Question

# Let$f(x)$ be a polynomial with positive degree satisfying the relation $f(x)f(y)=f(x)+f(y)+f(xy)-2$For all real x and y. Suppose$f(4)=65$ Then(A) ${{f}^{'}}(x)$ is a polynomial of degree two(B) roots of equation ${{f}^{'}}(x)=2x+1$ are real(C) $x{{f}^{'}}(x)=3\left[ f(x)-1 \right]$(D) ${{f}^{'}}(-1)=3$

Hint: First convert the equation in one form and assume$y=\dfrac{1}{x}$and solve it further.

Let $f(x)$ is a polynomial satisfying,
$f(x)f(y)=f(x)+f(y)+f(xy)-2$â€¦â€¦.. (1)
Now Let us consider $y=\dfrac{1}{x}$â€¦â€¦.(2)
Now let us substitute (2) in (1) we get,
$f(x)f\left( \dfrac{1}{x} \right)=f(x)+f\left( \dfrac{1}{x} \right)+f(1)-2$â€¦â€¦â€¦.. (3)
Let us take $x=1$ So substituting the value of $x$ in (3),
So we get ,
$\Rightarrow$ $f{{(1)}^{2}}=3f(1)-2$
Simplifying we get,
\begin{align} & f{{(1)}^{2}}-3f(1)+2=0 \\ & f{{(1)}^{2}}-2f(1)-f(1)+2=0 \\ & f(1)(f(1)-2)-(f(1)-2)=0 \\ & (f(1)-1)(f(1)-2)=0 \\ \end{align}
So by solving we get two values for$f(1)$,
So the values for$f(1)$ are as follows,
$f(1)=1,2$â€¦â€¦â€¦â€¦â€¦ (4)
Let us take $y=1$and substituting in (1),
So we get,
$\Rightarrow$ $f(x)f(1)=f(x)+f(1)+f(x)-2$
$\Rightarrow$ $f(x)f(1)=2f(x)+f(1)-2$
So rearranging the equation we get,
$\Rightarrow$ $(f(x)-1)(f(1)-2)=0$
So here $f(x)\ne 1$ and we can say that$f(1)=2$â€¦â€¦â€¦â€¦..(5)
So from equation (4) and (5) we get to know that $f(1)=2$,
So substituting $f(1)=2$in (3) we get,
So we get,
$\Rightarrow$ $f(x)f\left( \dfrac{1}{x} \right)=f(x)+f\left( \dfrac{1}{x} \right)+2-2$
$\Rightarrow$ $f(x)f\left( \dfrac{1}{x} \right)=f(x)+f\left( \dfrac{1}{x} \right)$
So $f(x)$ is a polynomial function, let us consider it as,
$\Rightarrow$ $f(x)=\pm {{x}^{n}}+1$
$\Rightarrow$ $f(4)=\pm {{4}^{n}}+1=65$â€¦â€¦â€¦â€¦.( Given in question that $f(4)=65$)
\begin{align} & \pm {{4}^{n}}=64 \\ & \pm {{4}^{n}}={{4}^{3}} \\ \end{align}â€¦â€¦â€¦â€¦â€¦â€¦ (as we know${{4}^{3}}=64$so writing${{4}^{3}}$instead of$64$)
So we get the value of $n$ as $3$,
So we get$f(x)$as,
$f(x)={{x}^{3}}+1$
So differentiating $f(x)$ We get,
So we get${{f}^{'}}(x)$as,
${{f}^{'}}(x)=3{{x}^{2}}$
So considering option (A),
${{f}^{'}}(x)=3{{x}^{2}}$
So it has a polynomial of degree two. Option (A) is correct,
Now for option (B) it is mentioned that${{f}^{'}}(x)$ is Real ,
So${{f}^{'}}(x)=3{{x}^{2}}$ so it is real, if we put any value we will get ${{f}^{'}}(x)$as real.
So option (B) is correct.
Now considering Option (C) We get
$x{{f}^{'}}(x)=3\left[ f(x)-1 \right]$
Let us take $x=1$
We get LHS$=$RHS
Option (C) is also correct.
For Option (D) it is given that${{f}^{'}}(-1)=3$
So we have found${{f}^{'}}(x)$above
So${{f}^{'}}(x)=3{{x}^{2}}$
So Substituting $x=-1$ in${{f}^{'}}(x)$ We get,
${{f}^{'}}(-1)=3$
Hence Option (D) is also correct.
So here all options are correct.
Option (A), (B), (C) and (D) are correct.

Note: While solving be careful of what you are supposed to substitute. Also donâ€™t jumble yourself and use proper signs and assumptions. Use the polynomial as given in question. Use proper substitution as we had used $y=\dfrac{1}{x}$. So be careful about solving all the options and proving it right or wrong. You should not make a mistake at simplifying this one$f(x)=\pm {{x}^{n}}+1$.