Let the domain of the function f(x)=cos -1 (4 x+5/3 x-7 ) be [α, β] and the domain of g(x)=log 2 (2-6 log 27 (2 x . +5) ) be (γ, δ) . Then 7(α+β)+4(γ+δ) is equal to \ \ \ \ .

Mathematics · Functions

JEE Main 2025 — 8 April, Evening Shift — Question 43

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

$g(x)=\log _{2}\left(2-6 \log _{27}(2 x\right.$ $+5)$ ) be $(\gamma, \delta)$ . Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to $\_\_\_\_$ .

Numerical answer — enter this value.

$f(x)=\cos ^{-1}\left(\frac{4 x+5}{3 x-7}\right)$

$-1 \leq \frac{4 x+5}{3 x-7} \leq 1$

$\frac{7 x-2}{3 x-7} \geq 0, \frac{x+12}{3 x-7} \leq 0$

$x \in\left(-\infty, \frac{2}{7}\right] \cup\left(\frac{7}{3}, \infty\right), x \in\left[-12, \frac{7}{3}\right)$

$\Rightarrow \quad x \in\left[-12, \frac{2}{7}\right] \Rightarrow \alpha=-12, \beta=\frac{2}{7}$

$g(x)=\log _{2}\left(2-6 \log _{27}(2 x+5)\right)$

\begin{aligned} & \Rightarrow \quad 2-6 \log _{27}(2 x+5)>0,2 x+5>0 \\& \log _{27}(2 x+5)<\frac{1}{3} \Rightarrow(2 x+5)^{3}<27 \\& \Rightarrow \quad 2 x+5<3 \Rightarrow x<-1 \\& \Rightarrow \quad x \in\left(-\frac{5}{2},-1\right) \Rightarrow \gamma=-\frac{5}{2}, \delta=-1 \\& \Rightarrow \quad|7(\alpha+\beta)+4(\gamma+\delta)|=96 \end{aligned}

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.