The number of integral solution x of log_(x+ 7 2) ( x-7 2x-3)^2≥ 0 is - JEE Main 2023 Mathematics PYQ

JEE Main 2023 — Mathematics Algebra

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The number of integral solution $x$ of ${\mathrm{log}}_{(x+\frac{7}{2})}{(\frac{x-7}{2x-3})}^{2}\geq 0$ is

Official previous-year question

Held on 11 Apr 2023 · Verified 1 Jul 2026.

Options

A
$7$
B
$8$
C
$6$
D
$5$

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Solution

We have been given that ${\mathrm{log}}_{(x+\frac{7}{2})}{(\frac{x-7}{2x-3})}^{2}\geq 0$
We need to find the domain.
For ${\mathrm{log}}_{g(x)}(f(x))$ is defined when $f(x)>0$ and $g(x)>0,g(x)\neq 1$ .
$\Rightarrow x+\frac{7}{2}>0$
$\Rightarrow x>\frac{-7}{2}$
Also, $x+\frac{7}{2}\neq 1$
$\Rightarrow x\neq \frac{-5}{2}$
And $\frac{x-7}{2x-3}\neq 0$
$\Rightarrow x\neq 7$ and $x\neq \frac{3}{2}$
Domain : $(-\frac{7}{2},\infty )-{\frac{-5}{2},0,\frac{3}{2}}$
Now let us take different cases.
Case I: $0<x+\frac{7}{2}<1$
$\Rightarrow -\frac{7}{2}<x<-\frac{5}{2}.....(1)$
$\Rightarrow {(\frac{x-7}{2x-3})}^{2}\leq 1$
$\Rightarrow -1\leq \frac{x-7}{2x-3}\leq 1$
$\Rightarrow \frac{x-7}{2x-3}+1\geq 0$
$\Rightarrow \frac{x-7+2x-3}{2x-3}\geq 0$
$\Rightarrow \frac{3x-10}{2x-3}\geq 0...(2)$
JEE Main 2023 Mathematics, Basic of Mathematics — solution figure 1
And $\frac{x-7}{2x-3}-1\leq 0$
$\Rightarrow \frac{x-7-2x+3}{2x-3}\leq 0$
$\Rightarrow \frac{-x-4}{2x-3}\leq 0$
$\Rightarrow \frac{x+4}{2x-3}\geq 0....(3)$

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