Self-inverse functions alternate by parity

A function ff is said to be a self-inverse function if f(x)=f−1(x) f(x)=f^{-1}(x). The function ff is defined by f(x)=ax+bcx−2 f(x)=\frac{ax+b}{cx-2}, x∈Rx\in\mathbb R, x≠2/cx\ne2/c, where a,b,c,da, b, c, d are non-zero constants.

(i) By finding f−1(x) f^{-1}(x), show that a=2 a=2 for ff to be self-inverse. marks

For the rest use a=2,b=3,c=5 a=2, b=3, c=5. (ii) Find f2(x) f^2(x). mark (iii) Evaluate f71(4) f^{71}(4). marks

The function gg is defined by g(x)=2x2−3 g(x)=2x^2-3, x∈Rx\in\mathbb R. (iv) Explain why composite fgfg does not exist. marks (v) Find exact solutions of gf(x)=5 gf(x)=5. marks

[2020/CJC/Promo/10]

Total marks: 10

Solution

Choose a part to attempt. Each page keeps the full question available and loads only that part’s working and exploration.

  1. (i) Self-inverse functions →
  2. (ii) Powers of a function →
  3. (iii) Reducing large powers →
  4. (iv) Existence of composites →
  5. (v) Solving rational equations →