JEE Main 2023 — Basic of Mathematics Question with Solution

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

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 $g\left(x\right)>0, g\left(x\right)\ne 1$.

$\Rightarrow x+\frac{7}{2}>0$

$\Rightarrow x>\frac{-7}{2}$

Also, $x+\frac{7}{2}\ne 1$

$\Rightarrow x\ne \frac{-5}{2}$

And $\frac{x-7}{2x-3}\ne 0$

$\Rightarrow x\ne 7$ and $x\ne \frac{3}{2}$

Domain : $\left(-\frac{7}{2},\infty \right)-\left\{\frac{-5}{2},0,\frac{3}{2}\right\}$

Now let us take different cases.

Case I: $0<x+\frac{7}{2}<1$
$\Rightarrow -\frac{7}{2}<x<-\frac{5}{2}.....\left(1\right)$

$\Rightarrow {\left(\frac{x-7}{2x-3}\right)}^2\le 1$

$\Rightarrow -1\le \frac{x-7}{2x-3}\le 1$

$\Rightarrow \frac{x-7}{2x-3}+1\ge 0$

$\Rightarrow \frac{x-7+2x-3}{2x-3}\ge 0$

$\Rightarrow \frac{3x-10}{2x-3}\ge 0...\left(2\right)$

And $\frac{x-7}{2x-3}-1\le 0$

$\Rightarrow \frac{x-7-2x+3}{2x-3}\le 0$

$\Rightarrow \frac{-x-4}{2x-3}\le 0$

$\Rightarrow \frac{x+4}{2x-3}\ge 0....\left(3\right)$

From $\left(1\right),\left(2\right),\left(3\right)$ we can tell that there is no intersection, no solution

Case II:

$\Rightarrow x+\frac{7}{2}>1$

$\Rightarrow x>-\frac{5}{2}$

$\Rightarrow {\left(\frac{x-7}{2x-3}\right)}^2\ge 1$

$\Rightarrow \left|\frac{x-7}{2x-3}\right|\ge 1$

$\Rightarrow \frac{x-7}{2x-3}\le -1$

$\Rightarrow x\in \left(\frac{3}{2},\frac{10}{3}\right]$

Or,$\frac{x-7}{2x-3}\ge 1$

$\Rightarrow \frac{x-7-2x+3}{2x-3}\ge 0$

$\Rightarrow \frac{x+4}{2x-3}\le 0$

$\Rightarrow x\in \left[-4,\frac{3}{2}\right)$

$x\in \left(-\frac{5}{2},\frac{3}{2}\right)\cup \left(\frac{3}{2},\frac{10}{3}\right]$

Hence there are total 6 integers $\left\{-2,-1,0,1,2,3\right\}$.

Hence this is the required option.