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Let `f(x)=sin(x/(n!))+cos((2x)/((n+1)!))`. Find the period of `f(x)`. |
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Answer» Period of `sin(x/n_1)=T_1` `T_1/(n!) =2pi` `T_1=n!(2pi)` Period of `cos((2x)/(n+1)!)=T_2` `(2T_2)/((n+1)!)=2pi` `T_2=(n+1)!pi` Period of`sin(x/n_1)+cos((2x)/((n+1)!))` is T such that`n_1T_1=n_2T_2=T` Such at `n_1n_2` are to each `(n!)2pi*n_1=n_2(n+1)!pi` `2n_1=(n+1)n_2` `n_1/n_2=(n+1)/2` when n is own,`n_1/n_2=(n+1)/2` `n_1=n+1,n_2=2` `T=2(n+1)!pi` When n is odd `n_1/n_2=(n+1)/2` `n_1=(n+1)/2,n_2=1` `T=(n+1)!pi=(n+1)!pi` `T=2(n+1)!pi` when n is even. `T=(n+1)!pi`, when n is odd. |
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