Exam

JEE Main

Year

2025

Date

8 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Functions

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Functions

Let the domain of the function f(x)= ^ -1 ((4 x+5)/(3 x-7)) be [α,...

Let the domain of the function
$f(x)=\cos ^{-1}\left(\frac{4 x+5}{3 x-7}\right)$ be $[\alpha, \beta]$ and the domain of $\mathrm{g}(\mathrm{x})=\log _2\left(2-6 \log _{27}(2 \mathrm{x}+5)\right)$ be $(\gamma, \delta)$.
Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to ________

Step-by-step solution

$$\begin{aligned} & f(x)=\cos ^{-1}\left(\frac{4 x+5}{3 x-7}\right) \\ & \Rightarrow-1 \leq\left(\frac{4 x+5}{3 x-7}\right) \leq 1 \\ & \left(\frac{4 x+5}{3 x-7}\right) \geq-1 \\ & \frac{4 x+5+3 x-7}{3 x-7} \geq 0 \\ & \Rightarrow \frac{7 x-2}{3 x-7} \geq 0\end{aligned}$$

$$\begin{aligned} & x \in\left(-\infty, \frac{2}{7}\right] \cup\left(\frac{7}{3}, \infty\right) \\ & & \frac{4 x+5}{3 x-7} \leq 1 \Rightarrow \frac{x+12}{3 x-7} \leq 0\end{aligned}$$
$\therefore$ Domain of $f(x)$ is
$$\begin{aligned}<br/>& {\left[-12, \frac{2}{7}\right] \alpha=-12, \beta=\frac{2}{7}} \\ & \mathrm{~g}(\mathrm{x})=\log _2\left(2-6 \log _{27}(2 \mathrm{x}+5)\right)<br/>\end{aligned}$$
Domain
$2-6 \log _{27}(2 x+5) \gt 0$
$$\begin{array}{ll}\Rightarrow & 6 \log _{27}(2 \mathrm{x}+5) \lt 2 \\ \Rightarrow & \log _{27}(2 \mathrm{x}+5) \lt \frac{1}{3} \\ \Rightarrow & 2 \mathrm{x}+5 \lt 3 \\ \Rightarrow & \mathrm{x} \lt -1\end{array}$$
$\& 2 x+5 \gt 0 \Rightarrow x \gt -\frac{5}{2}$
Domain is $\mathrm{x} \in\left(-\frac{5}{2},-1\right)$
$$\begin{aligned}<br/>& \gamma=-\frac{5}{2}, \delta=-1 \\ & |7(\alpha+\beta)+4(\gamma+\delta)|=\left\lvert\, 7\left(\left.-12+\frac{2}{7}+4\left(-\frac{5}{2}-1\right) \right\rvert\,\right.\right. \\ & |-82-14|=96<br/>\end{aligned}$$