QJEEM23Mae3c0: JEE Main 2023 Mathematics - Class 12 JEEM23Mae3c0 Solution: The number of integral solutions of is :

Question

Verified Solution • 2025–26 Session moderate

The number of integral solutions $x$ of $\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$ is :

Accepted Answer

\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0

Domain :

\begin{aligned}&amp; x+\frac{7}{2}&gt;0 \\\\&amp; x&gt;\frac{-7}{2} \\\\&amp; x+\frac{7}{2} \neq 1 \\\\&amp; x \neq \frac{-5}{2} \\\\&amp; \frac{x-7}{2 x-3} \neq 0 \\\\&amp; x \neq 7 \\\\&amp; x \neq \frac{3}{2}\end{aligned}

\text { Taking intersection : } x \in\left(\frac{-7}{2}, \infty\right)-\left\{-\frac{5}{2}, \frac{3}{2}, 7\right\}

Now

\log _{\mathrm{a}} \mathrm{b} \geq 0

if

\mathrm{a}&gt;1

and

\mathrm{b} \geq 1

\begin{gathered}\text { Or } \\\\a \in(0,1) \text { and } b \in(0,1)\end{gathered}

Case I :

x+\frac{7}{2}&gt;1 \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \geq 1

\therefore

x &gt; -\frac{5}{2}

and

\begin{aligned}&amp; (2 x-3)^2-(x-7)^2 \leq 0 \\\\&amp; (2 x-3+x-7)(2 x-3-x+7) \leq 0 \\\\&amp; (3 x-10)(x+4) \leq 0\end{aligned}

x \in\left[-4, \frac{10}{3}\right]

\text { Intersection : } \mathrm{x} \in\left(\frac{-5}{2}, \frac{10}{3}\right]

Case II :

x+\frac{7}{2} \in(0,1) \text { and }\left(\frac{x-7}{2 x-3}\right)^2 \in(0,1)

\begin{aligned}&amp; \therefore 0 &lt; x+\frac{7}{2}&lt;1 \\\\&amp; -\frac{7}{2} &lt; x &lt; \frac{-5}{2}\end{aligned}

and

\begin{aligned}&amp; \left(\frac{x-7}{2 x-3}\right)^2&lt;1 \\\\&amp; \Rightarrow (x-7)^2 &lt; (2 x-3)^2\end{aligned}

\therefore

x \in(-\infty,-4) \cup\left(\frac{10}{3}, \infty\right)

No common values of

\mathrm{x}

.

Hence intersection with feasible region.

We get

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

Integral value of

x

are

\{-2,-1,0,1,2,3\}

No. of integral values

=6

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