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