The correct options are
A f(20) = 382
B f(n) is always an even number
C f−1(92)=10
f(1) = 2
f(n) = f(n - 1) + 2(n - 1); ∀n≥2
⇒ f(n) = f(n – 1) + 2(n – 1)
f(n) – f(1) = 2(1 + 2 + 3+…..+ n – 1) = (n – 1)n
∴ f(20) = 382
n2−n+2=92⇒n=10
as n(n – 1) must be even, f(n) is always even.