Basic Mathematics & Logarithm Question - None

Question

If the function $g: (-1, 1) \to \left(-\frac{\pi}{2}, \frac{7\pi}{56}\right]$ is given by $g(u) = 2\tan^{-1}(e^u) - \frac{\pi}{2}$, then $g$ is

Accepted Answer

Step 1: Check if g is even or odd

We have g(u) = 2tan⁻¹(eᵘ) - π/2

Compute g(-u):

g(-u) = 2tan⁻¹(e⁻ᵘ) - π/2

Step 2: Use the inverse tangent identity

For x > 0: tan⁻¹(x) + tan⁻¹(1/x) = π/2

Therefore: tan⁻¹(eᵘ) + tan⁻¹(e⁻ᵘ) = π/2

This gives: tan⁻¹(e⁻ᵘ) = π/2 - tan⁻¹(eᵘ)

Step 3: Substitute into g(-u)

g(-u) = 2[π/2 - tan⁻¹(eᵘ)] - π/2

g(-u) = π - 2tan⁻¹(eᵘ) - π/2

g(-u) = π/2 - 2tan⁻¹(eᵘ)

g(-u) = -(2tan⁻¹(eᵘ) - π/2)

g(-u) = -g(u)

Therefore, g is an odd function.

Step 4: Check monotonicity by computing g'(u)

g'(u) = d/du[2tan⁻¹(eᵘ) - π/2]

g'(u) = 2 · 1/(1+(eᵘ)²) · eᵘ

g'(u) = 2eᵘ/(1 + e²ᵘ)

Step 5: Analyze the sign of g'(u)

For all u ∈ (-1, 1):

• eᵘ > 0 (exponential is always positive)

• 1 + e²ᵘ > 0 (sum of positive terms)

Therefore, g'(u) > 0 for all u ∈ (-1, 1)

Thus g is strictly increasing on (-1, 1).

∴ Answer: C (g is odd and strictly increasing in (-1, 1))

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