Question

If $f\left( x \right) + f\left( y \right) + f\left( {xy} \right) = 2 + f\left( x \right)f\left( y \right)\,\,\forall x,y \in R$ and $f\left( x \right)$ is polynomial with $f\left( 4 \right) = 17$. Find $\dfrac{{f\left( 5 \right)}}{{13}}$.

Answer 1

Hint: First, substitute $y = \dfrac{1}{x}$ and then substitute x=1 which give two values of f(1). After that put y=1 which gives the value of f(1)=2. Substitute the value of f(1) in the equation which will give the function $f\left( x \right) = \pm {x^n} + 1$. Now substitute the value f(4)=17 to find the exact function. After that calculate the value of $\dfrac{{f\left( 5 \right)}}{{13}}$.

Complete step-by-step answer:
Given:- $f\left( x \right) + f\left( y \right) + f\left( {xy} \right) = 2 + f\left( x \right)f\left( y \right)$ …..(1)
$f\left( 4 \right) = 17$ .
Substitute $y = \dfrac{1}{x}$,
$\Rightarrow$$f\left( x \right) + f\left( {\dfrac{1}{x}} \right) + f\left( 1 \right) = 2 + f\left( x \right)f\left( {\dfrac{1}{x}} \right)$ ….(2)
Now substitute x=1 in the above equation,
$\Rightarrow$$f\left( 1 \right) + f\left( 1 \right) + f\left( 1 \right) = 2 + f\left( 1 \right)f\left( 1 \right)$
Move all terms to one side,
$\Rightarrow$${f^2}\left( 1 \right) - 3f\left( 1 \right) + 2 = 0$
Resolve the equation,
$\Rightarrow$${f^2}\left( 1 \right) - 2f\left( 1 \right) - f\left( 1 \right) + 2 = 0$
Take out common factors,
$\Rightarrow$$f\left( 1 \right)\left[ {f\left( 1 \right) - 2} \right] - 1\left[ {f\left( 1 \right) - 2} \right] = 0$
Take the common factors,
$\Rightarrow$$\left[ {f\left( 1 \right) - 2} \right]\left[ {f\left( 1 \right) - 1} \right] = 0$
Equate $f\left( 1 \right) - 2$ to 0,
$\Rightarrow$$f\left( 1 \right) - 2 = 0$
Move 2 to other sides,
$\Rightarrow$$f\left( 1 \right) = 2$
Now, equate $f\left( 1 \right) - 1$ to 0,
$\Rightarrow$$f\left( 1 \right) - 1 = 0$
Move 1 to other sides,
$\Rightarrow$$f\left( 1 \right) = 1$
Thus, $f\left( 1 \right) = 1,2$
Now, put y=1 in equation (1),
$\Rightarrow$$f\left( x \right) + f\left( 1 \right) + f\left( x \right) = 2 + f\left( x \right)f\left( 1 \right)$
Add the like terms,
$\Rightarrow$$2f\left( x \right) + f\left( 1 \right) = 2 + f\left( x \right)f\left( 1 \right)$
Move all terms to one side,
$\Rightarrow$$f\left( x \right)f\left( 1 \right) - 2f\left( x \right) - f\left( 1 \right) + 2 = 0$
Take out common factors,
$\Rightarrow$$f\left( x \right)\left[ {f\left( 1 \right) - 2} \right] - 1\left[ {f\left( 1 \right) - 2} \right] = 0$
Take the common factors,
$\Rightarrow$$\left[ {f\left( 1 \right) - 2} \right]\left[ {f\left( x \right) - 1} \right] = 0$
Equate $f\left( 1 \right) - 2$ to 0,
$\Rightarrow$$f\left( 1 \right) - 2 = 0$
Move 2 to other sides,
$\Rightarrow$$f\left( 1 \right) = 2$
Now, equate $f\left( x \right) - 1$ to 0,
$\Rightarrow$$f\left( x \right) - 1 = 0$
Move 1 to other sides,
$\Rightarrow$$f\left( x \right) = 1$
It shows that the function $f\left( x \right)$ is constant. But, $f\left( 4 \right) = 17$.
So, it is not possible.
Thus, $f\left( 1 \right) = 2$.
Now, substitute the value of $f\left( 1 \right)$ in equation (2),
$f\left( x \right) + f\left( {\dfrac{1}{x}} \right) + 2 = 2 + f\left( x \right) + f\left( {\dfrac{1}{x}} \right)$
Cancel out 2 from both sides,
$\Rightarrow$$f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = f\left( x \right) + f\left( {\dfrac{1}{x}} \right)$
When the equation is in this form. Then,
$\Rightarrow$$f\left( x \right) = \pm {x^n} + 1$ …..(3)
As $f\left( 4 \right) = 17$. Put $f\left( x \right) = 17$ and x=4,
$17 = \pm {4^n} + 1$
Move 1 to the other side and subtract from 17.
$\Rightarrow$$ \pm {4^n} = 16$
Since, the negative sign is not possible. So, neglect the negative sign,
$\Rightarrow$${4^n} = {4^2}$
Since, the base is the same. So, equate the power,
$\Rightarrow$$n = 2$
Substitute the value of n in equation (3),
$\Rightarrow$$f\left( x \right) = {x^2} + 1$
Now, find $\dfrac{{f\left( 5 \right)}}{{13}}$.
$\Rightarrow$$\dfrac{{f\left( 5 \right)}}{{13}} = \dfrac{{{5^2} + 1}}{{13}}$
Square 5 and add 1 in the numerator,
$\Rightarrow$$\dfrac{{f\left( 5 \right)}}{{13}} = \dfrac{{26}}{{13}}$
Cancel out common factors from numerator and denominator,
$\Rightarrow$$\dfrac{{f\left( 5 \right)}}{{13}} = 2$.

Hence, the value of $\dfrac{{f\left( 5 \right)}}{{13}}$ is 2.

Note: The students might make mistake if they are unaware of the solution to the equation $f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = f\left( x \right) + f\left( {\dfrac{1}{x}} \right)$.
A function is a relation which describes that there should be only one output for each input. We can say that a special kind of relation (a set of ordered pairs) which follows a rule i.e every X-value should be associated with only one y-value is called a function.