Q51 / 120 · VITEEE 2008 · Mathematics
If $g(x)$ is a polynomial satisfying $g(x) g(y)=g(x)+g(y)+g(x y)-2$ for all real $x$ and $y$ and $g(2)=5$ then $\operatorname{Lt}_{x \rightarrow 3} g(x)$ is
- A9
- Correct answerB10
- C25
- D20
Solution
$\mathrm{g}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{y})=\mathrm{g}(\mathrm{x})+\mathrm{g}(\mathrm{y})+\mathrm{g}(\mathrm{x} \mathrm{y})-2 \ldots(1)$
Put $\mathrm{x}=1, \mathrm{y}=2$, then
$\mathrm{g}(1) \cdot \mathrm{g}(2)=\mathrm{g}(1)+\mathrm{g}(2)+\mathrm{g}(2)-2$
$5 \mathrm{~g}(1)=\mathrm{g}(1)+5+5-2$
$4 \mathrm{~g}(1)=8 \quad \therefore \mathrm{g}(1)=2$
Put $y=\frac{1}{x}$ in equation (1), we get
$\begin{array}{l}
g(x) \cdot g\left(\frac{1}{x}\right)=g(x)+g\left(\frac{1}{x}\right)+g(1)-2 \\
g(x) \cdot g\left(\frac{1}{x}\right)=g(x)+g\left(\frac{1}{x}\right)+2-2 \\
\quad[\because g(1)=2]
\end{array}$
This is valid only for the polynomial
$\therefore \quad g(x)=1 \pm x^{n}$ ...(2)
Now $g(2)=5$ (Given)
$\therefore 1 \pm 2^{\mathrm{n}}=5 \quad$ [Using equation $\left.(2)\right]$
$\pm 2^{\mathrm{n}}=4, \Rightarrow 2^{\mathrm{n}}=4,-4$
Since the value of $2^{\mathrm{n}}$ cannot be-Ve.
So, $2^{\mathrm{n}}=4, \Rightarrow \mathrm{n}=2$
Now, put $\mathrm{n}=2$ in equation $(2)$, we get $g(x)=1 \pm x^{2}$
$\begin{aligned}
\therefore \operatorname{Lt}_{x \rightarrow 3} g(x) &=\operatorname{Lt}_{x \rightarrow 3}\left(1 \pm x^{2}\right)=1 \pm(3)^{2} \\
&=1 \pm 9=10,-8
\end{aligned}$
Hence, the correct answer is option: 'B'.