Complete set of range of the function f(x)=1x6+|x|3−1 is equal to
−1,0
−∞,−1∪0,∞
−∞,−2∪0,∞
−∞,0∪1,∞
Given that f(x)=1x6+|x|3−1 Clearly x6+|x|3≥0⇒x6+|x|3−1≥−1 [ If x>a(a<0) then 1x∈(−∞,1a)∪(0,∞) So range of f(x)=(−∞,−1]∪(0,∞) .
Therefore, the correct answer is (2).