Question

If f(x) = $2f \bigg \lgroup \frac{1}{3x}\bigg \rgroup$, x ≠ 0 and S = {x ∈ R : f(x) = f(–x)}; then S:
(a) contains exactly two elements.
(b) contains more than two elements.
(c) is an empty set.
(d) contains exactly one element.

Answer 1

$f(x) = 2f( \frac{1}{3x} ) \\ \\ given \\ for \: S \\ \\ f(x) = f( - x) \\ \\ now \\ \\ f(x) - 2f( \frac{1}{3x}) = 0 ...... < 1 > \: \\ \\ put \: x = \frac{1}{3x} \\ \\ f( \frac{1}{3x} ) - 2f(x) = 0..... < 2 > \\ \\ from \: \: < 2 > \\ \\ f( \frac{1}{3x} ) = 2f(x) \\ \\ sub \: in \: < 1 > \\ \\ f(x) - 6(f(x)) = 0 \\ \\ f(x) = 0 \: for \: all \: x < ( - \infty \: \: \infty )$

Therefore S contains only one element for

$f(x) = f( - x)$

I.e. 0.

hope it helps