# 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) ${{f}^{'}}(-1)=3$

Last updated date: 24th Mar 2023

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Answer

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

Complete step-by-step answer:

Let $f(x)$ is a polynomial satisfying,

$f(x)f(y)=f(x)+f(y)+f(xy)-2$…….. (1)

Now Let us consider $y=\dfrac{1}{x}$…….(2)

Now let us substitute (2) in (1) we get,

$f(x)f\left( \dfrac{1}{x} \right)=f(x)+f\left( \dfrac{1}{x} \right)+f(1)-2$……….. (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,

$f(1)=1,2$…………… (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$f(1)=2$…………..(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,

$\Rightarrow$ $f(x)=\pm {{x}^{n}}+1$

$\Rightarrow$ $f(4)=\pm {{4}^{n}}+1=65$………….( Given in question that $f(4)=65$)

$\begin{align}

& \pm {{4}^{n}}=64 \\

& \pm {{4}^{n}}={{4}^{3}} \\

\end{align}$……………… (as we know${{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 don’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$f(x)=\pm {{x}^{n}}+1$.

Complete step-by-step answer:

Let $f(x)$ is a polynomial satisfying,

$f(x)f(y)=f(x)+f(y)+f(xy)-2$…….. (1)

Now Let us consider $y=\dfrac{1}{x}$…….(2)

Now let us substitute (2) in (1) we get,

$f(x)f\left( \dfrac{1}{x} \right)=f(x)+f\left( \dfrac{1}{x} \right)+f(1)-2$……….. (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,

$f(1)=1,2$…………… (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$f(1)=2$…………..(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,

$\Rightarrow$ $f(x)=\pm {{x}^{n}}+1$

$\Rightarrow$ $f(4)=\pm {{4}^{n}}+1=65$………….( Given in question that $f(4)=65$)

$\begin{align}

& \pm {{4}^{n}}=64 \\

& \pm {{4}^{n}}={{4}^{3}} \\

\end{align}$……………… (as we know${{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 don’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$f(x)=\pm {{x}^{n}}+1$.

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