Question

If f(x) is a polynomial satisfying $f\left( x \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ for all $x\in R$ and $f\left( 3 \right)=-26$ , then find $f\left( 4 \right)$ .
(a) -35
(b) -63
(c) 65
(d) None of these

Answer 1

Hint: Firstly, we have to find the polynomial that satisfies $f\left( x \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ . For this, we will rearrange this equation and find f(x) in terms of $f\left( \dfrac{1}{x} \right)$ and $f\left( \dfrac{1}{x} \right)$ in terms of f(x). Then, we will multiply these two equations and simplify. We will then substitute $g\left( x \right)=f\left( x \right)-1$ and simplify the equation accordingly which will result in an equation of the form $g\left( x \right)g\left( \dfrac{1}{x} \right)=1$ . We will assume $g\left( x \right)=\pm {{x}^{n}}$ and find the value of f(x). We will get two values for f(x) because of the $\pm $ sign. From the given result, $f\left( 3 \right)=-26$ , we will get the right value of the function f(x). Finally, we have to find the value of f(4) by substituting $x=4$ in this polynomial.

Complete step by step answer:
We are given that f(x) is a polynomial satisfying $f\left( x \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ . We have to find this polynomial. Let us consider $f\left( x \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)...\left( i \right)$ .
 We have to take f(x) from the RHS to the LHS.
$\Rightarrow f\left( x \right)f\left( \dfrac{1}{x} \right)-f\left( x \right)=f\left( \dfrac{1}{x} \right)$
We have to take the common term outside.
$\Rightarrow f\left( x \right)\left( f\left( \dfrac{1}{x} \right)-1 \right)=f\left( \dfrac{1}{x} \right)$
Let us take the coefficient of f(x) to the RHS.
$\Rightarrow f\left( x \right)=\dfrac{f\left( \dfrac{1}{x} \right)}{f\left( \dfrac{1}{x} \right)-1}...\left( ii \right)$
Now, let us take $f\left( \dfrac{1}{x} \right)$ from the RHS to the LHS in equation (i).
$\Rightarrow f\left( x \right)f\left( \dfrac{1}{x} \right)-f\left( \dfrac{1}{x} \right)=f\left( x \right)$
We have to take the common term outside.
$\Rightarrow \left( f\left( x \right)-1 \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)$
Let us take the coefficient of $f\left( \dfrac{1}{x} \right)$ to the RHS.
$\Rightarrow f\left( \dfrac{1}{x} \right)=\dfrac{f\left( x \right)}{f\left( x \right)-1}...\left( iii \right)$
Now, we have to multiply (ii) and (iii).

$\begin{align}
  & \Rightarrow f\left( x \right)f\left( \dfrac{1}{x} \right)=\dfrac{f\left( \dfrac{1}{x} \right)}{f\left( \dfrac{1}{x} \right)-1}\left( \dfrac{f\left( x \right)}{f\left( x \right)-1} \right) \\
 & \Rightarrow \require{cancel}\cancel{f\left( x \right)f\left( \dfrac{1}{x} \right)}=\dfrac{\require{cancel}\cancel{f\left( \dfrac{1}{x} \right)f\left( x \right)}}{\left[ f\left( \dfrac{1}{x} \right)-1 \right]\left( f\left( x \right)-1 \right)} \\
 & \Rightarrow 1=\dfrac{1}{\left[ f\left( \dfrac{1}{x} \right)-1 \right]\left( f\left( x \right)-1 \right)} \\
\end{align}$
Let us move the denominator of the RHS to the LHS.
$\Rightarrow \left[ f\left( x \right)-1 \right]\left[ f\left( \dfrac{1}{x} \right)-1 \right]=1...\left( iv \right)$
Let us assume $g\left( x \right)=f\left( x \right)-1...\left( v \right)$ .
Then, we can find $g\left( \dfrac{1}{x} \right)$ by substituting $x=\dfrac{1}{x}$ in this equation.
$\Rightarrow g\left( \dfrac{1}{x} \right)=f\left( \dfrac{1}{x} \right)-1...\left( vi \right)$
Let us substitute (v) and (vi) in equation (iv).
$\Rightarrow g\left( x \right)g\left( \dfrac{1}{x} \right)=1$
If we assume g(x) to be a polynomial, $g\left( x \right)=\pm {{x}^{n}}$ , the above condition will be satisfied. Let us substitute this value of g(x) in the equation (v).
$\begin{align}
  & \Rightarrow \pm {{x}^{n}}=f\left( x \right)-1 \\
 & \Rightarrow f\left( x \right)=1\pm {{x}^{n}}...\left( vii \right) \\
\end{align}$
Therefore, we obtained a polynomial f(x) satisfying $f\left( x \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ . We can consider two values for x as positive or negative. Therefore, the polynomial will be either $f\left( x \right)=1+{{x}^{n}}$ or $f\left( x \right)=1-{{x}^{n}}$ . We are given that $f\left( 3 \right)=-26$ . Let us substitute $x=3$ in each of this polynomial.
$\begin{align}
  & \Rightarrow f\left( x \right)=1+{{x}^{n}} \\
 & \Rightarrow f\left( 3 \right)=1+{{3}^{n}} \\
 & \Rightarrow -26=1+{{3}^{n}} \\
 & \Rightarrow -26-1={{3}^{n}} \\
 & \Rightarrow -27={{3}^{n}} \\
\end{align}$
We can write -27 in terms of powers of 3.
$\Rightarrow -{{3}^{9}}={{3}^{n}}$
We cannot compare the exponents of similar bases in the above equation since the LHS contains a negative sign. Therefore, $f\left( x \right)=1+{{x}^{n}}$ is not a polynomial.
Now, let us consider $f\left( x \right)=1-{{x}^{n}}$ .
$\begin{align}
  & \Rightarrow f\left( 3 \right)=1-{{3}^{n}} \\
 & \Rightarrow -26=1-{{3}^{n}} \\
 & \Rightarrow -26-1=-{{3}^{n}} \\
 & \Rightarrow \require{cancel}\cancel{-}27=\require{cancel}\cancel{-}{{3}^{n}} \\
 & \Rightarrow 27={{3}^{n}} \\
\end{align}$
We can write -27 in terms of powers of 3.
$\Rightarrow {{3}^{3}}={{3}^{n}}$
When, comparing the exponents of similar bases, we will get
$\Rightarrow n=3$
Therefore, the polynomial satisfying $f\left( x \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ is $f\left( x \right)=1-{{x}^{3}}$ . Now, let us find $f\left( 4 \right)$ by substituting for x as 4 in the above polynomial.
$\begin{align}
  & \Rightarrow f\left( 4 \right)=1-{{4}^{3}} \\
 & \Rightarrow f\left( 4 \right)=1-64 \\
 & \Rightarrow f\left( 4 \right)=-63 \\
\end{align}$

So, the correct answer is “Option b”.

Note: Students must be through with the concept of algebra and exponents. They must also know to find the value of a function for a given value of its variable. Students can also find the polynomial f(x) using trial and error method. This is shown below.
We are given with $f\left( 3 \right)=-26$ . Let us substitute $x=3$ in $f\left( x \right)f\left( \dfrac{1}{x} \right)=f\left( x \right)+f\left( \dfrac{1}{x} \right)$ .
$\begin{align}
  & \Rightarrow f\left( 3 \right)f\left( \dfrac{1}{3} \right)=f\left( 3 \right)+f\left( \dfrac{1}{3} \right) \\
 & \Rightarrow -26\times f\left( \dfrac{1}{3} \right)=-26+f\left( \dfrac{1}{3} \right) \\
\end{align}$
Let us take $f\left( \dfrac{1}{3} \right)$ to the LHS.
$\begin{align}
  & \Rightarrow -26\times f\left( \dfrac{1}{3} \right)-f\left( \dfrac{1}{3} \right)=-26 \\
 & \Rightarrow \left( -26-1 \right)f\left( \dfrac{1}{3} \right)=-26 \\
 & \Rightarrow -27\times f\left( \dfrac{1}{3} \right)=-26 \\
\end{align}$
Let us move the coefficient of $f\left( \dfrac{1}{x} \right)$ to the RHS.
$\begin{align}
  & \Rightarrow f\left( \dfrac{1}{3} \right)=\dfrac{-26}{-27} \\
 & \Rightarrow f\left( \dfrac{1}{3} \right)=\dfrac{26}{27}...\left( a \right) \\
\end{align}$
We can guess that $f\left( 3 \right)=-{{3}^{3}}+1=-26$ . Therefore, we can find $f\left( \dfrac{1}{3} \right)$ in similar way as
$\begin{align}
  & \Rightarrow f\left( \dfrac{1}{3} \right)=-{{\left( \dfrac{1}{3} \right)}^{3}}+1 \\
 & \Rightarrow f\left( \dfrac{1}{3} \right)=-\dfrac{1}{27}+1 \\
 & \Rightarrow f\left( \dfrac{1}{3} \right)=\dfrac{-1+27}{27} \\
 & \Rightarrow f\left( \dfrac{1}{3} \right)=\dfrac{26}{27}...\left( b \right) \\
\end{align}$
We can see that (a) and (b) are equal. Hence, the polynomial will be $f\left( x \right)=1-{{x}^{3}}$ .