If the domain of the function f(x) = ^ - 1 ((2x)/(5x + 3)) is [ α,...

We know that,

Domain of $\sec^{- 1}x$ is $( - \infty , - 1 ] \cup [ 1, \infty )$.

Now, $f\left(x\right)$ = \sec^{- 1}\left(\dfrac{2x}{5x + 3}\right)$</p><p>So,</p><p>$\dfrac{2x}{5x + 3}$\le - 1$\text{or}\dfrac{2x}{5x + 3} \ge 1$</p><p>$⇒ \dfrac{2x}{5x + 3} +1 \le 0 \text{or}\dfrac{2x}{5x + 3} -$1 \ge 0$

$⇒$\dfrac{7x + 3}{5x + 3}$\le 0$\text{or}\dfrac{- 3x - 3}{5x + 3} \ge 0$</p><p>$⇒ \dfrac{7x + 3}{5x + 3} \le 0 \text{or}\dfrac{x + 1}{5x + 3}$\le 0$

$⇒ x \in [ - 1, -$\dfrac{3}{5}$) \cup ( -$\dfrac{3}{5}$, -$\dfrac{3}{7}$]$

So,

$\alpha = - 1, \beta = -$\dfrac{3}{5}$, \gamma = -$\dfrac{3}{5}$, \delta = -$\dfrac{3}{7}

Now,

\left|3\alpha + 10\left(\beta + \gamma\right) + 21\delta\right| = \left|- 3 - 6 - 6 - 9\right| = 24

Hence this is the required answer.