If #f_1(x)=-((2x+7)/(x+3)), f_(n+1)(x)=f_1(f_n(x))# then #f_2001(2002)# is?

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Answer 1 · 2018-07-31T18:47:41

# 2002#.

We have, #f_1(x)=-((2x+7)/(x+3)), and, f_1(f_n(x))=f_(n+1)(x)#.

Observe that, #f_1(x)=-((2x+7)/(x+3))=-((2x+6+1)/(x+3))#

#=-((2x+6)/(x+3)+1/(x+3))#.

# rArr f_1(x)=-(2+1/(x+3))=-2-1/(x+3)#

#:. f_2(x)=f_1(f_1(x))#

#=f_1(-2-1/(x+3))=f_1(y)," say, "y=-2-1/(x+3)#,

#=-2-1/(y+3)=-2-1/(-2-1/(x+3)+3)#,

#=-2-1/(1-1/(x+3))=-2-(x+3)/(x+3-1)#,

#=-2-(x+3)/(x+2)=-2-(x+2+1)/(x+2)#,

#=-2-((x+2)/(x+2)+1/(x+2))#

# rArr f_2(x)=-3-1/(x+2)#.

#:. f_3(x)=f_1(f_2(x))#,

#=f_1(t)," say, where, "t=f_2(x)=-3-1/(x+2)#,

#=-2-1/(t+3)#,

#=-2-1/(-3-1/(x+2)+3)=-2-1/(-1/(x+2))=-2+x+2#.

#rArr f_3(x)=x#.

Clearly, then, #f_6(x)=f_9(x)=...=f_(3m)(x)=x, AA m in NN#.

Since, #2001=3(667)#, we have,

#f_2001(x)=x, & therefore, f_2001(2002)=2002#.

#color(magenta)("Enjoy Maths.!")#