Question

If f(x) is a polynomial function satisfying f\left( x \right)\times f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)\forall x\in R-\left\\{ 0 \right\\} and$f\left( 2 \right)=9$, then find$f\left( 3 \right)$?

Answer 1

In the given question, we have been asked to find the value of$f\left( 3 \right)$. In order to answer the question, first we need to write a polynomial which satisfies $f\left( x \right)\times f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ is$\pm {{x}^{n}}+1$. Then after putting the value in the standard form of polynomial we have written earlier, we will get the value of n. after substituting the value of ‘x’ and ‘n’ . We will get the value of$f\left( 3 \right)$.

Complete answer:
We have the given polynomial,
$\Rightarrow f\left( x \right)\times f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$
The polynomial or a function which satisfies$f\left( x \right)\times f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$,
$\Rightarrow \pm {{x}^{n}}+1$, this is the standard form.
It is given that,
$\Rightarrow f\left( 2 \right)=9$
Here x=2,
Writing it in standard form, we get
$\Rightarrow \pm {{2}^{n}}+1=9$
Simplifying the above, we get
$\Rightarrow {{2}^{n}}=8$ (Negative sign is not possible)
As, we know that ${{2}^{3}}=8$,
Replacing it in above equation, we get
$\Rightarrow {{2}^{n}}={{2}^{3}}$
Solving for the value of n, we get
$\Rightarrow n=3$
Therefore,
The function is $f\left( x \right)={{x}^{3}}+1$
For find the value of $f\left( 3 \right)$, we will substitute x=3
Thus,
$\Rightarrow f\left( 3 \right)={{3}^{3}}+1=27+1=28$
$\therefore f\left( 3 \right)=28$
It is the required answer.

Note:
While doing these types of questions, it may have seen many times that students are finding the wrong answer. Students often made mistakes in each step of the question as they were not able to substitute the correct value of ‘x’. They are not able to substitute the value of ‘x’ is the correct variable, they get confused and hence they will get the wrong answer. Students should carefully examine each and every step of the solution and do each calculation explicitly to avoid making errors.