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1.Relation and Function
hard
माना $\mathrm{f}^1(\mathrm{x})=\frac{3 \mathrm{x}+2}{2 \mathrm{x}+3}, \mathrm{x} \in \mathrm{R}-\left\{\frac{-3}{2}\right\}$ है $\mathrm{n} \geq 2$ के लिए $\mathrm{f}^{\mathrm{n}}(\mathrm{x})=\mathrm{f}^1 0 \mathrm{f}^{\mathrm{n}-1}(\mathrm{x})$ द्वारा परिभाषित कीजिए। यदि $\mathrm{f}^5(\mathrm{x})=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{bx}+\mathrm{a}}, \operatorname{gcd}(\mathrm{a}, \mathrm{b})=1$, है, तो $\mathrm{a}+\mathrm{b}$ बराबर है_________.
A
$3124$
B
$3123$
C
$3126$
D
$3125$
(JEE MAIN-2023)
Solution
$f^1(x)=\frac{3 x+2}{2 x+3}$
$\Rightarrow f^2(x)=\frac{13 x+12}{12 x+13}$
$\Rightarrow f^3(x)=\frac{63 x+62}{62 x+63}$
$\therefore f^5(x)=\frac{1563 x+1562}{1562 x+1563}$
$a+b=3125$
Standard 12
Mathematics
