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Match the items of List-I to the items of List-II
The correct match is
$\begin{array}{llll}A & B & C & D\end{array}$
Options:
The correct match is
$\begin{array}{llll}A & B & C & D\end{array}$
Solution:
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Verified Answer
The correct answer is:
V I II III
A. Period of $\sin ^2 x$ is $\pi$
$\begin{aligned}
f(x) & =\sin ^2 x \\
f(\pi+x) & =\sin ^2(\pi+x)=(-\sin x)^2=\sin ^2 x=f(x)
\end{aligned}$
B. $\frac{\pi}{3}(\sqrt{3} \cos 3 x+\sin 3 x)$
Maximum value $=\frac{\pi}{3} \sqrt{(\sqrt{3})^2+(1)^2}=\frac{\pi}{3} \times 2=\frac{2 \pi}{3}$
C. We have, $\sin \frac{x}{3}+\cos \frac{x}{2}$
Period of $\sin x$ is $6 \pi$ and period of $\cos \frac{x}{2}$ is $4 \pi$.
$\therefore$ Period of $\sin \frac{x}{3}+\cos \frac{x}{2}$ is LCM of $6 \pi$ and $4 \pi$ i.e., $12 \pi$.
D. We have, $y=|\sin x|$ and $y=1 \quad \Rightarrow \quad 1=|\sin x|$
$\Rightarrow \quad \sin x= \pm 1 \Rightarrow x=\frac{\pi}{2}$ in $[0, \pi]$
$\therefore$ Intersection part is $\frac{\pi}{2}$.
$\begin{aligned}
f(x) & =\sin ^2 x \\
f(\pi+x) & =\sin ^2(\pi+x)=(-\sin x)^2=\sin ^2 x=f(x)
\end{aligned}$
B. $\frac{\pi}{3}(\sqrt{3} \cos 3 x+\sin 3 x)$
Maximum value $=\frac{\pi}{3} \sqrt{(\sqrt{3})^2+(1)^2}=\frac{\pi}{3} \times 2=\frac{2 \pi}{3}$
C. We have, $\sin \frac{x}{3}+\cos \frac{x}{2}$
Period of $\sin x$ is $6 \pi$ and period of $\cos \frac{x}{2}$ is $4 \pi$.
$\therefore$ Period of $\sin \frac{x}{3}+\cos \frac{x}{2}$ is LCM of $6 \pi$ and $4 \pi$ i.e., $12 \pi$.
D. We have, $y=|\sin x|$ and $y=1 \quad \Rightarrow \quad 1=|\sin x|$
$\Rightarrow \quad \sin x= \pm 1 \Rightarrow x=\frac{\pi}{2}$ in $[0, \pi]$
$\therefore$ Intersection part is $\frac{\pi}{2}$.
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