Question

The number of integral values of $x$ satisfying the inequality ${\left(\frac{3}{4}\right)}^{6x+10-x^2}<\frac{27}{64}$ is

Answer 1

${\left(\frac{3}{4}\right)}^{6x+10-x^2}<\frac{27}{64}$
$\Rightarrow {\left(\frac{3}{4}\right)}^{6x+10-x^2}<{\left(\frac{3}{4}\right)}^3$
$\Rightarrow 6x+10-x^2>3(\because \frac{3}{4}<1)$
$\Rightarrow x^2-6x-7<0$
$\Rightarrow (x+1)(x-7)<0$
$\Rightarrow x\in (-1,7)$
Hence, $x=0,1,2,3,4,5,6$