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Question: Answered & Verified by Expert
Let f:RR be a function defined fx=2e2xe2x+e. Then f1100+f2100+f3100++f99100 is equal to ______.
MathematicsFunctionsJEE MainJEE Main 2022 (27 Jun Shift 1)
Solution:
2290 Upvotes Verified Answer
The correct answer is: 99

fx=2e2xe2x+e

f1x=2e21xe21x+e=2ee+e2x

Now, fx+f1x=2 

i.e. f1100+f99100=2

f1100+f2100+...f99100=2×49+f50100

=98+1=99

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