JEE MAIN - Mathematics (2023 - 30th January Morning Shift - No. 16)
Let $$f^{1}(x)=\frac{3 x+2}{2 x+3}, x \in \mathbf{R}-\left\{\frac{-3}{2}\right\}$$ For $$\mathrm{n} \geq 2$$, define $$f^{\mathrm{n}}(x)=f^{1} \mathrm{o} f^{\mathrm{n}-1}(x)$$. If $$f^{5}(x)=\frac{\mathrm{a} x+\mathrm{b}}{\mathrm{b} x+\mathrm{a}}, \operatorname{gcd}(\mathrm{a}, \mathrm{b})=1$$, then $$\mathrm{a}+\mathrm{b}$$ is equal to ____________.
Answer
3125
Explanation
$$f'(x) = {{3x + 2} \over {2x + 3}}x \in R - \left\{ { - {3 \over 2}} \right\}$$
$${f^5}(x) = {f_o}{f_o}{f_o}{f_o}f(x)$$
$${f_o}f(x) = {{13x + 12} \over {12x + 13}}$$
$${f_o}{f_o}{f_o}{f_o}f(x) = {{1563x + 1562} \over {1562x + 1563}}$$
$$ \equiv {{ax + b} \over {bx + a}}$$
$$\therefore$$ $$a = 1563,b = 1562$$
$$ = 3125$$
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