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 $\Lt_{x \to 3} g(x)$ is

• A. 9
• B. 10
• C. 25
• D. 20

Answer 1

Answer: (B)

10

Answer 2

$g (x). g(y) = g(x) + g (y) + g (x y) - 2$ ........(1) Put $x = 1 , y = 2,$ then $g (1). g(2) = g (1) + g (2) + g (2) - 2$ $5g (1) = g (1) + 5 + 5 - 2$ $4g (1) = 8$ $\therefore \, \, g(1) = 2$ Put $y = \frac{1}{x}$ in equation (1) , we get $g\left(x\right).g\left(\frac{1}{x}\right) = g\left(x\right) + g\left(\frac{1}{x}\right) + g\left(1\right) - 2$ $g\left(x\right).g\left(\frac{1}{x}\right) = g\left(x\right) + g\left(\frac{1}{x}\right) + 2-2$ \hspace50mm [\because \, g(1) = 2 ] This is valid only for the polynomial $\therefore \, \, g(x) = 1 \pm x^n$ ....(2) Now $g(2) = 5$ (Given) $\therefore \, \, 1 \pm 2^n = 5$ [Using equation (2)] $\pm 2^n = 4 , \Rightarrow \, 2^n = 4 , - 4$ Since the value of 2n cannot be -Ve. So, $2^n = 4 , \Rightarrow \, n = 2$ Now, put n = 2 in equation (2), we get $g(x) = 1 \pm x^2$ $Lt_{x \to 3} g(x) = Lt_{x \to 3} ( 1 \pm x^2) = 1 \pm (3) ^2$ $= 1 \pm 9 = 10, -8$