FunctionshardFree

Functions: Domain Value

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

Question

If the domain of

f(x) = \dfrac{\sin^{-1}(\sin x)}{\sqrt{-\log_{\frac{x+4}{2}}\log_2\!\left(\frac{2x-1}{3+x}\right)}}

(a,b)\cup(c,\infty)

, then the value of

a+b+3c

Solution

Answer: 5

Step 1: Conditions for the denominator to be real and positive We need

-\log_{\frac{x+4}{2}}\log_2\!\left(\frac{2x-1}{3+x}\right) > 0

, i.e.,:

\log_{\frac{x+4}{2}}\left[\log_2\!\left(\frac{2x-1}{3+x}\right)\right] < 0

Also require: base

\frac{x+4}{2} > 0

\neq 1

(so

x > -4

x \neq -2

), and argument of inner log

\frac{2x-1}{3+x} > 0

. Step 2: Case 1 — base in

(0,1)

, i.e.,

-4 < x < -2

Need

\log_2\!\left(\frac{2x-1}{3+x}\right) > 1

, i.e.,

\frac{2x-1}{3+x} > 2

\frac{2x-1}{3+x}-2 = \frac{-7}{3+x} > 0 \implies 3+x < 0 \implies x < -3

Combined with

-4 < x < -2

x \in (-4,-3)

. Step 3: Case 2 — base greater than 1, i.e.,

x > -2

Need

0 < \log_2\!\left(\frac{2x-1}{3+x}\right) < 1

, i.e.,

1 < \frac{2x-1}{3+x} < 2

. Lower bound:

\dfrac{2x-1}{3+x} > 1 \Rightarrow \dfrac{x-4}{3+x} > 0 \Rightarrow x > 4

(for

x > -2

). Upper bound:

\dfrac{-7}{3+x} < 0

, always true for

x > -3

. Combined:

x > 4

. Step 4: Combine and evaluate Domain

= (-4,-3)\cup(4,\infty)

. So

a = -4

b = -3

c = 4

a+b+3c = -4+(-3)+12 = 5

Answer: 5

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