Question

For real values of $ x $ and $ a $, if the expression $ \frac{x+a}{2x^2 - 3x + 1} $ assumes all real values, then:

• A. \( a-1 \) or \( a-\frac{1}{2} \)
• B. \( -1a-\frac{1}{2} \)
• C. \( \frac{1}{2}a1 \)
• D. \( a\frac{1}{2} \) or \( a1 \)

Hint:

To ensure a rational function assumes all real values, analyze the denominator’s zeros and numerator constraints.

Answer 1

The Correct Option is B

Step 1: Identify Restrictions The denominator \( 2x^2 - 3x + 1 \) should not be zero. Solve: \[ 2x^2 - 3x + 1 = 0 \] Using quadratic formula: \[ x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \] \[ x = 1, \quad x = \frac{1}{2} \] Step 2: Condition for All Real Values For the function to assume all real values, \( a \) should lie in the range: \[ -1a-\frac{1}{2} \] Thus, the correct answer is \( -1a-\frac{1}{2} \).