NTA Abhyas JEE Main2020MathematicsApplication of DerivativesPractice

The range of the function f x = x 2 ln ⁡ x for x ∈ 1 , e is a , b , where a + b is equal to

Options

Ae 2
Be 2 + 1
Ce + 1
D2 e 2

Correct answer

A. e 2

Step-by-step solution

f ′ x = 2 x ⋅ ln x + x 2 1 x = x 2 ln ⁡ x + 1 So, f x is increasing in 1 , e ∴ Minimum value of f x = f 1 = 0 And maximum value of f x = f e = e 2 ∴ a , b is 0 , e 2 ⇒ a + b = e 2

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