Question

The set of real values for which the expression ${\log }_{0.1}\left({\log }_2\left(\frac{x^2+1}{|x-1|}\right)\right)$ is defined,

• A. x ∈ R
• B. x ∈ R - {1}
• C. x ∈R - (-1, 0)
• D. x ∈ (-√2 -1, ∞)- {1}

Answer 1

The correct option is C

x ∈R - (-1, 0)

For ${\log }_{0.1}\left({\log }_2\left(\frac{x^2+1}{|x-1|}\right)\right)$ to be define

By definition of logarithm

Condition 1: $\frac{x^2+1}{|x-1|}>0$ Which is always true except x ≠ 1

Condition 2: ${\log }_2\frac{x^2+1}{|x-1|}>0$

Base of the log is greater than 1,

$\frac{x^2+1}{|x-1|}>2^{\circ }$

$\frac{x^2+1}{|x-1|}>1$

Case (i) when x > 1 $x^2$ + 1 > x-1

$x^2$ - x + 2 > 0

${(x-\frac{1}{2})}^2$ + $\frac{7}{4}$ > 0

This is always true. x ∈ R

Case (ii): when x < 1 $x^2$ + 1 > - (x-1)

$x^2$ + x > 0

x(x + 1) > 0

x ∈ ($-\infty$ -1) U (0,$\infty$)

From condition 2 of definition of log x ∈ R - (-1,0)

Above given logarithm inequality to be define when x should belongs to R - (-1,0)