Number of integers satisfying the inequality, x4−29x2+100≤0 is
2
4
6
8
We have x4−29x2+100≤0
⇒ x2−4x2−25≤0⇒ 4≤x2≤25⇒ x∈[−5,−2]∪[2,5]
Integers satisfying above are ±5,±4,±3,±2
So, there are eight integers.