Question

The solution set of the inequality $\frac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\le 0$ is

• A. $(-\infty ,0]\cup (4,\infty )$
• B. $(-2,0]\cup (4,\infty )$
• C. $(-1,0]\cup (4,\infty )$
• D. $(-2,-1)\cup (-1,0]\cup (4,\infty )$

Answer 1

The correct option is D $(-2,-1)\cup (-1,0]\cup (4,\infty )$

$\frac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\le 0\Rightarrow \frac{4^x({\left(\frac{3}{4}\right)}^x-1)\cdot \ln (x+2)}{(x-4)(x+1)}\le 0$

For $\log$ to be defined,
$x+2>0\Rightarrow x>-2\cdots \left(1\right)$

The critical points are $x=-1,0,4$


Hence, by wavy curve method,
$x\in (-2,-1)\cup (-1,0]\cup (4,\infty )$