Sets and Relations

$\sqrt{x^2+x-2}>1-x$

$ \Rightarrow \quad x^2+x-2 \geq 0 $

and $x^2+x-2>1+x^2-2 x$

$\Rightarrow(x+2)(x-1) \geq 0$ and $3 x>3$

$\Rightarrow x \leq-2$ or $x \geq 1 \quad$ and $x>1$

On combining the inequality, we get $x>1$ i.e., $(1, \infty)$

$ \begin{aligned} &\text { } A=\left\{x \in R / \sqrt{x^2-8 x+15} \in R\right\}\\ &\begin{array}{ll} \Rightarrow & x^2-8 x+15 \geq 0 \\ \Rightarrow & (x-5)(x-3) \geq 0 \\ \Rightarrow & x \in(-\infty, 3] \cup[5, \infty] \\ \therefore & A=(-\infty, 3] \cup[5, \infty] \\ & B=\left\{x \in R / \frac{x-3}{2 x-5}<\frac{x-6}{2 x-11}\right\} \\ \Rightarrow & \frac{x-3}{2 x-5}-\frac{x-6}{2 x-11}<0 \\ \Rightarrow & \frac{-10 x+3}{(2 x-5)(2 x-11)}<0 \\ \Rightarrow & \frac{10 x-3}{(2 x-5)(2 x-11)}>0 \end{array} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad & x \in\left(\frac{3}{10}, \frac{5}{2}\right) \cup\left(\frac{11}{2}, \infty\right) \\ \therefore \quad & B=\left(\frac{3}{10}, \frac{5}{2}\right) \cup\left(\frac{11}{2}, \infty\right) \\ & A \cap B=\left(\frac{5}{2}, 3\right] \cup\left[5, \frac{11}{2}\right) \end{aligned} $