Question

If the inequality ${\log }_a(x^2-x-2)>{\log }_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$, then find the interval in which the inequality is always valid.

• A. $(0,\frac{5}{2})$
• B. $(2,\frac{5}{2})$
  
• C. $(-2,\frac{5}{2})$
  
• D. $(-\frac{5}{2},2)$
  

Answer 1

The correct option is B $(2,\frac{5}{2})$

Log is defined when
$(x^2-x-2)>0\Rightarrow (x-2)(x+1)>0$
$\Rightarrow x\in (-\infty ,-1)\cup (2,\infty )$ ...(1)
And
$-x^2+2x+3>0\Rightarrow x^2-2x-3<0\Rightarrow x^2-3x+x-3<0\Rightarrow x(x-3)+(x-3)<0$
$\Rightarrow (x+1)(x-3)<0\Rightarrow x\in (-1,3)$ ...(2)

Now since given $x=\frac{9}{4}$

Substituting this in given inequality, we get

${\log }_a\left(\frac{13}{16}\right)>{\log }_a\left(\frac{39}{64}\right)$

which is possible when $0<a<1$

Given ${\log }_a(x^2-x-2)>{\log }_a({-x}^2+2x+3)$

$\Rightarrow x^2-x-2<{-x}^2+2x+3$

$\Rightarrow 2x^2-3x-5<0$

$\Rightarrow 2x^2-5x+2x-5<0$

$\Rightarrow x(2x-5)+(2x-5)<0$
$\Rightarrow (x+1)(2x-5)<0$

$\Rightarrow x\in (-1,\frac{5}{2})$
Therefore from this and (1), (2), we get
$x\in (2,\frac{5}{2})$