Question

If $g\left(x\right)$ is a polynomial such that $g\left(x\right)·g\left(y\right)=g\left(x\right)+g\left(y\right)+g\left(xy\right)-2$ for all real $x$ and $y$ and $g\left(2\right)=5$, then$\underset{x\to 2}{\lim }g\text{'}\left(x\right)$ is

• A. $2$
• B. $3$
• C. $4$
• D. none of these

Answer 1

The correct option is C

$4$

Explanation for correct option:

Step 1: Solve for the required polynomial:

Given: $g\left(x\right)·g\left(y\right)=g\left(x\right)+g\left(y\right)+g\left(xy\right)-2$ and $g\left(2\right)=5$

Put $x=2,y=1$

$$\begin{gathered} \Rightarrow g\left(2\right)·g\left(1\right)=g\left(2\right)+g\left(1\right)+g(2·1)-2 \\ \Rightarrow 5·g\left(1\right)=5+g\left(1\right)+g\left(2\right)-2 \\ \Rightarrow 5g\left(1\right)-g\left(1\right)=3+5 \\ \Rightarrow g\left(1\right)=2 \end{gathered}$$

put $y=\frac{1}{x}$

$$\begin{gathered} \Rightarrow g\left(x\right)·g\left(\frac{1}{x}\right)=g\left(x\right)+g\left(\frac{1}{x}\right)+g\left(1\right)-2 \\ \Rightarrow g\left(x\right)·g\left(\frac{1}{x}\right)=g\left(x\right)+g\left(\frac{1}{x}\right)+2-2 \\ \Rightarrow g\left(x\right)g\left(\frac{1}{x}\right)-g\left(x\right)-g\left(\frac{1}{x}\right)+1=1 \\ \Rightarrow g\left(x\right)\left[g\left(\frac{1}{x}\right)-1\right]-1\left[g\left(\frac{1}{x}\right)-1\right]=1 \\ \Rightarrow \left(g\left(x\right)-1\right)\left(g\left(\frac{1}{x}\right)-1\right)=1 \end{gathered}$$

Since, $g\left(x\right)$ is a polynomial so let $g\left(x\right)-1=x^n$

$\Rightarrow g\left(x\right)=x^n+1$

Step 2: Solve for the required value

$g\left(x\right)=x^n+1$

Now, $g\left(2\right)=5$

$$\begin{gathered} \Rightarrow 2^n+1=5 \\ \Rightarrow 2^n=4 \\ \Rightarrow 2^n=2^2 \\ \Rightarrow n=2 \end{gathered}$$

so, $g\left(x\right)=x^2+1$

therefore, $g\text{'}\left(x\right)=2x$

$$\begin{gathered} \underset{x\to 2}{\lim }g\text{'}\left(x\right)=2x \\ \Rightarrow \underset{x\to 2}{\lim }g\text{'}\left(x\right)=2\times 2=4 \end{gathered}$$

Hence, option(C) is correct.