Question

The equation $2({\log }_{3}x{)}^2-|{\log }_{3}x|+a=0,a\in R$ posses four solutions for all

• A. $a<\frac{1}{8}$
• B. $a<0$
• C. $a\in (0,\frac{1}{8})$
• D. $a\in (-1,0)$

Answer 1

Answer: (C)

$a\in (0,\frac{1}{8})$

Case $I:$ If ${\log }_{3}x\ge 0\Rightarrow x\ge 1$

From given equation

$2\left({\log }_{3}x\right)-{\log }_{3}x+a=0$

For real solution

$(-1{)}^2-\mathrm{4.2.}a>0$

$a<1/8....\left(i\right)$

Case $II:$ If ${\log }_{3}x<0$

$\therefore x<1$

From given equation

$2({\log }_{3}x{)}^2+{\log }_{3}c+a=0$

For real solution

$(1{)}^2-\mathrm{4.2.}a>0$

$\therefore a<1/\mathrm{8...}\left(ii\right)$

From equation

${\log }_{3}x=\frac{-1\pm \sqrt{1-8a}}{4}[have,{\log }_{3}a<0]$

$\Rightarrow \frac{-1\pm \sqrt{1-8a}}{4}<0\Rightarrow -1\pm \sqrt{1-8a}<0$

$\Rightarrow \sqrt{1-8a}<1$ [taking the sign]

$a>\mathrm{0....}\left(iii\right)$

Hence from equation $\left(i\right),\left(ii\right),\left(iii\right)$ we get $0<a<1/8$