Bonjour,
1) AR = (x - 1) et AP = (y - 1)
2) On sait que : AR x AP = 2
⇔ (x - 1)(y - 1) = 2
⇔ (y - 1) = 2/(x - 1)
⇔ y = 2/(x - 1) + 1
3) a) AP = y - 1 = 2/(x - 1) + 1 - 1 = 2/(x - 1)
b) AP ≥ 2
⇔ 2/(x - 1) ≥ 2
⇔ 2/(x - 1) - 2 ≥ 0
⇔ 2/(x - 1) - 2(x - 1)/(x - 1) ≥ 0
⇔ [2 - 2(x - 1)]/(x - 1) ≥ 0
⇔ (2 - 2x + 2)/(x - 1) ≥ 0
⇔ (-2x + 4)/(x - 1) ≥ 0
c)
x 0 1 2 +∞
(-2x + 4) + + 0 -
(x - 1) - 0 + +
Quotient - || + 0 -
donc (-2x + 4)/(x - 1) ≥ 0 ⇒ x ∈ ]1 ; 2]
⇒ AP ≥ 2 pour x ∈ ]1 ; 2]
