Animated Solution for Mathematics - Functions: If $f(x)+2f(\frac{1}{x})=3x,x\ne 0$ and $S=\{x\in R:f(x)=f(-x)\}$; then S:

Welcome, fellow travelers on the JEE journey. Today, we are going to peel back the layers of a functional equation. These problems often look intimidating, like a locked door, but they are actually elegant puzzles waiting for the right key.

Our problem today is:

We need to find the set $S=\{x\in \mathbb{R}:f(x)=f(-x)\}$. Think of this as a detective story where we must unmask the function $f(x)$ using the 'Mirror Trick'.

Substitution

The equation links $x$ and $\frac{1}{x}$. If we replace $x$ with $\frac{1}{x}$, we create a mirror image of the original equation.

Replacing $x$ with $\frac{1}{x}$ in the original equation gives us:

Now, we have a system of two linear equations with two unknowns, $f(x)$ and $f\left(\frac{1}{x}\right)$: 1) $f(x)+2f\left(\frac{1}{x}\right)=3x$ 2) $2f(x)+f\left(\frac{1}{x}\right)=\frac{3}{x}$

To isolate $f(x)$, we need to eliminate $f\left(\frac{1}{x}\right)$. Multiply equation (2) by $2$ to obtain:

Now, subtract equation (1) from this new equation. The $2f\left(\frac{1}{x}\right)$ terms vanish, leaving us with:

Simplifying this expression, we get $3f(x)=\frac{6}{x}-3x$. Dividing by $3$, we unmask the function:

Now, we return to the set $S$ where $f(x)=f(-x)$. First, we determine $f(-x)$ by substituting $-x$ into our derived function:

Setting $f(x)=f(-x)$, we have:

Rearranging the terms leads to:

Dividing by $2$ and multiplying by $x$ (noting $x eq 0$), we arrive at $x^2=2$. Thus, the solutions are $x=\sqrt{2}$ and $x=-\sqrt{2}$.

The final set is $S=\{\sqrt{2},-\sqrt{2}\}$. It contains exactly two elements.