Zadanie 1e, Paper 3, maj 2023, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2023, TZ2, Paper 3 (zadania wzorowane), pytanie 1(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
For x>0 and a>0 with a≠1, let fa(x) = loga x = ln x/ln a. This investigation compares its graph with the parabola y = x2 + e2, restricted to x>0. A point of contact is a common point with equal gradients. There is a unique base at which contact occurs.
Polecenie
The unique contact base lies between 1.05 and 1.1. Find the exact coordinates of the contact point P and the exact base a. You may use the fact that the resulting contact equation has exactly one solution with x>1; an exact candidate must still be verified in that equation.
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Odpowiedź z klucza
The two derivatives are 1/(x ln a) and 2x. Equal gradients imply ln a=1/(2x²). Substituting this into ln x/ln a=x²+e² gives x2(2ln x − 1) = e2. The value x=e satisfies this exactly because e²(2−1)=e². By the supplied uniqueness, x=e is the contact abscissa. Thus y=2e² and ln a=1/(2e²), giving a = exp(1/(2e2)) ≈ 1.0700096227…. For completeness, x≤1 cannot solve the positive-right-hand-side equation, and the left-hand derivative 4x ln x is positive for x>1.
Schemat punktowania
A1:correct derivative relation. M1:equate the two gradients. A1:ln a=1/(2x²). M1:combine with the common-point equation to obtain the contact equation. A1:x=e with exact substitution verifying it, using the given uniqueness. A1:y=2e². M1:solve for the base analytically. A1:a=exp(1/(2e²)). Decimal-only values do not earn the exact-coordinate/base marks.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Exact results require analytical working; graphs and numerical checks support but do not replace proofs.