Question

The domain of the real valued function \(f(x)=\dfrac{√(x2−7x+6)}{√(​x2−4​+1)}\)  is

• A. R-[-6,-2)
• B. R-[-6,-2)
• C. R-(-2,6]
• D. R-[-2,6)
• E. R-[-6,2)

Answer 1

The Correct Option is C

Step 1: Understand the problem and given function.

We are tasked with finding the domain of the real-valued function:

\[ f(x) = \sqrt{x^2 - 4} + \frac{1}{\sqrt{x^2 - 7x + 6}}. \]

The domain of \( f(x) \) is the set of all \( x \in \mathbb{R} \) for which the function is well-defined. This requires both terms in the function to be defined:

• \( \sqrt{x^2 - 4} \): The expression inside the square root must be non-negative, i.e., \( x^2 - 4 \geq 0 \).
• \( \frac{1}{\sqrt{x^2 - 7x + 6}} \): The denominator must not be zero, and the expression inside the square root must be positive, i.e., \( x^2 - 7x + 6 > 0 \).

Step 2: Analyze \( \sqrt{x^2 - 4} \).

The inequality \( x^2 - 4 \geq 0 \) can be rewritten as:

\[ x^2 \geq 4. \]

Taking the square root on both sides:

\[ |x| \geq 2. \]

This implies:

\[ x \in (-\infty, -2] \cup [2, \infty). \]

Step 3: Analyze \( \frac{1}{\sqrt{x^2 - 7x + 6}} \).

The denominator \( \sqrt{x^2 - 7x + 6} \) must be positive, so we solve:

\[ x^2 - 7x + 6 > 0. \]

Factorize the quadratic:

\[ x^2 - 7x + 6 = (x - 6)(x - 1). \]

The inequality becomes:

\[ (x - 6)(x - 1) > 0. \]

Using the sign chart method, the critical points are \( x = 1 \) and \( x = 6 \). Testing intervals around these points:

• For \( x \in (-\infty, 1) \), \( (x - 6)(x - 1) > 0 \) (positive).
• For \( x \in (1, 6) \), \( (x - 6)(x - 1) < 0 \) (negative).
• For \( x \in (6, \infty) \), \( (x - 6)(x - 1) > 0 \) (positive).

Thus:

\[ x \in (-\infty, 1) \cup (6, \infty). \]

Step 4: Combine the conditions.

To find the domain of \( f(x) \), we combine the two conditions:

• \( x \in (-\infty, -2] \cup [2, \infty) \) from \( \sqrt{x^2 - 4} \).
• \( x \in (-\infty, 1) \cup (6, \infty) \) from \( \frac{1}{\sqrt{x^2 - 7x + 6}} \).

The intersection of these two sets is:

\[ x \in (-\infty, -2] \cup (6, \infty). \]

Final Answer:

The domain of \( f(x) \) is:

\( R - [-2, 6) \).

Answer 2

Step 1: Analyze \( \sqrt{x^2 - 4} \)

The square root is defined when \( x^2 - 4 \geq 0 \). Factorizing:

\[ x^2 - 4 = (x - 2)(x + 2). \]

Solve the inequality:

\[ (x - 2)(x + 2) \geq 0. \]

Using a sign chart, the solution is:

\[ x \in (-\infty, -2] \cup [2, \infty). \]

Step 2: Analyze \( \frac{1}{\sqrt{x^2 - 7x + 6}} \)

The square root in the denominator requires \( x^2 - 7x + 6 > 0 \). Factorizing:

\[ x^2 - 7x + 6 = (x - 6)(x - 1). \]

Solve the inequality:

\[ (x - 6)(x - 1) > 0. \]

Using a sign chart, the solution is:

\[ x \in (-\infty, 1) \cup (6, \infty). \]

Additionally, the denominator cannot be zero, so:

\[ x \neq 1 \quad \text{and} \quad x \neq 6. \]

Step 3: Combine the Domains

The overall domain is the intersection of the two conditions:

From \( \sqrt{x^2 - 4} \):

\[ x \in (-\infty, -2] \cup [2, \infty). \]

From \( \frac{1}{\sqrt{x^2 - 7x + 6}} \):

\[ x \in (-\infty, 1) \cup (6, \infty). \]

Taking the intersection:

\[ x \in (6, \infty). \]

Step 4: Express in Terms of \( R \)

The domain of \( f(x) \) excludes all values except \( (6, \infty) \). In set notation:

\[ R - (-\infty, 6]. \]

From the options provided, the correct answer is:

\( R - (2, 6] \).