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Functions: Domain Value

JEE Maths question with a full step-by-step solution.

Question
If the domain of f(x)=sin⁡−1(sin⁡x)−log⁡x+42log⁡2 ⁣(2x−13+x)f(x) = \dfrac{\sin^{-1}(\sin x)}{\sqrt{-\log_{\frac{x+4}{2}}\log_2\!\left(\frac{2x-1}{3+x}\right)}} is (a,b)∪(c,∞)(a,b)\cup(c,\infty), then the value of a+b+3ca+b+3c.
Solution
Answer: 5
Step 1: Conditions for the denominator to be real and positive We need −log⁡x+42log⁡2 ⁣(2x−13+x)>0-\log_{\frac{x+4}{2}}\log_2\!\left(\frac{2x-1}{3+x}\right) > 0, i.e.,:
log⁡x+42[log⁡2 ⁣(2x−13+x)]<0\log_{\frac{x+4}{2}}\left[\log_2\!\left(\frac{2x-1}{3+x}\right)\right] < 0
 Also require: base x+42>0\frac{x+4}{2} > 0, ≠1\neq 1 (so x>−4x > -4, x≠−2x \neq -2), and argument of inner log 2x−13+x>0\frac{2x-1}{3+x} > 0. Step 2: Case 1 — base in (0,1)(0,1), i.e., −4<x<−2-4 < x < -2 Need log⁡2 ⁣(2x−13+x)>1\log_2\!\left(\frac{2x-1}{3+x}\right) > 1, i.e., 2x−13+x>2\frac{2x-1}{3+x} > 2.
2x−13+x−2=−73+x>0  ⟹  3+x<0  ⟹  x<−3\frac{2x-1}{3+x}-2 = \frac{-7}{3+x} > 0 \implies 3+x < 0 \implies x < -3
 Combined with −4<x<−2-4 < x < -2: x∈(−4,−3)x \in (-4,-3). Step 3: Case 2 — base greater than 1, i.e., x>−2x > -2 Need 0<log⁡2 ⁣(2x−13+x)<10 < \log_2\!\left(\frac{2x-1}{3+x}\right) < 1, i.e., 1<2x−13+x<21 < \frac{2x-1}{3+x} < 2. Lower bound: 2x−13+x>1⇒x−43+x>0⇒x>4\dfrac{2x-1}{3+x} > 1 \Rightarrow \dfrac{x-4}{3+x} > 0 \Rightarrow x > 4 (for x>−2x > -2). Upper bound: −73+x<0\dfrac{-7}{3+x} < 0, always true for x>−3x > -3. Combined: x>4x > 4. Step 4: Combine and evaluate Domain =(−4,−3)∪(4,∞)= (-4,-3)\cup(4,\infty). So a=−4a = -4, b=−3b = -3, c=4c = 4.
a+b+3c=−4+(−3)+12=5a+b+3c = -4+(-3)+12 = 5
 Answer: 5
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