Questions & Answers

Question

Answers

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

Answer
Verified

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,

$f(x)f(y)=f(x)+f(y)+f(xy)-2$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$f(x)=\pm {{x}^{n}}+1$.

Complete step-by-step answer:

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

×

Sorry!, This page is not available for now to bookmark.