Zadanie 1c, Paper 3, maj 2025, TZ3, Matematyka AA IB HL

At n=1 the expression is c^{(1−m)/(1−m)}x^m = cx^m = f(x), establishing the base case. Assume it is true for n=k, where k≥1. Then $f^{\circ(k+1)}(x)=c[c^{(1-m^k)/(1-m)}x^{m^k}]^m=c^{1+m(1-m^k)/(1-m)}x^{m^{k+1}}$. Combining exponents of c, 1 + m(1 − m^k)/(1 − m) = [1 − m + m − m^k+1]/(1 − m) = (1 − m^k+1)/(1 − m). This proves the formula for k+1. Since the base case holds and the implication from k to k+1 holds, the formula is true for all positive integers n.

R1: explicitly verify n=1. M1: assume the full formula true for an arbitrary positive integer k. M1: attempt the composition giving iterate k+1. A1: correct substitution of the assumed formula. M1: combine the exponent of c using a common denominator, or an equivalent geometric-sum argument. A1: correct combined numerator before cancellation. A1: establish the exact k+1 formula. R1: conclude by induction, referring to the base case and induction implication; award this final mark only if at least four of the preceding seven marks are earned. Subsequent intermediate marks are not automatically lost for a missing base verification or an imprecisely stated hypothesis.