Question

The domain of the function $\sqrt {\log \left( {{x^2} - 6x + 6} \right)} $ is
A) $\left( { - \infty ,\infty } \right)$
B) $\left( { - \infty ,3 - \sqrt 3 } \right) \cup \left( {3 + \sqrt 3 ,\infty } \right)$
C) $\left( { - \infty ,1} \right] \cup \left[ {5,\infty } \right)$
D) $\left[ {0,\infty } \right)$

Answer 1

Hint: First, start with the condition check for the square root i.e., the value in square root cannot be negative. After applying the condition, the value in the log cannot be less than 1. After that perform the operations to find the roots of the equation. Then, apply the condition for which the equation satisfies.

Complete step-by-step answer:
Given:- $\sqrt {\log \left( {{x^2} - 6x + 6} \right)} $
Since, $\sqrt {\log \left( {{x^2} - 6x + 6} \right)} $ is in square root. So, the value inside the square root cannot be negative. Then,
$\log \left( {{x^2} - 6x + 6} \right) \geqslant 0$
As we know that the value in the log cannot be less than 1. Because the number in log less than 1 gives the negative value of log which is not possible for square root.
Then,
${x^2} - 6x + 6 \geqslant 1$
Shift 1 to the LHS of the equation,
${x^2} - 6x + 5 \geqslant 0$
Resolve the equation to find the roots of the equation,
${x^2} - 5x - x + 5 \geqslant 0$
Take out common factors from the equation,
$x\left( {x - 5} \right) - 1\left( {x - 5} \right) \geqslant 0$
Take the common factors from the equation,
$\left( {x - 5} \right)\left( {x - 1} \right) \geqslant 0$
Equate $\left( {x - 5} \right)$ greater than 0. So that it can satisfies the equation,
$x - 5 \geqslant 0$
Move 5 to the right side,
$x \geqslant 5$
Now, equate $\left( {x - 1} \right)$ less than 0. So, that it can satisfies the equation,
$x - 1 \leqslant 0$
Move 1 to the left side,
$x \leqslant 1$
Here, $x \leqslant 1,x \geqslant 5$
Thus, the domain of the function is $\left( { - \infty ,1} \right] \cup \left[ {5,\infty } \right)$.

Hence, option (C) is the correct answer.

Note: The students might make mistakes in the log value as the log value cannot be less than 1. If the log value will be less than 1, it will return a negative value which is an error as the value in the square root must be greater than 0.
Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Product rule law,
$\log xy = \log x + \log y$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$