JEE MAIN - Mathematics (2025 - 3rd April Evening Shift - No. 17)
If the domain of the function $f(x)=\log _7\left(1-\log _4\left(x^2-9 x+18\right)\right)$ is $(\alpha, \beta) \cup(\gamma, o)$, then $\alpha+\beta+\gamma+\hat{o}$ is equal to
17
15
16
18
Explanation
$$\begin{aligned} & 1-\log _4\left(x^2-9 x+18\right)>0 \\ & \log _4\left(x^2-9 x+18\right)<1 \\ & x^2-9 x+18<4 \\ & x^2-9 x+14<0 \\ & x \in(2,7) \\ & x^2-9 x+18>0 \\ & x \in(-\infty, 3) \cup(6, \infty) \end{aligned}$$
_3rd_April_Evening_Shift_en_17_1.png)
$$\begin{aligned} & x \in(2,3) \cup(6,7) \\ & \alpha+\beta+\gamma+\delta=18 \end{aligned}$$
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