Mathematics - Functions Question with Solution | TestHub

(A) Domain of f and g both are ' R '. $\mathrm{f}(\mathrm{x})=\cos \left(2{\tan }^{-1}\mathrm{x}\right)=\frac{1-{\tan }^2\left({\tan }^{-1}\mathrm{x}\right)}{1+{\tan }^2\left({\tan }^{-1}\mathrm{x}\right)}=\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}=\mathrm{g}(\mathrm{x}).$ (B) Domain of f and g is R. g(x) = sin(2cot⁻¹x) = 2tan(cot⁻¹x)/(1+tan²(cot⁻¹x)) = (2/x)/(1+1/x²) = 2x/(1+x²) = f(x). (C) Domain of f and g is R − {nπ/2 : n ∈ I}. f(x) = tan x + cot x = (sin²x+cos²x)/(sin x cos x)n x cos x) = 2/sin 2x = 2 cosec 2x = g(x). (D) cot⁻¹x ∈ (0,π) for all x ∈ R, so sgn(cot⁻¹x) = 1 and f(x) = e^(ln 1) = 1; also [1+{x}] = 1 > 0 for all x ∈ R, so g(x) = e^(ln 1) = 1. Hence each pair is identical.