Question

The domain of the definition of the real function f(x) = √(log12 x² ) of the real variable x is


Answer it With explanation​

Answer 1

Answer:

$\Large\underline{\underline{\sf{ \color{magenta}{\qquad Given \qquad} }}}$

Answer: |x| ≥ 1

explanation

We have f(x) = √(log12 x²)

Since, loga k ≥ 0 if a > 1, k ≥ 1

or 0 < a < 1 and 0 < k ≤ 1

So, the function f(x) exists if

log12 x² ≥ 0

⇒ x² ≥ 1

⇒ |x| ≥ 1

Answer 2

Answer:

The domain of the function $f(x)=\sqrt{log_{12}x^{2}}$ is $|x|\geq 1$

Step-by-step explanation:

The given function is $f(x)=\sqrt{log_{12}x^{2}}$

To find the domain of the above function we need to find the values at which the above function exists.

The above functions exists only when the term under the root is either zero or positive.Therefore

$log_{12}x^{2}\geq 0\\= > x^{2}\geq 12^{0}= > x^{2}\geq 1\\= > x^{2}-1\geq 0$

Factorizing the above inequality we get

$(x-1)(x+1)\geq 0= > x\leq -1 ,x\geq 1\\= > |x|\geq 1$

Therefore,the domain of the function $f(x)=\sqrt{log_{12}x^{2}}$ is $|x|\geq 1$

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