Question

If x is real, the maximum value of $\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$ is

• A. $\frac{1}{4}$
• B. 41
• C. 1
• D. $\frac{17}{7}$

Answer 1

Answer: (B)

$y = \frac{3x^{2} + 9x + 17}{3x^{2} + 9x +7}$
$3x^{2} \left(y -1\right)+9x\left(y-1\right)+7y -17 = 0$
$D \ge 0 \,\,\,\because\, x$ 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 $41$