The set of all real values of x satisfying th | vedclass

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

Question

(B) Given the inequality: $\frac{7 x^2-5 x-18}{2 x^2+x-6} < 2$Subtract $2$ from both sides: $\frac{7 x^2-5 x-18}{2 x^2+x-6} - 2 < 0$$\Rightarrow \frac

Vedclass · Accepted Answer

$\left(-2, -\frac{2}{3}\right) \cup \left(\frac{3}{2}, 3 \right)$

Vedclass · Answer

(B) Given the inequality: $\frac{7 x^2-5 x-18}{2 x^2+x-6} Subtract $2$ from both sides: $\frac{7 x^2-5 x-18}{2 x^2+x-6} - 2 $\Rightarrow \frac{7 x^2-5 x-18 - 2(2 x^2+x-6)}{2 x^2+x-6} $\Rightarrow \frac{7 x^2-5 x-18 - 4 x^2-2 x+12}{(2 x-3)(x+2)} $\Rightarrow \frac{3 x^2-7 x-6}{(2 x-3)(x+2)} Factor the numerator: $3 x^2-7 x-6 = 3 x^2-9 x+2 x-6 = 3 x(x-3)+2(x-3) = (3 x+2)(x-3)$So,the inequality becomes: $\frac{(3 x+2)(x-3)}{(2 x-3)(x+2)} The critical points are $x = -2, -\frac{2}{3}, \frac{3}{2}, 3$.Using the wavy curve method,we test the intervals:For $x > 3$,the expression is positive.For $\frac{3}{2} For $-\frac{2}{3} For $-2 For $x Since we need the expression to be less than $0$,the solution is $x \in \left(-2, -\frac{2}{3}\right) \cup \left(\frac{3}{2}, 3 \right)$.

The set of all real values of $x$ satisfying the inequality $\frac{7 x^2-5 x-18}{2 x^2+x-6} < 2$ is

Similar Questions

The set of all real values of $x$ for which $\frac{x^2-1}{(x-4)(x-3)} \geq 1$ is

$\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4} < 0\right\}$ is equal to

Ravi obtained $70$ and $75$ marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least $60$ marks.

Find all pairs of consecutive even positive integers,both of which are larger than $5$,such that their sum is less than $23$.

The origin belongs to a region between the lines $x+2y-5=0$ and $3x-4y+5=0$. The number of points in that region of the form $((\alpha-1)^2, \alpha)$,where $\alpha \in \mathbb{Z}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module