A) \[1+2{{x}^{2}}\]
B) \[2+{{x}^{2}}\]
C) \[1+x\]
D) \[2+x\]
Correct Answer: B
Solution :
\[g(x)=1+\sqrt{x}\] and \[f(g(x))=3+2\sqrt{x}+x\] .....(i) Þ \[f(1+\sqrt{x})=3+2\sqrt{x}+x\] Put \[1+\sqrt{x}=y\] Þ \[x={{(y-1)}^{2}}\] then, \[f(y)=3+2(y-1)+{{(y-1)}^{2}}\]\[=2+{{y}^{2}}\] therefore, \[f(x)=2+{{x}^{2}}\].
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