Question:
\(f(x)=\frac{2x^2-3x+8}{2x^2+3x+8}\), Then sum of maximum and minimum values of f(x) is:
\(f(x)=\frac{2x^2-3x+8}{2x^2+3x+8}\), Then sum of maximum and minimum values of f(x) is:
Updated On: Apr 5, 2024
- \(\frac{136}{55}\)
- \(\frac{146}{55}\)
- \(\frac{146}{11}\)
- \(\frac{136}{11}\)
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The Correct Option is B
Solution and Explanation
The correct answer is option (B): \(\frac{146}{55}\)
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