The range of values of the function f(x) = fr | vedclass

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(C) We know that for any real $x$,the range of $\sin x$ is $[-1, 1]$.So,$-1 \leq \sin x \leq 1$.Multiplying by $-3$,we get $-3 \leq -3 \sin x \leq 3$.

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$( -\infty, -1 ] \cup [ 1/5, \infty )$

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(C) We know that for any real $x$,the range of $\sin x$ is $[-1, 1]$.So,$-1 \leq \sin x \leq 1$.Multiplying by $-3$,we get $-3 \leq -3 \sin x \leq 3$.Adding $2$ to all parts,we get $2 - 3 \leq 2 - 3 \sin x \leq 2 + 3$,which simplifies to $-1 \leq 2 - 3 \sin x \leq 5$.Now,we consider the function $f(x) = \frac{1}{2 - 3 \sin x}$.Since $2 - 3 \sin x$ takes values in $[-1, 5]$ excluding $0$ (where the function is undefined),we analyze the intervals.When $2 - 3 \sin x \in [-1, 0)$,the value of $\frac{1}{2 - 3 \sin x}$ lies in $(-\infty, -1]$.When $2 - 3 \sin x \in (0, 5]$,the value of $\frac{1}{2 - 3 \sin x}$ lies in $[1/5, \infty)$.Thus,the range is $(-\infty, -1] \cup [1/5, \infty)$.

The range of values of the function $f(x) = \frac{1}{2 - 3\sin x}$ is

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