If f1(x)=f(x)=12xf1(x)=f(x)=12x and fn(x)=f(fn−1(x)fn(x)=f(fn−1(x), then find f5(x)+f10(x)f5(x)+f10(x) where x = 1 Choose the correct choice from the following answers:
A) 3/9 B) 3/2 C) 3/1 D) 3/3 E) 3/4
f(x) = 12x12x and f(f(x)) = f(12x)f(12x) = 12(12x)12(12x) = x   Similarly f3(x)=f(f2(x))=f(x)=12xf3(x)=f(f2(x))=f(x)=12x and f4(x)=f(f3(x))=f(12x)=12×12x=xf4(x)=f(f3(x))=f(12x)=12×12x=x   Hence fn(x)=12xfn(x)=12x for n = odd and fn(x)=xfn(x)=x for n = even   f5(x)+f10(x)=12x+xf5(x)+f10(x)=12x+x   If x = 1, ⇒12×1+1=3/2   Answer:3/2
<result>B</result>
If f1(x)=f(x)=12xf1(x)=f(x)=12x and fn(x)=f(fn−1(x)fn(x)=f(fn−1(x), then find f5(x)+f10(x)f5(x)+f10(x) where x = 1