Let S = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Define f : S → S as
f(n) = { 2n, ,, if n = 1,2,3,4,5; 2n - 11, ,, if n = 6,7,8,9,10 .
Let g : S → S be a function such that fog(n) = { n + 1, ,, if n is odd; n - 1, ,, if n is even .
Then g(10)g(1) + g(2) + g(3) + g(4) + g(5)) is equal to _____________.

Solution

Maths sir · 2 steps

f is one-one and onto, so undo it: g(n) is the number that f sends to fog(n).

1. Undo f. For n = 1 to 5, f gives the even numbers 2, 4, 6, 8, 10. For n = 6 to 10, it gives the odd numbers 1, 3, 5, 7, 9. Every value in S is hit exactly once, so f has an inverse and g(n) = f⁻¹(f(g(n))).

   f⁻¹(k) = k/2, k even ; (k+11)/2, k odd

   n	f(g(n))	g(n) = f⁻¹(f(g(n)))
   1	1 + 1 = 2	2/2 = 1
   2	2 - 1 = 1	(1+11)/2 = 6
   3	3 + 1 = 4	4/2 = 2
   4	4 - 1 = 3	(3+11)/2 = 7
   5	5 + 1 = 6	6/2 = 3
   10	10 - 1 = 9	(9+11)/2 = 10

2. So. The expression is g(10)(g(1) + g(2) + g(3) + g(4) + g(5)) = 10 × (1 + 6 + 2 + 7 + 3) = 10 × 19, which is 190.

The common slip: 28. Multiplying g(10) by g(1) alone and then adding the other terms gives 10 + 6 + 2 + 7 + 3 = 28. The factor g(10) multiplies the whole sum.

The idea to keep. When you know f and f ° g, check whether f is a bijection. If it is, g = f⁻¹ ° (f ° g), so you find each g(n) by inverting f.

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