Q35 / 75 · JEE Main 2025 (4 Apr Shift 2) · Mathematics
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 $g(x)=\sin ^{-1}\left(\frac{7 x+10}{x-2}\right)$ be $(\alpha, \beta)$ and $[\gamma, \delta]$, respectively. Then $\alpha^2+\beta^2+\gamma^2+\delta^2$ is equal to :-
$\mathrm{f}(\mathrm{x})=\log _4 \log _3 \log _7\left(8-\log _2\left(\mathrm{x}^2+4 \mathrm{x}+5\right)\right)$ and $g(x)=\sin ^{-1}\left(\frac{7 x+10}{x-2}\right)$ be $(\alpha, \beta)$ and $[\gamma, \delta]$, respectively. Then $\alpha^2+\beta^2+\gamma^2+\delta^2$ is equal to :-
- Correct answerA$15$
- B$13$
- C$16$
- D$14$
Solution
$\begin{aligned} & \log _3\left(\log _7\left(8-\log _2\left(\mathrm{x}^2+4 \mathrm{x}+5\right)\right)\right) \gt 0 \\ & \log _2\left(\mathrm{x}^2+4 \mathrm{x}+5\right) \lt 1 \\ & \mathrm{x}^2+4 \mathrm{x}+3 \lt 0 \\ & \Rightarrow \mathrm{x} \in(-3,-1) \\ & -1 \leq \frac{7 \mathrm{x}+10}{\mathrm{x}-2} \leq 1 \\ & \Rightarrow \mathrm{x} \in[-2,-1] \\ & \alpha=-3, \beta=-1, \gamma=-2, \delta=-1 \\ & \alpha^2+\beta^2+\gamma^2+\delta^2=15 \\ & \text { option }(1)\end{aligned}$
Hence, the correct answer is option: 'A'.