If the domain of the function is , then is equal to, Inverse Trigonometric Functions

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Solution

Q.
If the domain of the function $\sin^{- 1} (\frac{3 x - 22}{2 x - 19}) + \log_{e} (\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10})$ is $(α, β]$ , then $3 α + 10 β$ is equal to [2024]

1 95

2 97

3 98

4 100

Ans.
(2)

$\sin^{- 1} (\frac{3 x - 22}{2 x - 19}) + \log_{e} (\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10})$

Since, $- 1 \leq \sin^{- 1} x \leq 1$

$∴$ $- 1 \leq \frac{3 x - 22}{2 x - 19} \leq 1$

$\frac{3 x - 22}{2 x - 19} + 1 \geq 0 and \frac{3 x - 22}{2 x - 19} - 1 \leq 0$

$\Rightarrow \frac{3 x - 22 + 2 x - 19}{2 x - 19} \geq 0 and \frac{3 x - 22 - 2 x + 19}{2 x - 19} \leq 0$

$\Rightarrow \frac{5 x - 41}{2 x - 19} \geq 0 and \frac{x - 3}{2 x - 19} \leq 0$

$x \in (- \infty, \frac{41}{5}] \cup (\frac{19}{2}, \infty) and x \in [3, \frac{19}{2})$

$\Rightarrow x \in [3, \frac{41}{5}]$ ...(i)

and $\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10} > 0$ $;$ $\frac{(3 x - 5) (x - 1)}{(x - 5) (x + 2)} > 0$

$\Rightarrow x \in (- \infty, - 2) \cup (1, \frac{5}{3}) \cup (5, \infty)$ ...(ii)

Taking intersection of individual domains

$x \in (5, \frac{41}{5}]$ $\Rightarrow α = 5 and β = \frac{41}{5}$

$Hence, 3 α + 10 β = 15 + 82 = 97$

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