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B
2
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C
4
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D
8
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Solution
The correct option is A0 Let I=4∫−4x20211+x2022dx
We know that, a∫−af(x)dx=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩2a∫0f(x)dx,if f(x) is even0,if f(x) is odd ⇒I=0 (∵x20211+x2022is an odd function.)