If x is real the expression \( \Large \frac{x+2}{2x^{2}+3x+6} \) takes all values in the interval:


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.

Correct Answer:
B) \( \Large \left(- \frac{1}{13}, \frac{1}{3}\right) \)

Description for Correct answer:

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.


Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations








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