A) \[a+{{b}^{r}}x\]
B) \[ar+{{b}^{r}}x\]
C) \[ar+b{{x}^{r}}\]
D) \[a({{b}^{r}}-1)+{{b}^{r}}x\]
E) \[a\left( \frac{{{b}^{r}}-1}{b-1} \right)+{{b}^{r}}x\]
Correct Answer: E
Solution :
\[f(x)=a+bx\] \[\therefore \] \[f\{f(x)\}=a+b(a+bx)\] \[=ab+a+{{b}^{2}}x\] \[=a(1+b)+{{b}^{2}}x\] \[\Rightarrow \]\[f[f\{f(x)\}]=f\{a(1+b)+{{b}^{2}}x\}\] \[=a+b[a(1+b)+{{b}^{2}}x]\] \[=a(1+b+{{b}^{2}})+{{b}^{3}}x\] \[\therefore \]\[{{f}^{r}}(x)=a(1+b+{{b}^{2}}+....+{{b}^{r-1}})+{{b}^{r}}x\] \[=a\left( \frac{{{b}^{r}}-1}{b-1} \right)+{{b}^{r}}x\]
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