JEE Math Practice Question
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.217
- B.182
- C.201
- D.195
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