JEE Main 2023DifferentiationDifferentiationEasyQuestionLet 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\}f1(x)=2x+33x+2,x∈R−{2−3} For n≥2\mathrm{n} \geq 2n≥2, define fn(x)=f1ofn−1(x)f^{\mathrm{n}}(x)=f^{1} \mathrm{o} f^{\mathrm{n}-1}(x)fn(x)=f1ofn−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})=1f5(x)=bx+aax+b,gcd(a,b)=1, then a+b\mathrm{a}+\mathrm{b}a+b is equal to ____________.Answer: 3Hide SolutionSolutionf′(x)=3x+22x+3x∈R−{−32}f'(x) = {{3x + 2} \over {2x + 3}}x \in R - \left\{ { - {3 \over 2}} \right\}f′(x)=2x+33x+2x∈R−{−23} f5(x)=fofofofof(x){f^5}(x) = {f_o}{f_o}{f_o}{f_o}f(x)f5(x)=fofofofof(x) fof(x)=13x+1212x+13{f_o}f(x) = {{13x + 12} \over {12x + 13}}fof(x)=12x+1313x+12 fofofofof(x)=1563x+15621562x+1563{f_o}{f_o}{f_o}{f_o}f(x) = {{1563x + 1562} \over {1562x + 1563}}fofofofof(x)=1562x+15631563x+1562 ≡ax+bbx+a \equiv {{ax + b} \over {bx + a}}≡bx+aax+b ∴\therefore∴ a=1563,b=1562a = 1563,b = 1562a=1563,b=1562 =3125 = 3125=3125