Question
If $$f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x,x \ne 0$$ and $$S = \left\{ {x\,I\,R:f\left( x \right) = f\left( { - x} \right)} \right\};$$ then $$S :$$
A.
contains exactly two elements.
B.
contains more than two elements.
C.
is an empty set.
D.
contains exactly one element.
Answer :
contains exactly two elements.
Solution :
$$\eqalign{
& f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x\,\,\,\,\,.....\left( 1 \right) \cr
& f\left( {\frac{1}{x}} \right) + 2f\left( x \right) = \frac{3}{x}\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Adding (1) and (2)
$$ \Rightarrow \,\,f\left( x \right) + f\left( {\frac{1}{x}} \right) = x + \frac{1}{x}$$
Substracting (1) from (2)
$$ \Rightarrow \,\,f\left( x \right) - f\left( {\frac{1}{x}} \right) = \frac{3}{x} - 3x$$
On adding the above equations
$$\eqalign{
& \Rightarrow \,\,f\left( x \right) = \frac{2}{x} - x \cr
& f\left( x \right) = f\left( { - x} \right) \cr
& \Rightarrow \,\,\frac{2}{x} - x = \frac{{ - 2}}{x} + x \cr
& \Rightarrow \,\,x = \frac{2}{x} \cr
& {x^2} = 2\,\,\,\,{\text{or }}\,x = \sqrt 2 , - \sqrt 2 \cr} $$