Question

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

Answer 1

We have to use the standard result that is if a polynomial $f\left( x \right)$ of degree ‘n’ satisfies the equation $f\left( x \right).f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ we will take the polynomial as $f\left( x \right)=1\pm {{x}^{n}}$. Now, we use $f\left( 2 \right)=9$ to obtain the value of ‘n’ and we find the value of$f\left( 3 \right)$.

Complete step-by-step solution
Let us assume that the given polynomial $f\left( x \right)$ as a polynomial as a degree of ‘n’ which satisfies the equation $f\left( x \right).f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ then we get
$f\left( x \right)=1\pm {{x}^{n}}$……………… equation (i)
Now let us consider $x=2$
By substituting $x=2$ in the equation (i) we get
\Rightarrow $$$$f\left( 2 \right)=1\pm {{2}^{n}}
By substituting $f\left( 2 \right)=9$ in above equation and first let us take negative sign and by solving we get

& \Rightarrow f\left( 2 \right)=1-{{2}^{n}} \\\ & \Rightarrow 9=1-{{2}^{n}} \\\ & \Rightarrow {{2}^{n}}=-8 \\\ \end{aligned}$$ We know that ‘n’ is a natural number and the above result is not valid for any value of ‘n’ negative case fails. Now, by solving the positive case we will get $$\begin{aligned} & \Rightarrow f\left( 2 \right)=1+{{2}^{n}} \\\ & \Rightarrow 9=1+{{2}^{n}} \\\ & \Rightarrow {{2}^{n}}=8 \\\ \end{aligned}$$ Here, by writing 8 as $${{2}^{3}}$$ and comparing the powers on both sides we will get $$\begin{aligned} & \Rightarrow {{2}^{n}}={{2}^{3}} \\\ & \Rightarrow n=3 \\\ \end{aligned}$$ Now, by substituting the value of ‘n’ in equation (i) we get $$\Rightarrow f\left( x \right)=1+{{x}^{3}}$$ ……………..equation (ii) Now, let us find the value of $$f\left( 3 \right)$$ By substituting $$x=3$$ in equation (ii) we get $$\begin{aligned} & \Rightarrow f\left( x \right)=1+{{x}^{3}} \\\ & \Rightarrow f\left( 3 \right)=1+{{3}^{3}} \\\ & \Rightarrow f\left( 3 \right)=1+27 \\\ & \Rightarrow f\left( 3 \right)=28 \\\ \end{aligned}$$ **Therefore the value of $$f\left( 3 \right)$$ is 28.** **Note:** Some students might not remember the general formula if the polynomial satisfies the given equation. Some students will miss taking $$f\left( x \right)=1\pm {{x}^{n}}$$. There are some more standard results that need to be remembered as follows If $$f\left( x \right)$$ satisfies $$f\left( x \right)+f\left( y \right)=f\left( x+y \right)$$ then $$f\left( x \right)=kx$$ where $$k$$ is arbitrary constant. If $$f\left( x \right)$$satisfies $$f\left( x \right).f\left( y \right)=f\left( x+y \right)$$ then $$f\left( x \right)={{a}^{k\left( x+y \right)}}$$ where $$a,k$$ are arbitrary constants. Also we have to take both negative and positive signs when we come across $$'\pm '$$ situations. Some students will miss one sign and proceed for the solution. That is the only point one needs to be taken care of.