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The period of $f(x)=\cos \left(\frac{x}{3}\right)+\sin \left(\frac{x}{2}\right)$ is
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The correct answer is:
$12 \pi$
Given, $f(x)=\cos \left(\frac{x}{3}\right)+\sin \left(\frac{x}{2}\right)$
Period of $\cos x$ and $\sin x$ are $2 \pi$.
$$
\begin{aligned}
\therefore \text { Period of } f(x) & =\text { Period of }\left[\cos \frac{x}{3}+\sin \frac{x}{2}\right] \\
& =\text { Period of } \cos \frac{x}{3}+\text { Period of } \sin \frac{x}{2} \\
= & \frac{2 \pi}{\left(\frac{1}{3}\right)}+\frac{2 \pi}{\left(\frac{1}{2}\right)}=\frac{6 \pi}{1}+\frac{4 \pi}{1} \\
= & \frac{\text { LCM of }(6 \pi, 4 \pi)}{\text { LCM of }(1,1)}=\frac{12 \pi}{1}=12 \pi
\end{aligned}
$$
Period of $\cos x$ and $\sin x$ are $2 \pi$.
$$
\begin{aligned}
\therefore \text { Period of } f(x) & =\text { Period of }\left[\cos \frac{x}{3}+\sin \frac{x}{2}\right] \\
& =\text { Period of } \cos \frac{x}{3}+\text { Period of } \sin \frac{x}{2} \\
= & \frac{2 \pi}{\left(\frac{1}{3}\right)}+\frac{2 \pi}{\left(\frac{1}{2}\right)}=\frac{6 \pi}{1}+\frac{4 \pi}{1} \\
= & \frac{\text { LCM of }(6 \pi, 4 \pi)}{\text { LCM of }(1,1)}=\frac{12 \pi}{1}=12 \pi
\end{aligned}
$$
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