Question

Find the domain of the function f(x)=log((3tan^(-1)x+pi)/(pi-4tan^(-1)x)).​

Answer 1

EXPLANATION.

Domain of the function.

⇒ f(x) = ㏒(3tan⁻¹x + π/ π - 4tan⁻¹x).

As we know that,

Equation must be > 0.

⇒ (3tan⁻¹x + π/ π - 4tan⁻¹x) > 0.

⇒ (3tan⁻¹x + π/ 4tan⁻¹x - π) < 0.

Find zeroes of the equation.

⇒ 3tan⁻¹x +  π = 0.

⇒ 3tan⁻¹x = - π.

⇒ tan⁻¹x = - π/3. = -√3.

⇒ 4tan⁻¹x -  π = 0.

⇒ 4tan⁻¹x =  π.

⇒ tan⁻¹x =  π/4. = 1.

Put this point on wavy curve method, we get.

⇒ x ∈ (-√3,1).

Domain of the function = (-√3,1).

                                                                                                                       

MORE INFORMATION.

Algebra of the function.

(1) = (fog) (x) = f[g(x)].

(2) = (fof) (x) = f[f(x)].

(3) = (gog) (x) = g[g(x)].

(4) = (fg) (x) = f(x).g(x).

(5) = (f ± g) (x) = f(x) ± g(x).

(6) = (f/g) (x) = f(x)/g(x), g(x) ≠ 0.

Answer 2

$\huge \mathbb\fcolorbox{black}{lavenderblush}{Solution : ♡}$

EXPLANATION.

• Domain of the function.

⇒ f(x) = ㏒(3tan⁻¹x + π/ π - 4tan⁻¹x).

As we know that,

Equation must be > 0.

⇒ (3tan⁻¹x + π/ π - 4tan⁻¹x) > 0.

⇒ (3tan⁻¹x + π/ 4tan⁻¹x - π) < 0.

• Find zeroes of the equation.

⇒ 3tan⁻¹x + π = 0.

⇒ 3tan⁻¹x = - π.

⇒ tan⁻¹x = - π/3. = -√3.

⇒ 4tan⁻¹x - π = 0.

⇒ 4tan⁻¹x = π.

⇒ tan⁻¹x = π/4. = 1.

• Put this point on wavy curve method, we get.

⇒ x ∈ (-√3,1).

Domain of the function = (-√3,1).

MORE INFORMATION.

Algebra of the function.

(1) = (fog) (x) = f[g(x)].

(2) = (fof) (x) = f[f(x)].

(3) = (gog) (x) = g[g(x)].

(4) = (fg) (x) = f(x).g(x).

(5) = (f ± g) (x) = f(x) ± g(x).

(6) = (f/g) (x) = f(x)/g(x), g(x) ≠ 0.