Exam

JEE Main

Year

2025

Date

29 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Functions

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Functions

If the domain of the function _5(18 x-x^2-77) is (α, β) and the...

If the domain of the function $\log _5\left(18 x-x^2-77\right)$ 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 :

Step-by-step solution

$$\begin{aligned} & f_1(x)=\log _5\left(18 x-x^2-77\right) \\ & \therefore \quad 18 x-x^2-77>0 \\ & \quad x^2-18 x+77 < 0 \\ & x \in(7,11) \\ & \alpha=7, \beta=11 \\ & f_2(x)=\log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right) \\ & x>1, x-1 \neq 1, \frac{2 x^2+3 x-2}{x^2-3 x-4}>0 \\ & x>1, x \neq 2, \frac{(2 x-1)(x+2)}{(x-4)(x+1)}>0\end{aligned}$$
$$\begin{aligned} & \therefore \quad x \in(4, \infty) \\ & \therefore \quad \gamma=4 \\ & \therefore \quad \alpha^2+\beta^2+\gamma^2=49+121+16 \\ & =186 \end{aligned}$$