Question

If g (x) is a polynomial satisfying
g (x) g(y) = g(x) + g(y) + g(xy) - 2
for all real x and y and g (2) = 5 then $\underset{\text{x $\rightarrow$ 3}}{\ce{Lt }}$ g(x)is

• A. 9
• B. 10
• C. 25
• D. 20

Answer 1

Answer: (B)

g (x). g(y) = g(x) + g (y) + g (x y) - 2 ...(1)
Put x = 1, y = 2, then
g (1). g(2) = g (1) + g (2) + g (2) - 2
5g (1) = g (1) + 5 + 5 - 2
4g (1) = 8 $\quad$$\quad$$\therefore $ g(1) = 2
Put y =$\frac{1}{x}$ in equation (1) , we get
g(x).g $\left(\frac{1}{x}\right)$ =g(x) +g$\left(\frac{1}{x}\right)$ g(1) -2
g(x).g $\left(\frac{1}{x}\right)$ =g(x) +g +2 -2
$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$[$\therefore $ g(1) = 2 ]
This is valid only for the polynomial
$\therefore $$\quad$$\quad$g (x) = 1 $\pm$ x$^n$$\quad$$\quad$ ... (2)
Now g (2) = 5$\quad$$\quad$$\quad$$\quad$(Given)
$\therefore $ 1 $\pm$2n = 5$\quad$$\quad$[Using equation (2)]
$\quad$$\quad$$\quad$$\quad$ $\pm$2$^n$ = 4, $\Rightarrow $ 2$^n$ = 4, -4
Since the value of 2$^n$ cannot be -Ve.
So, 2$^n$ = 4, $\Rightarrow $ n = 2
Now, put n = 2 in equation (2), we get
g (x) = 1 + x$^2$
$\therefore $ $\underset{\text{x $\rightarrow$ 3}}{\ce{Lt }}$ g(x) = $\underset{\text{x $\rightarrow$ 3}}{\ce{Lt }}$ (1$\pm$x$^2$) =1$\pm$(3)$^2$
$\quad$$\quad$$\quad$$\quad$=1$\pm$9 = 10, - 8