A) 9
B) 10
C) 25
D) 20
Correct Answer: B
Solution :
Since, \[g(x)g(y)=g(x)+g(y)+g(xy)-2\] Now, at \[x=0,\] \[y=2,\] we get \[g(0)g(2)=g(0)+g(2)+g(0)-2\] \[[\because g(2)=5]\] \[\Rightarrow \] \[5g(0)=5+2g(0)-2\] \[\Rightarrow \] \[3g(0)=3\] \[\Rightarrow \] \[g(0)=1\] \[g(x)\] is given in a polynomial and by the relation given\[g(x)\]cannot be linear. Let \[g(x)={{x}^{2}}+5\] \[\Rightarrow \] \[g(x)={{x}^{2}}+1\] \[[\because g(0)=1]\] From Eq. (i), \[({{x}^{2}}+1)({{y}^{2}}+1)={{x}^{2}}+1+{{y}^{2}}+1+{{x}^{2}}{{y}^{2}}+1-2\] \[\therefore \underset{x\to 3}{\mathop{\lim }}\,g(x)=g(3)={{3}^{2}}+1=10\]
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