Let f^1(x) = frac{3x + 2}{2x + 3},x in R - le | vedclass

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(D) Given $f^1(x) = \frac{3x + 2}{2x + 3}$.Calculate $f^2(x) = f^1(f^1(x)) = \frac{3(\frac{3x+2}{2x+3}) + 2}{2(\frac{3x+2}{2x+3}) + 3} = \frac{9x + 6

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$3125$

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(D) Given $f^1(x) = \frac{3x + 2}{2x + 3}$.Calculate $f^2(x) = f^1(f^1(x)) = \frac{3(\frac{3x+2}{2x+3}) + 2}{2(\frac{3x+2}{2x+3}) + 3} = \frac{9x + 6 + 4x + 6}{6x + 4 + 6x + 9} = \frac{13x + 12}{12x + 13}$.Calculate $f^3(x) = f^1(f^2(x)) = \frac{3(\frac{13x+12}{12x+13}) + 2}{2(\frac{13x+12}{12x+13}) + 3} = \frac{39x + 36 + 24x + 26}{26x + 24 + 36x + 39} = \frac{63x + 62}{62x + 63}$.Observe the pattern: $f^n(x) = \frac{(2^{2n}-1)x + (2^{2n}-2)}{2(2^{2n}-2)x + (2^{2n}-1)}$ is not quite right,let's look at the coefficients: $a_n = 3, 13, 63, \dots$ where $a_n = \frac{3^n + 2^n}{2}$? No,let $f^n(x) = \frac{A_n x + B_n}{B_n x + A_n}$.For $n=1, A_1=3, B_1=2$. For $n=2, A_2=13, B_2=12$. For $n=3, A_3=63, B_3=62$.The recurrence is $A_n = 3A_{n-1} + 2B_{n-1}$ and $B_n = 2A_{n-1} + 3B_{n-1}$.Adding these: $A_n + B_n = 5(A_{n-1} + B_{n-1})$.Since $A_1 + B_1 = 3 + 2 = 5$,then $A_n + B_n = 5^n$.For $n=5$,$A_5 + B_5 = 5^5 = 3125$.

Let $f^1(x) = \frac{3x + 2}{2x + 3}$,$x \in R - \left\{-\frac{3}{2}\right\}$. For $n \geq 2$,define $f^n(x) = f^1 \circ f^{n-1}(x)$. If $f^5(x) = \frac{ax + b}{bx + a}$ and $\gcd(a, b) = 1$,then $a + b$ is equal to $............$.

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