Question

If ${\log }_{2}4,{\log }_{\sqrt{2}}8$, and ${\log }_3{9}^{k-1}$ are consecutive terms of a geometric sequence, then the number of integers that satisfy the system of inequalities $x^2-x>6$ and $|x|<k^2$ is

• A. 193
• B. 194
• C. 195
• D. 196

Answer 1

Answer: (A)

$2,6,2(k-1)\text{ are in G.P. }\Rightarrow k=10.$
$\text{ now }{x}^2-x-6>0\Rightarrow (x-3)(x+2)>0$ ......(1)
$\text{ and }|x|<100\Rightarrow -100<x<100$ ....(2)
$\text{ from (1) and (2) }\Rightarrow x\in (-100,-2)\cup (3,100)$
$\therefore \text{ number of integers }$
$-99\text{ to }-3\text{ and }4\text{ to }99$
$\text{ i.e. }97+96=193$