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(D) The domain of $f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$ requires that the argument of $\sin^{-1}$ lies in the interval $[-1, 1]$ and the den

Vedclass · Accepted Answer

$32$

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(D) The domain of $f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}ight)$ requires that the argument of $\sin^{-1}$ lies in the interval $[-1, 1]$ and the denominator is non-zero.$1$. Denominator condition: $2x + 3 
eq 0 \implies x 
eq -\frac{3}{2}$.$2$. Inequality condition: $\left|\frac{x-1}{2x+3}ight| \leq 1$.Since $2x+3 
eq 0$,we have $|x-1| \leq |2x+3|$.Squaring both sides: $(x-1)^2 \leq (2x+3)^2$.$x^2 - 2x + 1 \leq 4x^2 + 12x + 9$.$3x^2 + 14x + 8 \geq 0$.Factoring the quadratic: $(3x + 2)(x + 4) \geq 0$.This inequality holds for $x \in (-\infty, -4] \cup [-\frac{2}{3}, \infty)$.$3$. Combining with the condition $x 
eq -\frac{3}{2}$:The domain is $(-\infty, -4] \cup [-\frac{2}{3}, \infty) \setminus \{-\frac{3}{2}\}$.However,the problem states the domain is $R - (\alpha, \beta)$. This implies the excluded region is an open interval. Looking at the complement,the excluded values are $(-4, -\frac{2}{3})$.Thus,$\alpha = -4$ and $\beta = -\frac{2}{3}$.$4$. Calculating $12\alpha\beta$:$12 	imes (-4) 	imes (-\frac{2}{3}) = 12 	imes \frac{8}{3} = 4 	imes 8 = 32$.

If the domain of the function $f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$ is $R - (\alpha, \beta)$,then $12\alpha\beta$ is equal to:

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