Question:
If \(x\) is real number, then \( \frac{x}{x^2 - 5x + 9} \) must lie between
If \(x\) is real number, then \( \frac{x}{x^2 - 5x + 9} \) must lie between
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For rational functions, ensure denominator positivity and use derivative for range.
Updated On: May 8, 2026
- \( \frac{1}{11} \) and \(1\)
- \(-1\) and \( \frac{1}{11} \)
- \(-11\) and \(1\)
- \(-1\) and \(11\)
- \(-\frac{1}{11}\) and \(1\)
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The Correct Option is
Solution and Explanation
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
\[ f(3)=\frac{3}{9-15+9}=1 \] \[ f(-3)=\frac{-3}{9+15+9}=-\frac{1}{11} \] \[ \boxed{-\frac{1}{11} \leq f(x) \leq 1} \]
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
\[ f(3)=\frac{3}{9-15+9}=1 \] \[ f(-3)=\frac{-3}{9+15+9}=-\frac{1}{11} \] \[ \boxed{-\frac{1}{11} \leq f(x) \leq 1} \]
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