+4)}{(x-4)(x+5)}\leq0$

$Here \: x^2+16 > 0\ \ \forall x\in\mathbb{R}.$

$\small\displaystyle\Longrightarrow\dfrac{x+4}{(x-4)(x+5)}\leq0$

Performing wavy curve method,

$\setlength{\unitlength}{1mm}\begin{picture}(0,0)\thicklines\put(10,0){\line(1,0){60}}\multiput(20,0)(40,0){2}{\circle{2}}\put(40,0){\circle*{2}}\multiput(0,0)(50,10){2}{\qbezier(10,-10)(15,-5)(20,0)}\qbezier(20,0)(30,10)(40,0)\qbezier(40,0)(50,-10)(60,0)\put(15,-5){ -5 }\put(37,-5){ -4 }\put(60,-5){ 4 }\end{picture}$

Therefore,

$\small\displaystyle\longrightarrow x\in(-\infty,\ -5)\cup[-4,\ 4)\quad\quad\dots(2)$

Hence the answer is given by the union of (1) and (2).

$\small\displaystyle\longrightarrow x\in\big[(-\infty,\ -4)\cup(12,\ \infty)\big]\cup\big[(-\infty,\ -5)\cup[-4,\ 4)\big]$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{x\in(-\infty,\ 4)\cup(12,\ \infty)}} }</p><p>$