Question

Solve the inequality: ${{\log }_{0.1}}\left( {{\log }_{2}}\dfrac{{{x}^{2}}+1}{\left| x-1 \right|} \right)< 0$.

Answer 1

Hint: We need to change the logarithmic form to indices form using logarithmic theorems. Then we need to break the quadratic equation into two parts where modulus is in the denominator. We solve the inequalities individually to find the solution of the problem as a domain of x.

Complete step-by-step answer:
We have been given an inequation ${{\log }_{0.1}}\left( {{\log }_{2}}\dfrac{{{x}^{2}}+1}{\left| x-1 \right|} \right)< 0$. We need to break it into two parts and apply rules of logarithm to find the quadratic equation of x.
We know that ${{\log }_{a}}b=c\Rightarrow {{a}^{c}}=b$. This relation holds for inequality also.
We apply the rule twice to eliminate all the logarithms.
\[\begin{align}
  & {{\log }_{0.1}}\left( {{\log }_{2}}\dfrac{{{x}^{2}}+1}{\left| x-1 \right|} \right)< 0 \\
 & \Rightarrow {{\log }_{2}}\dfrac{{{x}^{2}}+1}{\left| x-1 \right|}< \left( {{0.1}^{0}}=1 \right) \\
 & \Rightarrow \dfrac{{{x}^{2}}+1}{\left| x-1 \right|}< \left( {{2}^{1}}=2 \right) \\
\end{align}\]
We get the quadratic form as \[\dfrac{{{x}^{2}}+1}{\left| x-1 \right|}< 2\].
As a modulus form is present in the denominator, we break the domain of x as $x< 1$ and $x> 1$. Value of x can’t be 1 as then the denominator value becomes 0 which is impossible.
For $x< 1$, the modulus form of $\left| x-1 \right|$ becomes $-\left( x-1 \right)=1-x$ and $x> 1$, the modulus form of $\left| x-1 \right|$ becomes $\left( x-1 \right)$.
We solve them individually. For $x< 1$ the equation is \[\dfrac{{{x}^{2}}+1}{1-x}< 2\].
We solve to get
\[\begin{align}
  & \dfrac{{{x}^{2}}+1}{1-x}< 2 \\
 & \Rightarrow {{x}^{2}}+1< 2-2x \\
 & \Rightarrow {{x}^{2}}+2x-1< 0 \\
\end{align}\]
We solve the quadratic to find the domain of the x as
\[\begin{align}
  & {{x}^{2}}+2x-1< 0 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}< 2 \\
 & \Rightarrow \left| x+1 \right|< \sqrt{2} \\
 & \Rightarrow x\in \left( -1-\sqrt{2},-1+\sqrt{2} \right) \\
\end{align}\]
We got two inequalities as $x< 1$ and \[x\in \left( -1-\sqrt{2},-1+\sqrt{2} \right)\].
Together they imply \[x\in \left( -1-\sqrt{2},-1 \right)\].
For $x> 1$ the equation is \[\dfrac{{{x}^{2}}+1}{x-1}< 2\].
We solve to get
\[\begin{align}
  & \dfrac{{{x}^{2}}+1}{x-1}< 2 \\
 & \Rightarrow {{x}^{2}}+1< 2x-2 \\
 & \Rightarrow {{x}^{2}}-2x+3< 0 \\
\end{align}\]
We find the discriminant of the quadratic which is $d=\sqrt{{{\left( -2 \right)}^{2}}-4\times 1\times 3}=\sqrt{-8}\in \mathbb{C}$
So, there is no solution for the second part.
The only solution for the inequality ${{\log }_{0.1}}\left( {{\log }_{2}}\dfrac{{{x}^{2}}+1}{\left| x-1 \right|} \right)< 0$ is \[x\in \left( -1-\sqrt{2},-1 \right)\].

Note: The domain of x is always real, that’s why the imaginary values of x were eliminated. We need to always cross check a value from the domain of the solution to make sure the solution is right.