Let $f(x)=\frac{1}{7-\sin 5 x}$ be a function defined on $\mathbf{R}$. Then the range of the function $f(x)$ is equal to :
Solution
<p>$$\begin{aligned}
& f(x)=\frac{1}{7-\sin 5 x} \\\\
& -1 \leq \sin 5 x \leq 1 \\\\
& -1 \leq-\sin 5 x \leq 1 \\\\
& -1+7 \leq 7-\sin 5 x \leq 1+7 \\\\
& 6 \leq 7-\sin 5 x \leq 8 \\\\
& \frac{1}{8} \leq \frac{1}{7-\sin 5 x} \leq \frac{1}{6} \\\\
& \frac{1}{8} \leq f(x) \leq \frac{1}{6} \\\\
& \text { Range }=\left[\frac{1}{8}, \frac{1}{6}\right]
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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