Let f(x) = 2x + tan⁻¹x and g(x) = logₑ(√(1 + x²) + x), x ∈ [0, 3]. Then (1) min f(x) = 1 + max g(x) (2) there exist 0

Question

Let f(x) = 2x + tan⁻¹x and g(x) = logₑ(√(1 + x²) + x), x ∈ [0, 3]. Then (1) min f(x) = 1 + max g(x) (2) there exist 0 < x₁ < x₂ < 3 such that f(x₁) < g(x₁), ∀ x ∈ (x₁, x₂) (3) there exists x̂ ∈ [0, 3] such that f′(x̂) < g′(x̂) (4) max f(x) > max g(x)

Accepted Answer

(4). The correct option is (4) because the function f(x) = 2x + tan⁻¹x is a linear function with a positive slope, which increases without bound as x increases, while g(x) = logₑ(√(1 + x²) + x) grows slower than f(x) in the interval [0, 3]. Therefore, the maximum value of f(x) will exceed the maximum value of g(x) in this range.

Let f(x) = 2x + tan⁻¹x and g(x) = logₑ(√(1 + x²) + x), x ∈ [0, 3]. Then (1) min f(x) = 1 + max g(x) (2) there exist 0 < x₁ < x₂ < 3 such that f(x₁) < g(x₁), ∀ x ∈ (x₁, x₂) (3) there exists x̂ ∈ [0, 3] such that f′(x̂) < g′(x̂) (4) max f(x) > max g(x)

Solution

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