The correct option is C (−2,−√2)∪(√2,2)
log5(x2−2)<log5(32|x|−1) is defined if
x2−2>0 and 32|x|−1>0
⇒x∈(−∞,−√2)∪(√2,∞) ⋯(1)
and x∈(−∞,−23)∪(23,∞) ⋯(2)
Now, log5(x2−2)<log5(32|x|−1)
⇒x2−2<32|x|−1⇒x2−32|x|−1<0⇒2x2−3|x|−2<0
⇒2|x|2−3|x|−2<0 (∵|x|2=x2, ∀ x∈R)
⇒(2|x|+1)(|x|−2)<0
⇒−12<|x|<2
But |x|≥0
So, 0≤|x|<2
⇒x∈(−2,2) ⋯(3)
From (1),(2) and (3), we have
x∈(−2,−√2)∪(√2,2)