Question

If \(f(x) = \frac{x+2}{x^2-3x+1}\), then the values of \(x\) for which \(f(x)\) is not defined are

• A. \(x = \frac{3+\sqrt{5}}{2}, x = \frac{3-\sqrt{5}}{2}\)
• B. \(x = \frac{-3+\sqrt{5}}{2}, x = \frac{-3-\sqrt{5}}{2}\)
• C. \(x = \frac{3+\sqrt{3}}{2}, x = \frac{3-\sqrt{3}}{2}\)
• D. \(x = \frac{2+\sqrt{5}}{2}, x = \frac{2-\sqrt{5}}{2}\)

Hint:

The function is undefined where the denominator is zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
A rational function is undefined at the zeros of its denominator.

Step 2: Key Formula or Approach
Solve \(x^2-3x+1=0\) with the quadratic formula.

Step 3: Detailed Explanation
\[ x=\frac{3\pm\sqrt{9-4}}{2}=\frac{3\pm\sqrt5}{2} \]
The numerator \(x+2\) is not zero at either value, so both are genuine points where \(f\) is undefined.

Final Answer:
The function is undefined at \(x=\frac{3\pm\sqrt5}{2}\), option (A). \[ \boxed{x=\dfrac{3\pm\sqrt5}{2}\ \text{(A)}} \]