QJEEM16Ma17cb: JEE Main 2016 Mathematics - Class 11 JEEM16Ma17cb Solution: For x R , x 0, Let f 0 (x) = and f n+1 (x) = f 0 (f n (x)), n = …

As   f_n+1(x) = f₀(f_n(x))

$\therefore$    f₁(x) = f₀₊₁(x) = f₀(f₀(x)) = ${1 \over {1 - {1 \over {1 - x}}}}$ = ${{x - 1} \over x}$

f₂(x) = f₁₊₁(x) = f₀(f₁(x)) = ${1 \over {1 - {{x - 1} \over x}}}$ = x

f₃(x) = f₂₊₁(x) = f₀(f₂(x)) = ${1 \over {1 - x}}$

f₄(x) = f₃₊₁(x) = f₀(f₃(x)) = ${1 \over {1 - {1 \over {1 - x}}}}$ = ${{x - 1} \over x}$

$\therefore$ f₀(x) = f₃(x) = f₆(x) = . . . . . = ${1 \over {1 - x}}$

f₁(x) = f₄(x) = f₇(x) = . . . . . .= ${{x - 1} \over x}$

f₂(x) = f₅(x) = f₈(x) = . . . . . . = x

So, f₁₀₀(3) = ${{3 - 1} \over 3}$ = ${2 \over 3}$

f₁ $\left( {{2 \over 3}} \right)$ = ${{{2 \over 3} - 1} \over {{2 \over 3}}}$ = $-$ ${1 \over 2}$

f₂ $\left( {{3 \over 2}} \right)$ = ${{3 \over 2}}$

$\therefore$    f₁₀₀(3) + f₁ $\left( {{2 \over 3}} \right)$ + f₂ $\left( {{3 \over 2}} \right)$

= ${2 \over 3}$ $-$ ${1 \over 2}$ + ${3 \over 2}$

= ${5 \over 3}$