A) \( \Large \left(\frac{1}{13}, \frac{1}{3}\right) \) |
B) \( \Large \left(- \frac{1}{13}, \frac{1}{3}\right) \) |
C) \( \Large \left(- \frac{1}{3}, \frac{1}{13}\right) \) |
D) none of these. |
B) \( \Large \left(- \frac{1}{13}, \frac{1}{3}\right) \) |
Let given expression be y
i.e., \( \Large 4=\frac{x+2}{2x^{2}+3x+6} \)
=> \( \Large 2x^{2}y+ \left(3y-1\right)x+ \left(6y-2\right)=0 \)
If y ≠ 0 then \( \Large \triangle \ge 0 \) for real x.
i.e., \( \Large b^{2}-4ac\ge 0 \)
\( \Large \therefore \left(3y-1\right)^{2}-8y \left(6y-2\right)\ge 0 \)
=> \( \Large -39y^{2}+10y+1\ge 0 \)
=> \( \Large \left(13y+1\right) \left(3y-1\right) \le 0 \)
=> \( \Large -\frac{1}{13} \le y \le \frac{1}{3} \)
If y = 0, then x = -2 which is real and this value of y is included in above range.