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

Hint: 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
\[\begin{align}
  & \Rightarrow f\left( 2 \right)=1-{{2}^{n}} \\
 & \Rightarrow 9=1-{{2}^{n}} \\
 & \Rightarrow {{2}^{n}}=-8 \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow f\left( 2 \right)=1+{{2}^{n}} \\
 & \Rightarrow 9=1+{{2}^{n}} \\
 & \Rightarrow {{2}^{n}}=8 \\
\end{align}\]
Here, by writing 8 as \[{{2}^{3}}\] and comparing the powers on both sides we will get
\[\begin{align}
  & \Rightarrow {{2}^{n}}={{2}^{3}} \\
 & \Rightarrow n=3 \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow f\left( x \right)=1+{{x}^{3}} \\
 & \Rightarrow f\left( 3 \right)=1+{{3}^{3}} \\
 & \Rightarrow f\left( 3 \right)=1+27 \\
 & \Rightarrow f\left( 3 \right)=28 \\
\end{align}\]
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.