Let f be a real valued continuous function on [0, 1] and f(x) = x + ∫₀¹ (x - t)f(t)dt.
Then, which of the following points (x, y) lies on the curve y = f(x) ?

Solution

Maths sir · 4 steps

Split the integral and pull x outside. What is left is two constants, so f is a straight line.

1. Split the integral. ∫₀¹ (x-t)f(t) dt = x∫₀¹ f(t) dt - ∫₀¹ t f(t) dt. Both integrals have fixed limits, so each is just a number. Call them A and B. Then f(x) = (1+A)x - B.
2. Feed f back in. Write p = 1 + A, so f(t) = pt - B. Put this into the definitions: A = ∫₀¹ (pt - B) dt = p/2 - B and B = ∫₀¹ t(pt - B) dt = p/3 - B/2.

   p = 1 + p/2 - B ; 3B/2 = p/3

3. Solve. The second equation gives B = 2p/9. Putting this into the first gives p/2 = 1 - 2p/9, so 13p/18 = 1. This gives p = 18/13 and B = 4/13.
4. So. f(x) = (18x - 4)/13. At x = 6 this is 104/13 = 8, so (6, 8) lies on the curve. The other options fail: f(2) = 32/13, f(1) = 14/13 and f(4) = 68/13. The answer is (d).

The idea to keep. When f sits inside a definite integral with constant limits, that integral is only a number. Name each such integral, write f in terms of these constants, then substitute back to find them.

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