Concept:Even and odd functions: f is even if f(-x)=f(x); f is odd if f(-x)=-f(x). A function is strictly increasing if its derivative is positive for all x. The inverse tangent function, tan⁻¹, is odd and increasing on its domain.
Explanation:Given f(x) = 2(tan⁻¹(eˣ) – π/4).
First check parity:
f(-x) = 2(tan⁻¹(e⁻ˣ) – π/4).
Since eˣ > 0 for all real x, use the identity tan⁻¹(e⁻ˣ) + tan⁻¹(eˣ) = π/2.
Thus tan⁻¹(e⁻ˣ) = π/2 – tan⁻¹(eˣ).
Substituting:
f(-x) = 2(π/2 – tan⁻¹(eˣ) – π/4) = 2(π/4 – tan⁻¹(eˣ)) = π/2 – 2tan⁻¹(eˣ).
Now compute –f(x):
–f(x) = –2(tan⁻¹(eˣ) – π/4) = 2(π/4 – tan⁻¹(eˣ)) = π/2 – 2tan⁻¹(eˣ).
So f(-x) = –f(x), therefore f is an odd function.
Now check monotonicity by differentiating:
f'(x) = 2 × [d/dx tan⁻¹(eˣ)] = 2 × [1/(1+e²ˣ) × eˣ] = 2eˣ/(1+e²ˣ).
For all real x, eˣ > 0 and 1+e²ˣ > 0, so f'(x) > 0. Hence f is strictly increasing on (–∞, ∞).
Answer:Option C: odd and is strictly increasing in (–∞, ∞).