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Question: Answered & Verified by Expert
If the value of the integral I=02πsgnexdx is equal to kπ, then the smallest prime number greater than 2k is (where, sgnx represents the signum function of x)
MathematicsDefinite IntegrationJEE Main
Options:
  • A 3
  • B 5
  • C 7
  • D 11
Solution:
2269 Upvotes Verified Answer
The correct answer is: 5
As ex>0 xR
sgnex=1
I=02π1dx=2π
k=2
2k=4
Hence, the required prime number =5

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