Question

If $f\left( {x + \dfrac{1}{y}} \right) + f\left( {x - \dfrac{1}{y}} \right) = 2f\left( x \right)f\left( {\dfrac{1}{y}} \right)$ for all $x \in \mathbb{R},y \in \mathbb{R} - \\{ 0\\}$ and $f(0) = \dfrac{1}{2}$, then $f(4)$ is:
A) 0
B) 4
C) -4
D) 2

Answer 1

We will put in such particular values of x and y such that, we have f(4) and f(-4) in LHS and RHS in such a form that f(-4) cancels out from both the sides and thus we have the required answer.

Complete step-by-step answer:
Since, we are given the domain as $x \in \mathbb{R},y \in \mathbb{R} - \\{ 0\\}$. Therefore, we can put in any real value of x and any real value of y except 0.
Now, let us put in $x = 0$ and $y = - \dfrac{1}{4}$.
Then we will have the LHS as equal to:-
$\Rightarrow LHS = f\left( {0 + \dfrac{1}{{\left( { - \dfrac{1}{4}} \right)}}} \right) + f\left( {0 - \dfrac{1}{{\left( { - \dfrac{1}{4}} \right)}}} \right)$
Simplifying the denominators to get the following expression:-
$\Rightarrow LHS = f\left( {0 - 4} \right) + f\left( {0 + 4} \right)$
Simplifying the LHS further to get the following expression:-
$\Rightarrow LHS = f\left( { - 4} \right) + f\left( 4 \right)$ ………………(1)
Now, we will get the RHS as equal to:
$\Rightarrow RHS = 2f\left( 0 \right)f\left( {\dfrac{1}{{\left( { - \dfrac{1}{4}} \right)}}} \right)$
Simplifying the denominator to get the following expression:-
$\Rightarrow RHS = 2f\left( 0 \right)f\left( { - 4} \right)$
Simplifying the LHS by putting in $f(0) = \dfrac{1}{2}$ to get the following expression:-
$\Rightarrow RHS = 2 \times \dfrac{1}{2} \times f\left( { - 4} \right)$
Simplifying the expression further to obtain the following:-
$\Rightarrow RHS = f\left( { - 4} \right)$ ………………(2)
Now, equating (1) and (2), we will then get:-
$\Rightarrow f\left( { - 4} \right) + f\left( 4 \right) = f\left( { - 4} \right)$
Now, cutting off f(-4) from both the sides, we will get:-
$\Rightarrow f\left( 4 \right) = 0$
$\therefore$ the value of f(4) is 0.

$\therefore$ the correct option is (A).

Note: The students might think of any other way of solving this question. Every way is fine until you keep in mind the domain of x and y because you cannot put any element which is not in the domain of x and y.
The students might be intrigued and want to put in $x = 4$ and $y = \infty$. We will still get the answer and the answer will be verified as well without any issues. But you must note that the domain of y is given to be all real numbers except 0. And, $y = \infty$ is not a part of real numbers but it is the part of extended real numbers.