JEE Main 2023 — Mathematics Algebra

We have been given that${\mathrm{log}}_{(x+\frac{7}{2})}{(\frac{x-7}{2x-3})}^{2}\geq 0$log(x+27​)​(2x−3x−7​)2≥0
We need to find the domain.
For ${\mathrm{log}}_{g(x)}(f(x))$logg(x)​(f(x)) is defined when $f(x)>0$f(x)>0 and $g(x)>0,g(x)\neq 1$g(x)>0,g(x)=1.
$\Rightarrow x+\frac{7}{2}>0$⇒x+27​>0
$\Rightarrow x>\frac{-7}{2}$⇒x>2−7​
Also, $x+\frac{7}{2}\neq 1$x+27​=1
$\Rightarrow x\neq \frac{-5}{2}$⇒x=2−5​
And $\frac{x-7}{2x-3}\neq 0$2x−3x−7​=0
$\Rightarrow x\neq 7$⇒x=7 and $x\neq \frac{3}{2}$x=23​
Domain : $(-\frac{7}{2},\infty )-{\frac{-5}{2},0,\frac{3}{2}}$(−27​,∞)−2−5​,0,23​
Now let us take different cases.
Case I: $0<x+\frac{7}{2}<1$0<x+27​<1
$\Rightarrow -\frac{7}{2}<x<-\frac{5}{2}.....(1)$⇒−27​<x<−25​.....(1)
$\Rightarrow {(\frac{x-7}{2x-3})}^{2}\leq 1$⇒(2x−3x−7​)2≤1
$\Rightarrow -1\leq \frac{x-7}{2x-3}\leq 1$⇒−1≤2x−3x−7​≤1
$\Rightarrow \frac{x-7}{2x-3}+1\geq 0$⇒2x−3x−7​+1≥0
$\Rightarrow \frac{x-7+2x-3}{2x-3}\geq 0$⇒2x−3x−7+2x−3​≥0
$\Rightarrow \frac{3x-10}{2x-3}\geq 0...(2)$⇒2x−33x−10​≥0...(2)

And $\frac{x-7}{2x-3}-1\leq 0$2x−3x−7​−1≤0
$\Rightarrow \frac{x-7-2x+3}{2x-3}\leq 0$⇒2x−3x−7−2x+3​≤0
$\Rightarrow \frac{-x-4}{2x-3}\leq 0$⇒2x−3−x−4​≤0
$\Rightarrow \frac{x+4}{2x-3}\geq 0....(3)$⇒2x−3x+4​≥0....(3)

From $(1),(2),(3)$(1),(2),(3) we can tell that there is no intersection, no solution
Case II:
$\Rightarrow x+\frac{7}{2}>1$⇒x+27​>1
$\Rightarrow x>-\frac{5}{2}$⇒x>−25​
$\Rightarrow {(\frac{x-7}{2x-3})}^{2}\geq 1$⇒(2x−3x−7​)2≥1
$\Rightarrow |\frac{x-7}{2x-3}|\geq 1$⇒∣2x−3x−7​∣≥1
$\Rightarrow \frac{x-7}{2x-3}\leq -1$⇒2x−3x−7​≤−1
$\Rightarrow x\in (\frac{3}{2},\frac{10}{3}]$⇒x∈(23​,310​]
Or,$\frac{x-7}{2x-3}\geq 1$2x−3x−7​≥1
$\Rightarrow \frac{x-7-2x+3}{2x-3}\geq 0$⇒2x−3x−7−2x+3​≥0
$\Rightarrow \frac{x+4}{2x-3}\leq 0$⇒2x−3x+4​≤0
$\Rightarrow x\in [-4,\frac{3}{2})$⇒x∈[−4,23​)
$x\in (-\frac{5}{2},\frac{3}{2})\cup (\frac{3}{2},\frac{10}{3}]$x∈(−25​,23​)∪(23​,310​]
Hence there are total 6 integers ${-2,-1,0,1,2,3}$−2,−1,0,1,2,3.
Hence this is the required option.