JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let the sum of the maximum and the minimum values of the function \(f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}\) be \(\frac{m}{n}\), where \(\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1\). Then \(\mathrm{m}+\mathrm{n}\) is equal to :
- A \(182\)
- B \(217\)
- C \(195\)
- D \(201\)
Answer & Solution
Correct Answer
(D) \(201\)
Step-by-step Solution
Detailed explanation
\( y=\frac{2 x^2-3 x+8}{2 x^2+3 x+8} \) \( x^2(2 y-2)+x(3 y+3)+8 y-8=0 \) \( \text { use } D \geq 0 \) \( (3 y+3)^2-4(2 y-2)(8 y-8) \geq 0 \) \( (11 y-5)(5 y-11) \leq 0 \) \( \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}\right] \) \( y=1\) is also included
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