JEE Main 2014 — Mathematics Algebra

Given $f(x)=|\mathrm{sin}4x|+|\mathrm{cos}2x|$f(x)=∣sin4x∣+∣cos2x∣

We know that the period of $|sinx|$∣sinx∣ is $\pi$π and also if $f(x)$f(x) is a periodic function with fundamental period $T,$T, then the fundamental period of $f(kx)$f(kx) is $\frac{T}{|k|}.$∣k∣T​.

$\Rightarrow {T}_{1}=$⇒T1​=period of $|\text{sin4}\text{x}|=\frac{\pi }{4}$∣sin4x∣=4π​

Also, we know that the period of $|\mathrm{cos}x|$∣cosx∣ is $\pi$π

$\Rightarrow {T}_{2}=$⇒T2​=Period of $|\mathrm{cos}2x|=\frac{\pi }{2}$∣cos2x∣=2π​

Again, we know that the period of the sum or difference of two periodic functions is the $L.C.M.$L.C.M. of their periods.

So, the period of $f(x)$f(x) is $L.C.M.$L.C.M. of the periods of $|\mathrm{sin}4x|$∣sin4x∣ and $|\mathrm{cos}2x|$∣cos2x∣ i.e. the $L.C.M.({T}_{1},{T}_{2})$L.C.M.(T1​,T2​)

Now, $L.C.M.(\frac{\pi }{4},\frac{\pi }{2})=\frac{L.C.M.(\pi ,\pi )}{H.C.F.(4,2)}=\frac{\pi }{2}.$L.C.M.(4π​,2π​)=H.C.F.(4,2)L.C.M.(π,π)​=2π​.