Question

If \(x\) is real number, then \( \frac{x}{x^2 - 5x + 9} \) must lie between

• A. \( \frac{1}{11} \) and \(1\)
• B. \(-1\) and \( \frac{1}{11} \)
• C. \(-11\) and \(1\)
• D. \(-1\) and \(11\)
• E. \(-\frac{1}{11}\) and \(1\)

Hint:

For rational functions, ensure denominator positivity and use derivative for range.

Answer 1

Answer: (E)

Understanding the Concept: Use derivative or inequality approach to find range.

Step 1: Let function
\[ f(x) = \frac{x}{x^2 - 5x + 9} \]

Step 2: Denominator always positive
\[ x^2 -5x +9 = (x-\tfrac{5}{2})^2 + \tfrac{11}{4} > 0 \]

Step 3: Find extrema using derivative
Differentiate: \[ f'(x) = \frac{(x^2-5x+9) - x(2x-5)}{(x^2-5x+9)^2} \] Simplify numerator: \[ = x^2 -5x +9 -2x^2 +5x = -x^2 + 9 \]

Step 4: Solve critical points
\[ -x^2 +9 =0 \Rightarrow x=\pm3 \]

Step 5: Evaluate function
\[ f(3)=\frac{3}{9-15+9}=1 \] \[ f(-3)=\frac{-3}{9+15+9}=-\frac{1}{11} \] \[ \boxed{-\frac{1}{11} \leq f(x) \leq 1} \]