If the domain of the function f(x) = sec^{-1} | vedclass

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(C) The function $f(x) = \sec^{-1}\left(\frac{2x}{5x+3}ight)$ is defined when $\left|\frac{2x}{5x+3}ight| \geq 1$ and $5x+3 
eq 0$.This implies $

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$24$

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(C) The function $f(x) = \sec^{-1}\left(\frac{2x}{5x+3}ight)$ is defined when $\left|\frac{2x}{5x+3}ight| \geq 1$ and $5x+3 
eq 0$.This implies $\left|\frac{2x}{5x+3}ight| \geq 1$,which means $(2x)^2 \geq (5x+3)^2$.$(2x)^2 - (5x+3)^2 \geq 0$$(2x - 5x - 3)(2x + 5x + 3) \geq 0$$(-3x - 3)(7x + 3) \geq 0$$-(3x + 3)(7x + 3) \geq 0 \Rightarrow (x + 1)(7x + 3) \leq 0$.The solution to this inequality is $x \in [-1, -3/7]$.Additionally,the denominator $5x+3 
eq 0$ implies $x 
eq -3/5$.Thus,the domain is $[-1, -3/5) \cup (-3/5, -3/7]$.Comparing this with $[\alpha, \beta) \cup (\gamma, \delta]$,we get $\alpha = -1, \beta = -3/5, \gamma = -3/5, \delta = -3/7$.Now,calculate $|3\alpha + 10(\beta + \gamma) + 21\delta| = |3(-1) + 10(-3/5 - 3/5) + 21(-3/7)|$.$= |-3 + 10(-6/5) + 3(-3)| = |-3 - 12 - 9| = |-24| = 24$.

If the domain of the function $f(x) = \sec^{-1}\left(\frac{2x}{5x+3}\right)$ is $[\alpha, \beta) \cup (\gamma, \delta]$,then $|3\alpha + 10(\beta + \gamma) + 21\delta|$ is equal to $.......$.

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