If the domain of the function $\log _{5}\left(18 \mathrm{x}-\mathrm{x}^{2}-77\right)$ is $(\alpha, \beta)$ and the domain of the function $\log _{(x-1)}\left(\frac{2 x^{2}+3 x-2}{x^{2}-3 x-4}\right)$ is $(\gamma, \delta)$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is equal to :

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Sets Relations and Functions

If the domain of the function

lo g_{5} (18 x - x^{2} - 77)

(α, β)

and the domain of the function

lo g_{(x - 1)} (\frac{2 x ^{2} + 3 x - 2}{x ^{2} - 3 x - 4})

(γ, δ)

, then

α^{2} + β^{2} + γ^{2}

is equal to :

Answer: (C)

lo g_{5} (18 x - x^{2} - 77)

needs

18 x - x^{2} - 77 > 0 \Rightarrow x^{2} - 18 x + 77 < 0 \Rightarrow (x - 7) (x - 11) < 0 \Rightarrow x \in (7, 11)

, so

α = 7, β = 11

.

For

lo g_{(x - 1)} (\frac{2 x ^{2} + 3 x - 2}{x ^{2} - 3 x - 4})

, the base requires

x > 1, x \neq = 2

, and the argument, factored as

\frac{( 2 x - 1 ) ( x + 2 )}{( x - 4 ) ( x + 1 )}

must be positive. A sign chart over the critical points

- 2, - 1, \frac{1}{2}, 4

shows this holds on

(- \infty, - 2) \cup (- 1, \frac{1}{2}) \cup (4, \infty)

; intersecting with

x > 1, x \neq = 2

leaves only

(4, \infty)

, so

γ = 4

α^{2} + β^{2} + γ^{2} = 49 + 121 + 16 = 186

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option