The number of integral solutions x of log _{l | vedclass

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Question

(A) The given inequality is $\log _{x+\frac{7}{2}}\left(\frac{x-7}{2 x-3}\right)^2 \geq 0$.Feasible region:$1) \ x+\frac{7}{2} > 0 \Rightarrow x > -\f

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The given inequality is $\log _{x+\frac{7}{2}}\left(\frac{x-7}{2 x-3}ight)^2 \geq 0$.Feasible region:$1) \ x+\frac{7}{2} > 0 \Rightarrow x > -\frac{7}{2}$$2) \ x+\frac{7}{2} 
eq 1 \Rightarrow x 
eq -\frac{5}{2}$$3) \ \frac{x-7}{2x-3} 
eq 0 \Rightarrow x 
eq 7$$4) \ 2x-3 
eq 0 \Rightarrow x 
eq \frac{3}{2}$Intersection: $x \in \left(-\frac{7}{2}, \inftyight) \setminus \left\{-\frac{5}{2}, \frac{3}{2}, 7ight\}$.Case $I$: $x+\frac{7}{2} > 1$ and $\left(\frac{x-7}{2x-3}ight)^2 \geq 1$$x > -\frac{5}{2}$ and $(2x-3)^2 - (x-7)^2 \leq 0$$(2x-3-x+7)(2x-3+x-7) \leq 0$$(x+4)(3x-10) \leq 0 \Rightarrow x \in [-4, \frac{10}{3}]$Intersection with $x > -\frac{5}{2}$ gives $x \in \left(-\frac{5}{2}, \frac{10}{3}ight]$.Case $II$: $0 $-\frac{7}{2} $(2x-3-x+7)(2x-3+x-7) > 0$$(x+4)(3x-10) > 0 \Rightarrow x \in (-\infty, -4) \cup (\frac{10}{3}, \infty)$Intersection with $(-\frac{7}{2}, -\frac{5}{2})$ is empty.Combining results,$x \in \left(-\frac{5}{2}, \frac{10}{3}ight] \setminus \left\{\frac{3}{2}ight\}$.Integral values are $\{-2, -1, 0, 1, 2, 3\}$.Total number of integral solutions is $6$.

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

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