Question

Find the maximum value of $\left[x\right]$ for which the inequality $\frac{x^2+6x-11}{x+3}<1$ is satisfied.

Answer 1

$\left(\frac{x^2+6x-11}{x+3}\right)-1<0\left(\frac{x^2+6x-11-(x+3)}{x+3}\right)<0\left(\frac{x^2+5x-14}{x+3}\right)<0\left(\frac{(x+7)(x-2)}{x+3}\right)<0\therefore x\in (-\infty ,-7)\cup (-3,2)somaxvalueof\left[x\right]=1$