Question

$\displaystyle \text{Give answer if you know only}\\\\\\\displaystyle \text{Number of integers satisfying the inequalities }\\\\\\\displaystyle{\sqrt{\log_3x-1}+\left(\dfrac{\dfrac{1}{2}\log_3x^3 }{\log_3\dfrac{1}{3}}\right)+2 \ \ \textgreater \ 0 \ is}\\\\\\\displaystyle \text{Options are :}\\\\\\\displaystyle \text{a. \ 5}\\\\\\\displaystyle \text{b. \ 6}\\\\\\\displaystyle \text{ c. \ 7}\\\\\\\displaystyle \text{d. \ 8}\\\\\\\displaystyle \text{Kindly give good and contain quality answer with great explanation.}$

Answer 1

$\begin{aligned}&\sqrt{\log_3(x)-1}+\left(\dfrac{\dfrac{1}{2}\log_3(x^3)}{\log\left(\dfrac{1}{3}\right)}\right)+2>0\\ \\ \Longrightarrow\ \ &\sqrt{\log_3(x)-1}+\left(\dfrac{\dfrac{1}{2}\cdot3\log_3(x)}{\log\left(3^{^{-1}}\right)}\right)+2>0\\ \\ \Longrightarrow\ \ &\sqrt{\log_3(x)-1}+\left(\dfrac{\dfrac{3}{2}\log_3(x)}{-1}\right)+2>0\\ \\ \Longrightarrow\ \ &\sqrt{\log_3(x)-1}-\dfrac{3}{2}\log_3(x)+2>0\end{aligned}$

$\textsf{Let}\ \ \log_3(x)=k.\\ \\ \\ \begin{aligned}\Longrightarrow\ \ &\sqrt{k-1}-\dfrac{3}{2}k+2>0\\ \\ \Longrightarrow\ \ &\sqrt{k-1}>\dfrac{3}{2}k-2\\ \\ \Longrightarrow\ \ &k-1>\left(\dfrac{3}{2}k-2\right)^2\\ \\ \Longrightarrow\ \ &k-1>\dfrac{9}{4}k^2-6k+4\\ \\ \Longrightarrow\ \ &\dfrac{9}{4}k^2-6k+4-k+1<0\\ \\ \Longrightarrow\ \ &\dfrac{9}{4}k^2-7k+5<0\\ \\ \Longrightarrow\ \ &9k^2-28k+20<0\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &9k^2-18k-10k+20<0\\ \\ \Longrightarrow\ \ &9k(k-2)-10(k-2)<0\\ \\ \Longrightarrow\ \ &(k-2)(9k-10)<0\end{aligned}$

$\textsf{Since LHS <0 ,}\\ \\ \\ \Longrightarrow\ \ \dfrac{10}{9}<k<2\ \implies\ \dfrac{10}{9}<\log_3(x)<2\\ \\ \\ \Longrightarrow\ \ 3^{^{\displaystyle\small\frac{10}{9}}}<3^{^{\displaystyle\small\log_3(x)}}<3^{^{\displaystyle\small\text{2}}}\ \implies\ 3.389<x<9\\ \\ \\ \textsf{Since \ x\in\mathbb{Z} , \ we get \ x\in\{4,\ 5,\ 6,\ 7,\ 8\} .}\\ \\ \\ \textsf{Hence the answer is}\ \ \textbf{a.\ 5}$

Answer 2

Answer:

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