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KVPY2020MathematicsApplication of Derivatives

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

Options

  1. Ae 2 + ln 2 + 1
  2. Be 2 + 2 ln 2
  3. Ce π / 2 + π 2 ln π 2
  4. De 3 / 2 + 3 2 ln 3 2

Correct answer

B. e 2 + 2 ln 2

Step-by-step solution

f ' ( x ) = e x + 1 + ln x > 0 ( as   x ∈ [ 1 , 2 ] ) ⇒ f ( x ) increases in [ 1 , 2 ] ⇒ f max = f ( 2 ) = e 2 + 2 ℓ n 2

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