Let the domains of the functions \(\mathrm{f}(\mathrm{x})=\log _{4} \log _{3} \log _{7}\left(8-\log _{2}\left(\mathrm{x}^{2}+4 \mathrm{x}+5\right)\right)\) and \(\mathrm{g}(\mathrm{x})=\sin ^{-1}\left(\frac{7 \mathrm{x}+10}{\mathrm{x}-2}\right)\) be \((\alpha, \beta)\) and \([\gamma, \delta]\), respectively. Then \(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\) is equal to :-

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Sets Relations and Functions

Let the domains of the functions

f (x) = lo g_{4} lo g_{3} lo g_{7} (8 - lo g_{2} (x^{2} + 4 x + 5))

and

g (x) = sin^{- 1} (\frac{7 x + 10}{x - 2})

(α, β)

and

[γ, δ]

, respectively. Then

α^{2} + β^{2} + γ^{2} + δ^{2}

is equal to :-

Answer: (A)

For

f

, the outer logarithm forces

lo g_{3} lo g_{7} (8 - lo g_{2} (x^{2} + 4 x + 5)) > 0

, which unwinds to

8 - lo g_{2} (x^{2} + 4 x + 5) > 7

, i.e.

x^{2} + 4 x + 5 < 2

. Thus

(x + 1) (x + 3) < 0

and the domain is

(- 3, - 1)

, giving

α = - 3, β = - 1

. For

g

, the inverse sine requires

- 1 \leq \frac{7 x + 10}{x - 2} \leq 1

; the two inequalities give

- 2 \leq x < 2

and

x \leq - 1

, whose intersection is

[- 2, - 1]

, so

γ = - 2, δ = - 1

. Therefore

α^{2} + β^{2} + γ^{2} + δ^{2} = 9 + 1 + 4 + 1 = 15

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option