Question

The function \( f(x) = \frac{\sqrt{3x^2 - 5x - 2}}{2x^2 - 7x + 5} \) has discontinuous points at \( x = \dots \)

• A. \( \frac{5}{2}, 2 \)
• B. \( -\frac{1}{3}, 2 \)
• C. \( 1, \frac{5}{2} \)
• D. \( -\frac{1}{3}, 1 \)

Hint:

For rational functions with square roots, check both the denominator (for division by zero) and the expression inside the square root (for non-negative values).

Answer 1

The Correct Option is A

For the function \( f(x) = \frac{\sqrt{3x^2 - 5x - 2}}{2x^2 - 7x + 5} \), we need to find the values of \( x \) where the function is undefined or discontinuous. Step 1: The function will be undefined if the denominator \( 2x^2 - 7x + 5 = 0 \). Solving for \( x \): \[ 2x^2 - 7x + 5 = 0 \] Using the quadratic formula: \[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(5)}}{2(2)} = \frac{7 \pm \sqrt{49 - 40}}{4} = \frac{7 \pm \sqrt{9}}{4} \] \[ x = \frac{7 \pm 3}{4} \quad \Rightarrow \quad x = \frac{10}{4} = \frac{5}{2} \quad \text{or} \quad x = \frac{4}{4} = 1 \] Step 2: The function will also be discontinuous if the expression inside the square root is negative, i.e., \( 3x^2 - 5x - 2 \geq 0 \). After solving the inequalities and considering the points where the function becomes undefined, we find that the function is discontinuous at \( x = \frac{5}{2} \) and \( x = 2 \). % Final Answer The discontinuous points are \( \frac{5}{2} \) and \( 2 \).

Answer 2

Step 1: Analyze the structure of the function
The function is:
f(x) = √(3x² − 5x − 2) / (2x² − 7x + 5)

A function can be discontinuous at points where:
(a) The denominator is zero (undefined), or
(b) The expression inside the square root is negative (imaginary value for real function)

Step 2: Find where the denominator is zero
Denominator = 2x² − 7x + 5
Set it equal to zero:
2x² − 7x + 5 = 0
Solve using the quadratic formula:
x = [7 ± √(49 − 4·2·5)] / 2·2 = [7 ± √(49 − 40)] / 4 = [7 ± √9] / 4 = [7 ± 3] / 4
⇒ x = (7 + 3)/4 = 10/4 = 2.5, and (7 − 3)/4 = 4/4 = 1
So the denominator is zero at x = 1 and x = 5/2

Step 3: Ensure the square root remains defined (non-negative)
Check the domain of the numerator: √(3x² − 5x − 2) must be ≥ 0
Let’s solve the inequality:
3x² − 5x − 2 ≥ 0
Factor or use quadratic formula:
x = [5 ± √(25 + 24)] / 6 = [5 ± √49] / 6 = [5 ± 7] / 6
x = 12/6 = 2, x = −2/6 = −1/3
So inequality holds for x ≤ −1/3 or x ≥ 2
Hence, domain of the function is x ∈ (−∞, −1/3] ∪ [2, ∞), excluding any points where denominator = 0

Step 4: Determine the discontinuous points
From Step 2: x = 1 and x = 5/2 make the denominator zero
From Step 3: x = 1 is not in the domain (since 1 < 2)
So we only consider x = 5/2 which is ≥ 2 and in the domain of the square root

Thus, both x = 2.5 and x = 2 are points of discontinuity:
- At x = 2, the numerator becomes zero and the denominator becomes zero: check for removable discontinuity
- At x = 5/2, denominator is zero and function is undefined ⇒ discontinuity

Final Answer:
The points of discontinuity are: x = 2 and x = 5/2