The number of integral solution x of _ (x + (7)/(2)) ((x - 7)/(2x -...

We have been given that$\log_{\left(x + \dfrac{7}{2}\right)}{\left(\dfrac{x - 7}{2x - 3}\right)}^{2} \ge 0$

We need to find the domain.

For $\log_{g\left(x\right)}\left(f\left(x\right)\right)$ is defined when $f\left(x\right) > 0$ and $g\left(x\right) > 0, g\left(x\right) \ne 1$.

$⇒ x + \dfrac{7}{2} > 0$

$⇒ x > \dfrac{- 7}{2}$

Also, $x + \dfrac{7}{2} \ne 1$

$⇒ x \ne \dfrac{- 5}{2}$

And $\dfrac{x - 7}{2x - 3} \ne 0$

$⇒ x \ne 7$ and $x \ne \dfrac{3}{2}$

Domain : \left(- \dfrac{7}{2}, \infty\right) - \left\{\dfrac{- 5}{2}, 0, \dfrac{3}{2}\right\}$</p><p>Now let us take different cases.</p><p>Case I:$0 < x + \dfrac{7}{2} < 1$<br />$⇒ - \dfrac{7}{2} < x < - \dfrac{5}{2} . . . . . ( 1 )$</p><p>$⇒ {\left(\dfrac{x - 7}{2x - 3}\right)}^{2} \le 1$</p><p>$⇒ - 1 \le \dfrac{x - 7}{2x - 3} \le 1$</p><p>$⇒ \dfrac{x - 7}{2x - 3} + 1 \ge 0$</p><p>$⇒ \dfrac{x - 7 + 2x - 3}{2x - 3} \ge 0$</p><p>$⇒ \dfrac{3x - 10}{2x - 3} \ge 0 . . . ( 2 )$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/f30cb77d-6d61-4df2-9dae-21d79ad0e070-image.png" style="width: 200px; height: 108px;" /></p><p>And$\dfrac{x - 7}{2x - 3} - 1 \le 0$</p><p>$⇒ \dfrac{x - 7 - 2x + 3}{2x - 3} \le 0$</p><p>$⇒ \dfrac{- x - 4}{2x - 3} \le 0$</p><p>$⇒ \dfrac{x + 4}{2x - 3} \ge 0 . . . . ( 3 )$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/13daeaf2-89a8-49a3-8a01-d61a28251a39-image.png" style="width: 200px; height: 95px;" /></p><p>From$\left(1\right), $\left(2\right),$\left(3\right)$we can tell that there is no intersection, no solution</p><p>Case II:</p><p>$⇒ x + \dfrac{7}{2} > 1$</p><p>$⇒ x > - \dfrac{5}{2}$</p><p>$⇒ {\left(\dfrac{x - 7}{2x - 3}\right)}^{2} \ge 1$</p><p>$⇒ \left|\dfrac{x - 7}{2x - 3}\right| \ge 1$</p><p>$⇒ \dfrac{x - 7}{2x - 3} \le - 1$</p><p>$⇒ x \in \left(\dfrac{3}{2}, \dfrac{10}{3}\right]$</p><p>Or,$\dfrac{x - 7}{2x - 3} \ge 1$</p><p>$⇒ \dfrac{x - 7 - 2x + 3}{2x - 3} \ge 0$</p><p>$⇒ \dfrac{x + 4}{2x - 3} \le 0$</p><p>$⇒ x \in \left[- 4, \dfrac{3}{2}\right)$</p><p>$x \in \left(- \dfrac{5}{2}, \dfrac{3}{2}\right) \cup \left(\dfrac{3}{2}, \dfrac{10}{3}\right]$</p><p>Hence there are total 6 integers \left\{- 2, - 1, 0, 1, 2, 3\right\}$.

Hence this is the required option.