Q. If the function $f(x)$ is defined by $f(x) = a + bx$ and $f^{r} = fff$.... (repeated $r$ times), then $f^{r}(x)$ is equal to
A) $a + b^{r}x$
B) $ar + b^{r}x$
C) $ar + bx^{r}$
D) $a\left( \frac{b^{r} - 1}{b - 1} \right) + b^{r}x$
Explanation
(d)
$f(x) = a + bx$
$\therefore\ \ f\left\{ f(x) \right\} = a + b(a + bx) = a(1 + b)b^{2}x$
$\Rightarrow \ \ \ \ \ f\left\lbrack f\left\{ f(x) \right\} \right\rbrack = f\left\{ a(1 + b) + b^{2}x \right\} = a\left( 1 + b + b^{2} \right) + b^{3}x$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f^{r}(x) = a\left( 1 + b + b^{2} + \ldots + b^{r - 1} \right) + b^{r}x$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = a\left( \frac{b^{r} - 1}{b - 1} \right) + b'x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.