Let D be the domain of the function f(x) = si | vedclass

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(B) For the function $f(x)$ to be defined,we require the argument of $\sin^{-1}$ to be in $[-1, 1]$ and the base of the logarithm to be positive and n

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(B) For the function $f(x)$ to be defined,we require the argument of $\sin^{-1}$ to be in $[-1, 1]$ and the base of the logarithm to be positive and not equal to $1$.First,$\frac{6+2 \log_3 x}{-5x} > 0$ and $x > 0, x 
eq \frac{1}{3}$. Since $x > 0$,we need $6+2 \log_3 x Next,$-1 \leq \log_{3x} \left(\frac{6+2 \log_3 x}{-5x}ight) \leq 1$. Since $x Solving $15x^2 + 6 + 2 \log_3 x \geq 0$ and $6 + 2 \log_3 x + \frac{5}{3} \geq 0$ leads to $x \in [3^{-23/6}, \frac{1}{27})$.Thus,the domain $D = [3^{-23/6}, \frac{1}{27})$.Since $3^{-23/6} Then $\alpha = 3^{-23/6}$ and $\beta = \frac{1}{27}$.Calculating $\alpha^2 + \frac{5}{\beta} = (3^{-23/6})^2 + 5(27) = 3^{-23/3} + 135$. Since $3^{-23/3}$ is very small,the value is approximately $135$.

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