Discussion Forum

1)

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  $Lt_{x \rightarrow 3}g(x)$  is 

Answer:

Option B

Explanation:

 g(x).g(y) =g(x)+g(y)+g(xy)-2  ...(i)

 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  (i) , we get

 g(x) , g($\frac{1}{x}$ )= g(x) + g( $\frac{1}{x}$)+g(1)-2

g(x) . g($\frac{1}{x}$) = g(x) + g($\frac{1}{x}$)+2-2

                                                      [$\because$ g(1)=2]

 This valid only for the polynomial 

  $\therefore$ g(x) = 1$\pm$xⁿ .................(2)

 Now g(2)=5      (Given)

 $\therefore$   1$\pm$ 2ⁿ= 5     [Using equation(2)]

 $\pm$  2ⁿ =4, $\Rightarrow$ 2ⁿ =4,-4

 Since , the  value of 2ⁿ cannot be -Ve

 So , 2ⁿ =4 . $\Rightarrow$  n=2

 Now, put  n=2 in equation (2) , we get

 g(x) =1  $\pm$ x²

 $\therefore Lt_{x \rightarrow 3}g(x)=Lt_{x \rightarrow 3} (1\pm x^{2})= 1\pm (3)^{2}$

      =  $ 1\pm 9=10,-8$