Functions Question 547
Question: If $ f(x)= \begin{cases} x^{3}+1,x<0 \\ x^{2}+1,x\ge 0 \\ \end{cases},g(x)= \begin{cases} {{(x-1)}^{1/3}},x<1 \\ {{(x-1)}^{1/3}},x\ge 1 \\ \end{cases}, . . $ Then (gof) (x) is equal to
Options:
A) $ x,\forall x\in R $
B) $ x-1,\forall x\in R $
C) $ x+1,\forall x\in R $
D) None of these
Show Answer
Answer:
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}}=| x |=x $ $ (\because x\ge 0) $
$ \therefore (gof)(x)=x\forall x\in R $
BETA
