JEE MAIN - Mathematics (2006 - No. 20)
If $$x$$ is real, the maximum value of $${{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$ is
$${1 \over 4}$$
$$41$$
$$1$$
$${17 \over 7}$$
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 $$41$$
$$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$$
$$D \ge 0$$ 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 $$41$$
Comments (0)
