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|

• 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 $