Q47 / 91 · KVPY 2020 (SB/SX) · Mathematics

The maximum value of the function f(x)=ex+xlnx on the interval 1≤x≤2 is

  • A
    e2+ln2+1
  • Correct answer
    B
    e2+2ln2
  • C
    eπ/2+π2lnπ2
  • D
    e3/2+32ln32

Solution

f'(x)=ex+1+lnx>0

(as x∈[1,2])

⇒f(x) increases in [1,2]
⇒fmax=f(2)=e2+2ℓn2

Hence, the correct answer is option: 'B'.

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