Question

if (x-1)(x+1)(x-2)^2(e^3x-1)/(π^x-1)(3-x)^3(x-1)^2(x+3) <=0,then complete solution set of values of x is

Answer 1

We are given the inequality,

$\longrightarrow\sf{\dfrac{(x-1)(x+1)(x-2)^2(e^{3x}-1)}{(\pi^x-1)(3-x)^3(x-1)^2(x+3)}\leq0}$

$\longrightarrow\sf{\dfrac{(x+1)(x-2)^2(e^{3x}-1)}{(\pi^x-1)(3-x)^3(x-1)(x+3)}\leq0}$

Here the denominator should be non-zero. By this we get the following.

• $\sf{\pi^x-1\neq0\quad\implies\quad x\neq0}$

• $\sf{(3-x)^3\neq0\quad\implies\quad x\neq3}$

• $\sf{x-1\neq0\quad\implies\quad x\neq1}$

• $\sf{x+3\neq0\quad\implies\quad x\neq-3}$

On considering the equality we have,

• $\sf{x+1=0\quad\implies\quad x=-1}$

• $\sf{(x-2)^2=0\quad\implies\quad x=2}$

Since $\sf{x\neq0,\ e^{3x}-1\neq0.}$

Consider the number line given below.

$\setlength{\unitlength}{1cm}\begin{picture}(5,5)\put(0,0){\vector(-1,0){6}}\put(0,0){\vector(1,0){6}}\multiput(-5,0)(1,0){11}{\circle*{0.1}}\multiput(-1,-0.1)(3,0){2}{\line(0,1){0.2}}\put(-3.3,-0.4){ \sf{-3} }\put(-1.3,-0.4){ \sf{-1} }\put(-0.08,-0.4){ \sf{0} }\put(0.92,-0.4){ \sf{1} }\put(1.92,-0.4){ \sf{2} }\put(2.92,-0.4){ \sf{3} }\put(5.8,-0.4){ \sf{x} }\end{picture}$

Let $\sf{f(x)=\dfrac{(x+1)(x-2)^2(e^{3x}-1)}{(\pi^x-1)(3-x)^3(x-1)(x+3)}.}$

As the term $\sf{(3-x)^3}$ is in $\sf{f(x),}$ it has a value less than 0 for $\sf{x\ \textgreater\ 3}$ though other terms are positive. Thus $\sf{f(x)\ \textless\ 0}$ for $\sf{x\in(3,\ \infty).}$

Since the exponent of $\sf{(3-x)^3}$ is odd, sign of $\sf{f(x)}$ changes at $\sf{x=3.}$ Hence $\sf{f(x)\ \textgreater\ 0}$ for $\sf{x\in(2,\ 3).}$

Exponent of $\sf{(x-2)^2}$ is even so there's no sign change for $\sf{f(x)}$ at $\sf{x=2.}$ Hence $\sf{f(x)\ \textgreater\ 0}$ for $\sf{x\in(1,\ 2).}$

Exponent of $\sf{x-1}$ is odd so sign changes occurs at $\sf{x=1.}$ Hence $\sf{f(x)\ \textless\ 0}$ for $\sf{x\in(0,\ 1).}$

There are two (even) terms in $\sf{f(x)}$ which has the value 0 for $\sf{x=0;\ e^{3x}-1}$ and $\sf{\pi^x-1.}$ Thus no sign change occurs at $\sf{x=0.}$ Hence $\sf{f(x)\ \textless\ 0}$ at $\sf{x\in(-1,\ 1).}$

Exponent of $\sf{x+1}$ is odd so sign of $\sf{f(x)}$ changes at $\sf{x=-1.}$ Hence $\sf{f(x)\ \textgreater\ 0}$ for $\sf{x\in(-3,\ -1).}$

Exponent of $\sf{x+3}$ is also odd so sign changes occurs at $\sf{x=-3.}$ Hence $\sf{f(x)\ \textless\ 0}$ for $\sf{x\in(-\infty,\ -3).}$

Finally, sign of $\sf{f(x)}$ with $\sf{x}$ appears as the following:

$\setlength{\unitlength}{1cm}\begin{picture}(5,5)\put(0,0){\vector(-1,0){6}}\put(0,0){\vector(1,0){6}}\multiput(-5,0)(1,0){11}{\circle*{0.1}}\put(-3.3,-0.4){ \sf{-3} }\put(-1.3,-0.4){ \sf{-1} }\put(-0.08,-0.4){ \sf{0} }\put(0.92,-0.4){ \sf{1} }\put(1.92,-0.4){ \sf{2} }\put(2.92,-0.4){ \sf{3} }\put(3.36,0.2){ - }\put(2.36,0.2){ + }\put(1.36,0.2){ + }\put(0.36,0.2){ - }\put(-0.64,0.2){ - }\put(-2.14,0.2){ + }\put(-3.64,0.2){ - }\multiput(-1,-0.1)(3,0){2}{\line(0,1){0.2}}\put(5.8,-0.4){ \sf{x} }\end{picture}$

Hence the complete solution set of values of $\sf{x}$ for our inequality is,

$\longrightarrow\large\text{ \underline{\underline{\sf{x\in(-\infty,\ -3)\cup[-1,\ 0)\cup(0,\ 1)\cup\{2\}\cup(3,\ \infty)}}} }$