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Question: Answered & Verified by Expert
If limx01+px+qx2cosec x=2048, then the value of p11 is equal to (take ln2=0.69)
MathematicsLimitsJEE Main
Solution:
1366 Upvotes Verified Answer
The correct answer is: 0.69

The given limit is of the form 1

elimx0cosec x1+px+qx2-1=2048

limx0px+qx2sinx=ln2048

limx0px+qx2xxsinx=11ln2

limx0p+qx1=11ln2p=11ln2

Hence, p11=ln2=0.69

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