JEE Main 2025 — 29 January, Evening Shift — Question 45

$\mathrm{f}_{1}(\mathrm{x})=\log _{5}\left(18 \mathrm{x}-\mathrm{x}^{2}-77\right)$

$\therefore 18 \mathrm{x}-\mathrm{x}^{2}-77>0$

$x^{2}-18 x+77<0$

$x \in(7,11), \alpha=7, \beta=11$

$f_{2}(x)=\log _{(x-1)}\left(\frac{2 x^{2}+3 x-2}{x^{2}-3 x-4}\right)$

$\therefore \quad \mathrm{x}-1>0, \mathrm{x}-1 \neq 1, \frac{2 \mathrm{x}^{2}+3 \mathrm{x}-2}{\mathrm{x}^{2}-3 \mathrm{x}-4}>0$

$x>1, x \neq 2, \frac{(2 x-1)(x+2)}{(x-4)(x+1)}>0$

$x>1, x \neq 2$,

$\therefore \mathrm{x} \in(4, \infty)$

$\therefore \gamma=4$

$\therefore \alpha^{2}+\beta^{2}+\gamma^{2}=49+121+16$

$=186$