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

For $\cos^{-1}(t)$ to be defined, $-1 \le t \le 1$ So we need $-1 \le \frac{4x+5}{3x-7} \le 1, \qquad 3x-7\ne 0$

We solve the two inequalities separately.

(i) $\dfrac{4x+5}{3x-7} \ge -1$

$\frac{4x+5}{3x-7}+1 \ge 0$ $\frac{4x+5+3x-7}{3x-7} \ge 0$ $\frac{7x-2}{3x-7} \ge 0$ Critical points: $x=\dfrac27,\ \dfrac73$.

Sign analysis gives $\frac{7x-2}{3x-7} \ge 0 \iff x\in (-\infty,\tfrac27] \cup (\tfrac73,\infty)$

(ii) $\dfrac{4x+5}{3x-7} \le 1$

$\frac{4x+5}{3x-7}-1 \le 0$ $\frac{4x+5-3x+7}{3x-7} \le 0$ $\frac{x+12}{3x-7} \le 0$ Critical points: $x=-12,\ \dfrac73$.

Sign analysis gives $\frac{x+12}{3x-7} \le 0 \iff x\in [-12,\tfrac73)$

Intersection

Hence domain of $f$ is $[-12,\tfrac73)\cap \left(( -\infty,\tfrac27] \cup (\tfrac73,\infty)\right)= [-12,\tfrac27]$ So, $\alpha=-12,\quad \beta=\frac27$

2. Domain of $g(x)=\log_2\left(2-6\log_{27}(2x+5)\right)$

For logarithm to be defined:

• inner argument of $\log_{27}$ must be positive: $2x+5>0 \implies x> -\frac52$
• argument of outer log must be positive: $2-6\log_{27}(2x+5)>0$

So, $\log_{27}(2x+5)<\frac13$ Since base $27>1$, $2x+5<27^{1/3}=3$ $2x< -2$ $x< -1$ Combining with $x> -\dfrac52$, $x\in \left(-\frac52,-1\right)$ Thus, $\gamma=-\frac52,\quad \delta=-1$

3. Required value

We need $\left|7(\alpha+\beta)+4(\gamma+\delta)\right|$

First, $\alpha+\beta=-12+\frac27=\frac{-84+2}{7}= -\frac{82}{7}$ Hence, $7(\alpha+\beta)= -82$

Also, $\gamma+\delta=-\frac52-1=-\frac72$ Hence, $4(\gamma+\delta)=4\left(-\frac72\right)=-14$

Therefore, $7(\alpha+\beta)+4(\gamma+\delta)=-82-14=-96$ So, $\left|7(\alpha+\beta)+4(\gamma+\delta)\right|=96$

4. Comparison with stored answer

Derived answer = $96$. Stored correct answer = $96$. They agree.