Saved Bookmarks
| 1. |
Find the period of (a) (|sin4x|+|cos 4x|)/(|sin 4x-cos 4x|+|sin 4x+cos 4x|) (b) f(x)="sin"(pi x)/(n!)-"cos"(pi x)/((n+1)!) (c ) f(x)=sin x +"tan"(x)/(2)+"sin"(x)/(2^(2))+"tan"(x)/(2^(3))+ ... +"sin"(x)/(2^(n-1))+"tan"(x)/(2^(n)) |
|
Answer» Period of `|sin 4x-cos 4x|+|sin 4x+cos 4x|=(pi)/(8)` So, the period of GIVEN function is `(pi)/(8).` (B) `f(X)="sin"(pi)/(n!)-"cos"(pi x)/((n+1)!)` Period of `"sin"(pi)/(n!) " is " (2pi)/((pi)/(n!))=2n!` and period of` "cos"(pi x)/((n+1)!) " is " (2pi)/((pi )/((n+1)!))=2(n+1)!` HENCE, period of `f(x)=L.C.M" of "{2n!,29n+1)!}=2(n+1)!` (c ) `f(x)=sin x +"tan"(x)/(2)+"sin"(x)/(2^(2))+"tan"(x)/(2^(3))+ ... +"sin"(x)/(2^(n-1))+"tan"(x)/(2^(n))` Period of `sin x " is " 2pi.` Period of ` "tan"(x)/(2) " is " 2pi.` Period of ` "sin"(x)/(2^(2)) " is " 8pi.` Period of ` "tan"(x)/(2^(3)) " is " 8pi`. Period of `"tan"(x)/(2^(n)) " is " 2^(n) pi.` Hence, period of `f(x)=L.C.M. " of " (2pi,8pi, ..., 2^(n) pi)=2^(n) pi` |
|