Verified Solution • 2025–26 Session moderate

QJEEM23Mabfe7: JEE Main 2023 Mathematics - Class 12 JEEM23Mabfe7 Solution: Let For , define . If , then is equal to ____________.

Let f1(x)=3x+22x+3,x∈R−{−32}f^{1}(x)=\frac{3 x+2}{2 x+3}, x \in \mathbf{R}-\left\{\frac{-3}{2}\right\} For n≥2\mathrm{n} \geq 2 , define fn(x)=f1ofn−1(x)f^{\mathrm{n}}(x)=f^{1} \mathrm{o} f^{\mathrm{n}-1}(x) . If f5(x)=ax+bbx+a,gcd⁡(a,b)=1f^{5}(x)=\frac{\mathrm{a} x+\mathrm{b}}{\mathrm{b} x+\mathrm{a}}, \operatorname{gcd}(\mathrm{a}, \mathrm{b})=1 , then a+b\mathrm{a}+\mathrm{b} is equal to ____________.

Step-by-Step Verified Solution

f′(x)=3x+22x+3x∈R−{−32}f'(x) = {{3x + 2} \over {2x + 3}}x \in R - \left\{ { - {3 \over 2}} \right\}

f5(x)=fofofofof(x){f^5}(x) = {f_o}{f_o}{f_o}{f_o}f(x)

fof(x)=13x+1212x+13{f_o}f(x) = {{13x + 12} \over {12x + 13}}

fofofofof(x)=1563x+15621562x+1563{f_o}{f_o}{f_o}{f_o}f(x) = {{1563x + 1562} \over {1562x + 1563}}

≡ax+bbx+a\equiv {{ax + b} \over {bx + a}}

∴\therefore a=1563,b=1562a = 1563,b = 1562

=3125= 3125

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring