Question

Arrange the following in decreasing order of their periods :-
A= $\cos \left(\frac{2\pi }{5}x\right)-\sin \left(\frac{2\pi }{7}x\right)$
B= $\sin \left(\frac{\pi x}{2}\right)+\cos \left(\frac{\pi x}{3}\right)$
C=$\sin \left(\frac{\pi x}{5}\right)+\tan \left(\frac{\pi x}{3}\right)$
D=$x-\left[x\right]$

• A. $A,C,B,D$
• B. $A,B,C,D$
• C. $D,B,C,A$
• D. $B,A,D,C$

Answer 1

The correct option is A $A,C,B,D$

(A) Period of $\cos \frac{2\pi }{5}x$ is $\frac{2\pi }{2\pi /5}=5$

And period of $\sin \left(\frac{2\pi }{7}x\right)$ is $\frac{2\pi }{2\pi /7}=7.$

$\therefore L.C.M.$ of $5$ and $7$ is $35$

hence, period of $A$ is $35$

(B) Period of $\sin \left(\frac{\pi }{2}x\right)$ is $\frac{2\pi }{\pi /2}=4$

Period of $\cos \left(\frac{\pi }{3}x\right)$ is $\frac{2\pi }{\pi /3}=6$

$\therefore L.C.M.$ of $4$ and $6$ is $12$

Hence, period of $B$is $12$

(C) Period of $\sin \left(\frac{\pi }{5}x\right)$ is $\frac{2\pi }{\pi /5}=10$

Period of $\tan \left(\frac{\pi }{3}x\right)$ is $\frac{\pi }{\pi /3}=3$

$\therefore L.C.M.$ of $10$ and $3$ is $30$

Hence, period of $C$ is $30$

(D) $x-\left[x\right]=\left\{x\right\}$

As we know period of fractional function is $1$

Hence, period of $D$ is $1$

$\therefore$ In decseasing order it can be given as $A,C,B,D.$