Question

If $x=\frac{4}{9}$ satisfy the equation $\log a(x^2-x+2)>\log a(-x^2+2x+3)$ hen sum of all possible distinct value of [x] is where [.] represents the greatest integer function

Answer 1

$\log a(x^2-x+2)>\log a(-x^2+2x+3)$

$\Rightarrow \log a(x^2-x+2)-\log a(-x^2+2x+3)>0$

$\Rightarrow \log \frac{(x^2-x+2)}{(-x^2+2x+3)}>0$

$\Rightarrow \left(\frac{x^2-x+2}{-x^2+2x+3}\right)>1$ or $<1$ (depending on base of $\log$)

$\Rightarrow 2x^2-3x-1>0$ or $<0$

Checking it by the given $x=\frac{4}{9}$

$\Rightarrow 2\left(\frac{16}{81}\right)-3\left(\frac{4}{9}\right)-1=\frac{28-108-81}{81}<0$

So, $2x^2-3x-1<0$

$\Rightarrow (x-\frac{1}{2})(x-1)<0$

$\Rightarrow x\in [\frac{1}{2},1]$

$\left[x\right]\in \varphi$

The answer is $0$.