The domain of the function f(x)=(1/√ sin x) +√[3] sin x+ log 10 (x-5/x2-10 x+24) is

Question

The domain of the function f(x)=(1/√ sin x) +√[3] sin x+ log 10 (x-5/x2-10 x+24) is (A)  (2 k π,(2 k+1) π): k ∈ I  (B)  (2 k π,(2 k+1) π): k ∈ N  (C) (6, ∞) ∪(4,5) (D)  (2 k π,(2 k+1) π): k ∈ I ∪(6, ∞)

Tardigrade Admin · Accepted Answer

The domain of √[3] sin x is R. The domain of (1/√ sin x)= x: sin x>0 = (2 k π,(2 k+1) π): k ∈ I The domain of log 10 (x-5/x2-10 x+24)= log 10 (x-5/(x-6)(x-4)) = x: x ≠ 5, x >5, x >6, x>4 text or =(4,5) ∩(6, ∞) Thus the domain of f(x) = (2 k π,(2 k+1) π): k ∈ I ∩ (4,5) ∪(6, ∞) = (2 k π,(2 k+1) π): k ∈ N (since (4,5) does not intersect (2 k π, 2 k+1) π: k ∈ I )

Q. The domain of the function $f (x) = \frac{1}{s i n x}$ $+ 3 sin x + lo g_{10} \frac{x - 5}{x ^{2} - 10 x + 24}$ is

${(2 kπ, (2 k + 1) π) : k \in I}$

${(2 kπ, (2 k + 1) π) : k \in N}$

$(6, \infty) \cup (4, 5)$

${(2 kπ, (2 k + 1) π) : k \in I} \cup (6, \infty)$

Q. The domain of the function f(x)=sinx​1​ +3sinx​+log10​x2−10x+24x−5​ is

{(2kπ,(2k+1)π):k∈I}

{(2kπ,(2k+1)π):k∈N}

(6,∞)∪(4,5)

{(2kπ,(2k+1)π):k∈I}∪(6,∞)

Q. The domain of the function $f (x) = \frac{1}{s i n x}$ $+ 3 sin x + lo g_{10} \frac{x - 5}{x ^{2} - 10 x + 24}$ is

${(2 kπ, (2 k + 1) π) : k \in I}$

${(2 kπ, (2 k + 1) π) : k \in N}$

$(6, \infty) \cup (4, 5)$

${(2 kπ, (2 k + 1) π) : k \in I} \cup (6, \infty)$