A) \[x,\forall x\in R\]
B) \[x-1,\forall x\in R\]
C) \[x+1,\forall x\in R\]
D) None of these
Correct Answer: A
Solution :
[a] Let\[x<0\]. \[\therefore (gof)(x)=g(f(x))=g({{x}^{3}}+1)={{[({{x}^{3}}+1)-1]}^{1/3}}\] \[(\because x<0\Rightarrow {{x}^{3}}+1<1)\] \[={{({{x}^{3}})}^{1/3}}=x\] Let \[x\ge 0.\] \[\therefore (gof)(x)=g(f(x))=g({{x}^{2}}+1)={{(({{x}^{3}}+1)-1)}^{1/2}}\] \[(\because x\ge 0\Rightarrow {{x}^{2}}+1\ge 1)\] \[={{({{x}^{2}})}^{1/2}}=\left| x \right|=x\] \[(\because x\ge 0)\] \[\therefore (gof)(x)=x\forall x\in R\]You need to login to perform this action.
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