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Number of integers satisfying the inequality, x429x2+1000 is

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By Expert Faculty of Sri Chaitanya
a
2
b
4
c
6
d
8
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detailed solution

Correct option is D

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,±2So, there are eight integers.


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