Question

Number of integers satisfying the inequalities $\sqrt{{\log }_{3}x-1}+\frac{\frac{1}{2}{\log }_3{x}^3}{{\log }_3\frac{1}{3}}+2>0$, is

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

Answer: (A)

$5$

Given $\sqrt{{\log }_{3}x-1}+\frac{\frac{1}{2}{\log }_3{x}^3}{{\log }_3\frac{1}{3}}+2>0$

$⟹\sqrt{{\log }_{3}x-1}+\frac{\frac{3}{2}{\log }_{3}x}{{\log }_{3}1-{\log }_{3}3}+2>0$

$⟹\sqrt{{\log }_{3}x-1}+\frac{\frac{3}{2}{\log }_{3}x}{0-1}+2>0$

$⟹\sqrt{{\log }_{3}x-1}-\frac{3}{2}{\log }_{3}x+2>0$

$⟹\sqrt{{\log }_{3}x-1}>\frac{3}{2}{\log }_{3}x-2$

Squaring on both sides

$⟹{\log }_{3}x-1>{(\frac{3}{2}{\log }_{3}x-2)}^2$

Let ${\log }_{3}x=t$

$⟹t-1>{(\frac{3}{2}t-2)}^2$

$⟹t-1>\frac{9}{4}{t}^2+4-2\left(\frac{3}{2}\right)2t$

$⟹t-1>\frac{9}{4}{t}^2+4-6t$

$⟹\frac{9}{4}{t}^2+4-6t-t+1<0$

$⟹\frac{9}{4}{t}^2-7t+5<0$

Roots of the quadratic equation $a{x}^2+bx+c=0$ is given by $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Therefore, roots are $\frac{-(-7)\pm \sqrt{7^2-4\left(\frac{9}{4}\right)5}}{2\left(\frac{9}{4}\right)}$

$⟹\frac{7\pm \sqrt{49-45}}{\left(\frac{9}{2}\right)}$

$⟹\frac{7\pm 2}{\left(\frac{9}{2}\right)}$

$⟹\frac{7+2}{\left(\frac{9}{2}\right)}$ and $\frac{7-2}{\left(\frac{9}{2}\right)}$

$⟹2$ and $\frac{10}{9}$ are the roots

therefore $\frac{9}{4}{t}^2-7t+5<0⟹(t-2)(t-\frac{10}{9})<0$

$⟹(t-2)<0$ and $(t-\frac{10}{9})>0$ or

$(t-2)>0$ and $(t-\frac{10}{9})<0$ doesn't have a solution

therefore $t<2$ and $t>\frac{10}{9}$

$⟹{\log }_{3}x<2$ and ${\log }_{3}x>\frac{10}{9}$

$⟹x<3^2$ and $x>3^{\frac{10}{9}}$

$⟹x<9$ and $x>3.39$

$⟹x\in (3.39,9)$

$\{4,5,6,7,8\}$ are integers which belong to the set $(3.39,9)$

Therefore the total number of integers $=5$