If frac{2x}{2x^2 + 5x + 2} frac{1}{x + 1},t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the inequality: $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$Factor the denominator: $\frac{2x}{(2x + 1)(x + 2)} > \frac{1}{x + 1}$Subtract $

Vedclass · Accepted Answer

$-2 < x < -1$

Vedclass · Answer

(C) Given the inequality: $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$Factor the denominator: $\frac{2x}{(2x + 1)(x + 2)} > \frac{1}{x + 1}$Subtract $\frac{1}{x + 1}$ from both sides: $\frac{2x}{(2x + 1)(x + 2)} - \frac{1}{x + 1} > 0$Find a common denominator: $\frac{2x(x + 1) - (2x + 1)(x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$Simplify the numerator: $\frac{2x^2 + 2x - (2x^2 + 5x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$$\frac{-3x - 2}{(2x + 1)(x + 2)(x + 1)} > 0$Multiply by $-1$ and reverse the inequality: $\frac{3x + 2}{(2x + 1)(x + 2)(x + 1)} The critical points are $x = -2, -1, -\frac{2}{3}, -\frac{1}{2}$.Testing the intervals,the inequality holds for $x \in (-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2})$.

If $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$,then

Similar Questions

$\left\{x \in R : \frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}\right\}=$

Solve for $x$ the following inequality: $\frac{1}{|x|-3} \leq \frac{1}{2}$

If $\sqrt{9x^2+6x+1} < (2-x)$,then:

If $\theta$ is an acute angle and $\sin\theta = \frac{p - 6}{8 - p}$,then $p$ must satisfy

There are $30$ questions in a multiple-choice test. $A$ student gets $1$ mark for each unattempted question,$0$ mark for each wrong answer,and $4$ marks for each correct answer. If a student answered $x$ questions correctly and scored $60$,then the number of possible values of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module