VITEEE 2016 Paper

g (x). g(y) = g(x) + g (y) + g (xy)-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 ∴ 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 $\left(\frac{1}{x}\right)$ + 2 - 2 [Since g (1) = 2] This is valid only for the polynomial ∴ g (x) = 1 ± $x^{n}$ ... (2) Now g (2) = 5 (Given) ∴ 1 ± $2^{n}$ - 5 [Using equation (2)] ± $2^{n}$ = 4 ⇒ $2^{n}$ = 4 , - 4 Since the value of $2^{n}$ cannot be -Ve. So, $2^{n}$ = 4 ⇒ n = 2 Now, put n = 2 in equation (2), we get g (x) = 1 ± $x^{2}$ ∴ $\lim\limits_{x\rightarrow3}$ $\left(1\pm x^{2}\right)$ = 1 ± $(3)^{2}$ = 1 ± 9 = 10 , - 8