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Question #6ed7e

1 Answer

Jan 12, 2017

Given #f(x) = 1/logx and g(x) = 1/8 +x#

We have

#(fxxg)(x) = (1/logx)xx(1/8+x)#

#=>(fxxg)(x) = (1/8+x)/logx#

#(fxxg)(x)# will be defined when

#logx!=0=>x!=1#
and #x>0 and x in R#

So option (c) is valid here.

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