Question

For x ∈ R, x ≠ 0, let f₀(x) = `1/(1 - x)` and f_n+1 (x) = f₀(f_n(x)), n = 0, 1, 2, .... Then the value of `f_100(3) + f_1(2/3) + f_2(3/2)` is equal to ______.

Options

• `8/3`

• `4/3`

• `5/3`

• `1/3`

Answer 1

For x ∈ R, x ≠ 0, let f₀(x) = `1/(1 - x)` and f_n+1 (x) = f₀(f_n(x)), n = 0, 1, 2, .... Then the value of `f_100(3) + f_1(2/3) + f_2(3/2)` is equal to `underlinebb(5/3)`.

Explanation:

f₁ (x) = f₀₊₁ (x) = f₀(f₀ (x)) = `1/(1 - 1/(1 - x)) = (x - 1)/x`

f₂(x) = f₁₊₁ (x) = f₀(f₁ (x)) = `1/(1 - (x - 1)/x)` = x

f₃(x) = f₂₊₁ (x) = f₀(f₂ (x)) = f₀(x) = `1/(1 - x)`

f₄(x) = f₃₊₁ (x) = f₀(f₃ (x)) = `(x - 1)/x`

∴ f₀ = f₃ = f₆ = .......... = `1/(1 - x)`

f₁ = f₄ = f₇ = .......... = `(x - 1)/x`

f₂ = f₅ = f₈ = .......... = x

f₁₀₀(3) = `(3 - 1)/3`

= `2/3f_1(2/3)`

= `(2/3 - 1)/(2/3)`

= `-1/2`

`f_2(3/2) = 3/2`

∴ `f_100(3) + f_1(2/3) + f_2(3/2) = 5/3`