If $$x$$ is real, the maximum value of $${{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$ is
JEE · Math · previous-year question
- A.$${1 \over 4}$$
- B.$$41$$correct
- C.$$1$$
- D.$${17 \over 7}$$
Answer
B. $$41$$
Explanation
$$y = {{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$ $$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$$ $$D \ge 0$$ as $$x$$ is real $$81{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0$$ $$ \Rightarrow \left( {y - 1} \right)\left( {y - 41} \right) \le 0$$ $$ \Rightarrow 1 \le y \le 41$$ $$\therefore$$ Max value of $$y$$ is $$41$$
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