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# If $f\left( x \right) + 2f\left( {\dfrac{1}{x}} \right) = 3x,x \ne 0$ and $S = \left\{ {x \in R:f\left( x \right) = f\left( { - x} \right)} \right\}$, then $S$A) is an empty setB) contains exactly one elementC) contains exactly two elementsD) contains more than two elements

Last updated date: 09th Aug 2024
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Hint:If $f\left( x \right) + 2f\left( {\dfrac{1}{x}} \right) = 3x,x \ne 0$ is given, then substitute $x$ such that we get two equations and can be solved easily like replace $x$ by $\dfrac{1}{x}$. We get
$f\left( {\dfrac{1}{x}} \right) + 2f\left( x \right) = \dfrac{3}{x}$ and now solve the two equations to get the value of $f(x)$ then substitute in given set $S$ and get the required answer.

Here we are given an equation $f\left( x \right) + 2f\left( {\dfrac{1}{x}} \right) = 3x -----(1)$, where $x \ne 0$.
Now to make a similar equation, we can replace $x$ by $\dfrac{1}{x}$ in the above equation.
We will get $f\left( {\dfrac{1}{x}} \right) + 2f\left( x \right) = \dfrac{3}{x} -----(2)$
Now from (2) we get $f\left( {\dfrac{1}{x}} \right) = \dfrac{3}{x} - 2f\left( x \right)$
Now, substituting this value to equation (1)
We know, $f\left( x \right) + 2f\left( {\dfrac{1}{x}} \right) = 3x$
Now we know $f\left( {\dfrac{1}{x}} \right) = \dfrac{3}{x} - 2f\left( x \right)$
So,
$f\left( x \right) + 2\left( {\dfrac{3}{x} - 2f\left( x \right)} \right) = 3x \\ = f\left( x \right) + \dfrac{6}{x} - 4f\left( x \right) = 3x \\ f\left( x \right) = \dfrac{2}{x} - x \\$
Now according to question,
In the set $S$, it is given that $f\left( x \right) = f\left( { - x} \right)$. So we have to find how many elements are in $S$
So, upon solving,
$f\left( x \right) = f\left( { - x} \right) \\ \dfrac{2}{x} - x = \dfrac{2}{{ - x}} - \left( { - x} \right) \\ \dfrac{2}{x} - x = - \dfrac{2}{x} + x \\ \dfrac{4}{x} = 2x \\ {x^2} = 2 \\ x = \pm \sqrt 2 \\$
We got two values of $x$, that is $+ \sqrt 2 , - \sqrt 2$.
So, $S$ contains exactly two elements.

So, the correct answer is “Option C”.

Note:If $f\left( x \right)$ is given any polynomial let $a{x^2} + bx + c$. If we replace $x$ by $\dfrac{1}{x}$, then , $f\left( {\dfrac{1}{x}} \right) = a{\left( {\dfrac{1}{x}} \right)^2} + b\left( {\dfrac{1}{x}} \right) + c$. So, $x$ must be replaced from every place.