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Question: Answered & Verified by Expert
The range of the function fx=x2lnx for x1,e is a,b, where a+b is equal to
MathematicsApplication of DerivativesJEE Main
Options:
  • A e2
  • B e2+1
  • C e+1
  • D 2e2
Solution:
1719 Upvotes Verified Answer
The correct answer is: e2
f x =2xlnx+ x 2 1 x
=x2lnx+1
So, fx is increasing in 1,e
Minimum value of fx=f1=0
And maximum value of fx=fe=e2
a,b is 0,e2
a+b=e2

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