Let f(x)=12[f(xy)+f(xy)] for x,y∈R+ such that f(1)=0;f′(1)=2
f(e) is equal to
A
2
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B
1
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C
3
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D
4
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Solution
The correct option is C2 f(x) is of the form alog(x), where 'a' is a constant. Hence, 12[f(xy)+f(xy)] =12[alog(xy)+alog(xy)] =12[alogx+alogy+alogx−alogy] =alog(x). Now f(1)=0 And f′(1)=2 ⇒axx=1=2 ⇒a=2 Hence, f(x)=2log(x) Hence, f(e)=2log(e)=2