Q105 / 155 · WBJEE 2017 · Mathematics
For real $x$, the greatest value of $\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$ is
- A1
- B-1
- Correct answerC$\frac{1}{2}$
- D$\frac{1}{4}$
Solution
Let $y=\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$
$\Rightarrow 2 x^{2} y+4 x y+9 y=x^{2}+2 x+4$
$\Rightarrow(2 y-1) x^{2}+(4 y-2) x+9 y-4=0$
$\Rightarrow x=\frac{-(4 y-2) \pm \sqrt{(4 y-2)^{2}-4(2 y-1)(9 y-4)}}{2(2 y-1)}$
Since, $x$ is real number.
$\therefore \quad(4 y-2)^{2}-4(2 y-1)(9 y-4) \geq 0$
$\Rightarrow \quad 4(2 y-1)^{2}-4(2 y-1)(9 y-4) \geq 0$
$\Rightarrow \quad 4(2 y-1)(2 y-1-9 y+4) \geq 0$
$\Rightarrow \quad \quad 4(2 y-1)(3-7 y) \geq 0$
$\Rightarrow \quad(2 y-1)(7 y-3) \leq 0$
$\Rightarrow \quad \frac{3}{7} \leq y \leq \frac{1}{2}$
$\therefore$ Maximum value of $y=\frac{1}{2}$
Hence, the correct answer is option: 'C'.