#f(x)=[2x]sin3pix# and #f'(k^+)=lamdakpi(-1)^k#, where [.] denoted greater integer function and #kinN#, then what is the value of #lambda# ?

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Feb 21, 2018

Answer:

#lambda = 6#

Explanation:

Considering #d/dx[2x]=0# we have

#(df)/dx = 3pi[2x]cos(3pi x)#

and now if #k in NN# we have #[2k]=2k# so

#3pi 2k cos(3pi k)=lambda k pi(-1)^k# or

#6k pi cos(3k pi)=lambda k pi(-1)^k# and

#cos(3k pi) = (-1)^k# and then

#6k pi(-1)^k = lambda k pi(-1)^k#

and the solution is for #lambda = 6#

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