Question

Solution set of inequality ${\log }_x(2x^2+x-1)>{\log }_x\left(2\right)-1$ is

• A. $(\frac{1}{2},1)$
• B. $(\frac{1}{2},1)\cup (1,\infty )$
• C. $(1,\infty )$
• D. $(0,1)$

Answer 1

The correct option is B $(\frac{1}{2},1)\cup (1,\infty )$

${\log }_x(2x^2+x-1)>{\log }_x\left(2\right)-1.....\left(1\right)$
For (1) to hold, we must have
$x>0,x\ne 1;2x^2+x-1>0$
$\Rightarrow x>0,x\ne 1;(2x-1)(x+1)>0$
$\Rightarrow x>\frac{1}{2}$ only the solution.
We can write (1) as
${\log }_x\left(\frac{2x^2+x-1}{2}\right)>-1[\because \log a-\log b=\log \frac{a}{b}]......\left(2\right)$
For $\frac{1}{2}<x<1$, (2) can be written as
$\frac{2x^2+x-1}{2}<\frac{1}{x}$
$\Rightarrow 2x^3+x^2-x<2$
$\Rightarrow 2(x^3-1)+x(x-1)<0$
$\Rightarrow (x-1)(2x^2+3x+2)<0$
$\Rightarrow x<1[\because 2x^2+3x+2>0forallx>0]$
For $x>1$, (2) can be written as
$\frac{2x^2+x-1}{2}>\frac{1}{x}$
$\Rightarrow (x-1)(2x^2+3x+2)>0$
This is true for each $x>1$
Thus, (1) holds for $\frac{1}{2}<x<1,x>1$.
$\therefore x\in (\frac{1}{2},1)\cup (1,\infty )$
Ans: B