Let f¹(x)=(3 x+2)/(2 x+3), x ∈ R-{-3/2} For n ≥ 2, define fⁿ(x)=f¹ o fⁿ⁻¹(x). If f⁵(x)=(a x+b)/(b x+a), gcd(a, b)=1, then a+b is equal to ____________.

Solution

Maths sir · 4 steps

Track a+b and a-b, not a and b: one is multiplied by 5 each time, the other never changes.

1. One composition. Suppose fⁿ(x) = (ax+b)/(bx+a). Put this into f¹ and multiply top and bottom by bx+a. You get fⁿ⁺¹(x) = ((3a+2b)x + (2a+3b))/((2a+3b)x + (3a+2b)), which has the same shape.

   a' = 3a + 2b, b' = 2a + 3b

2. Sum and difference. Adding gives a' + b' = 5(a+b). Subtracting gives a' - b' = a - b. For f¹, a = 3 and b = 2, so the sum starts at 5 and the difference starts at 1.
3. Reach the fifth. After four more compositions, a + b = 5 × 5⁴ = 5⁵ = 3125 and a - b = 1. So a = 1563 and b = 1562. Check with f²: the rule gives (13x+12)/(12x+13), and 13 + 12 = 25 = 5².
4. So. 1563 and 1562 are consecutive integers, so gcd(a, b) = 1 and no cancelling is needed. Therefore a + b = 3125.

The common slip: 625. Counting f⁵ as four multiplications by 5 and forgetting that f¹ already has sum 5 gives 5⁴ = 625, not 5⁵.

The idea to keep. When (ax+b)/(bx+a) is composed with itself again and again, do not chase a and b separately. Their sum and difference each change by a fixed factor, so the n-th term follows at once.

Answer from the NTA final answer key. Solution by PrepWiser.