Let f:ℝ→ℝ be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)-1,∀ x,y∈ℝ. If f'(0)=2, then |f(-2)| is equal to ___________.

Solution

Maths sir · 4 steps

Put x = y = 0 to find f(0), then show f'(x) is the same constant everywhere.

1. Find f(0). Put x = y = 0: f(0) = 2f(0) - 1, so f(0) = 1.
2. Derivative everywhere. Use the first-principles definition with y = h. The relation gives f(x+h) - f(x) = f(h) - 1 = f(h) - f(0).

   f'(x) = lim(h → 0)(f(x+h)-f(x))/h = lim(h → 0)(f(h)-f(0))/h = f'(0) = 2

3. Integrate. Since f'(x) = 2 for all x, f(x) = 2x + c. From f(0) = 1 we get c = 1, so f(x) = 2x + 1. Check: 2(x+y) + 1 = (2x+1) + (2y+1) - 1, which holds.
4. So. f(-2) = 2(-2) + 1 = -3, so |f(-2)| = 3.

The common slip: 4. Assuming f(0) = 0 instead of finding it from the relation gives f(x) = 2x and the answer 4. Here f(0) = 1.

The idea to keep. For f(x+y) = f(x) + f(y) + k, first put x = y = 0 to get f(0). Then first principles shows f'(x) = f'(0), so f is a straight line.

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