Range of the function frac{1}{2 - sin 3x} is | vedclass

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(B) Let $f(x) = \frac{1}{2 - \sin 3x}$.We know that for any real $x$,the range of $\sin 3x$ is $[-1, 1]$.So,$-1 \le \sin 3x \le 1$.Multiplying by $-1$

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$[\frac{1}{3}, 1]$

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(B) Let $f(x) = \frac{1}{2 - \sin 3x}$.We know that for any real $x$,the range of $\sin 3x$ is $[-1, 1]$.So,$-1 \le \sin 3x \le 1$.Multiplying by $-1$,we get $-1 \le -\sin 3x \le 1$.Adding $2$ to all parts,we get $2 - 1 \le 2 - \sin 3x \le 2 + 1$,which simplifies to $1 \le 2 - \sin 3x \le 3$.Taking the reciprocal,the inequality signs reverse: $\frac{1}{1} \ge \frac{1}{2 - \sin 3x} \ge \frac{1}{3}$.Thus,$\frac{1}{3} \le f(x) \le 1$.Therefore,the range of the function is $[\frac{1}{3}, 1]$.

Range of the function $\frac{1}{2 - \sin 3x}$ is

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