Question

The domain of the function $f(x) = \ln\left(\frac{1}{\sqrt{x^2-4x+4}}\right) + \sin^{-1}(x^2-2)$ is

• A. $[1,3]$
• B. $[1,3)$
• C. $[1,\sqrt{3}]$
• D. $[1,\sqrt{3})$

Hint:

Combine domain restrictions for composite functions: ensure the logarithm’s argument is positive and the inverse sine’s argument is in $[-1, 1]$. Simplify expressions like $\sqrt{(x - a)^2} = |x - a|$.

Answer 1

The Correct Option is C

To determine the domain of the function \( f(x) = \ln\left(\frac{1}{\sqrt{x^2-4x+4}}\right) + \sin^{-1}(x^2-2) \), we must analyze the conditions under which each part of the function is defined.

Step 1: Consider \( \ln\left(\frac{1}{\sqrt{x^2-4x+4}}\right) \). The argument of the logarithm must be positive:

\( \frac{1}{\sqrt{x^2-4x+4}} > 0 \), where \( \sqrt{x^2-4x+4} \neq 0 \).

This implies \( x^2-4x+4 \neq 0 \). Observing that \( x^2-4x+4 = (x-2)^2 \), it simplifies to \( x \neq 2 \).

Additionally, since we require the expression under the square root to be non-negative:

\( x^2-4x+4 \geq 0 \).

Simplifying, \( (x-2)^2 \geq 0 \) is always true.

Step 2: Consider \( \sin^{-1}(x^2-2) \). For the inverse sine function, we must have:

\(-1 \leq x^2-2 \leq 1.\)

Simplify the inequalities:

1. \( x^2-2 \geq -1 \) gives \( x^2 \geq 1 \), so \( x \leq -1 \) or \( x \geq 1 \).

2. \( x^2-2 \leq 1 \) gives \( x^2 \leq 3 \), so \(-\sqrt{3} \leq x \leq \sqrt{3}\).

The intersection of conditions from Step 1 and Step 2 is:

\(-\sqrt{3} \leq x \leq \sqrt{3}\) excluding \( x = 2 \), but along with the constraints from the square root and logarithm, only the positive domain is relevant:

\(1 \leq x \leq \sqrt{3}\).

Thus, the domain of the function is \([1, \sqrt{3}]\).