Discussion Forum

If $f(x0 +2f(\frac{1}{x})=3x, x\neq  0$ and  $S= [x\epsilon R:f(x)=f(-x)]:$ then S 

Answer:

Explanation:

We have , $f(x)+2f(\frac{1}{x})=3x,x\neq 0.....(i)$

On replaceing x by 1/x in the above eq (i)

we get  $f(\frac{1}{x})+2f(x)=\frac{3}{x}$

  $\Rightarrow 2f(x)+2f(\frac{1}{x})=\frac{3}{x}$   .......(ii)

On multiplying Eq(ii)  by 2 and substracting Eq(i) from Wq(ii), we get

$4f(x)+2f(\frac{1}{x})=\frac{6}{x}$

 $\frac{f(x)+2f(\frac{1}{x})=3x}{3f(x)=\frac{6}{x}-3x}$

$\Rightarrow   f(x)=\frac{2}{x}-x$

now consider, f(x)=f(-x)

$\Rightarrow   \frac{2}{x}-4=-\frac{2}{x}+x$

$\Rightarrow   \frac{4}{x}=2x$

$\Rightarrow  2x^{2}=4\Rightarrow x^{2}=2$

$\Rightarrow  x=\pm \sqrt{2}$

 Hence, S contains exactly two elements