FunctionHard

Question

If the domain of the function $f(x) = \sin^{- 1}\left( \frac{5 - x}{3 + 2x} \right) + \frac{1}{\log_{e}g( + 0}$ is $( - \infty,\alpha\rbrack \cup \lbrack\beta,\gamma) - \{\delta\}$, then $6(\alpha + \beta + \gamma + \delta)$ is equal to

Options

A.70
B.66
C.67
D.68

Solution

$- 1 \leq \frac{5 - x}{2x + 3} \leq \& 10 - x > 0,10 - x \neq 1$

$${\left| \frac{5 - x}{2x + 3} \right| \geq 1\& x < 10\& x \neq 9 }{(5 - x)^{2} - (2x + 3)^{2} \leq 0\& x < 10\& 4x \neq 9 }{(x + 8)(3x - 2) \geq 0\& x < 10\& x \neq 9 }{\Rightarrow ( - \infty, - 8\rbrack \cup \left\lbrack \frac{2}{3},10 \right) - \{ 9\} }{\Rightarrow (\alpha + \beta + \gamma + \delta) = 6\left( - 8 + \frac{2}{3} + 10 + 9 \right) = 70}$$

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