Question

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

• A. (2, 4)
• B. (1, 2)
• C. (4, 17)
• D. (6, 8)

Answer 1

The Correct Option is D

The correct answer is (D) : (6,8)
\(f(x) = x \int_{0}^{1} (x - t) f(t) \,dt\)
\(f(x) = x + x\int_{0}^{1} f(t) \,dt - \int_{0}^{1} t \cdot f(t) \,dt\)
\(f(x) = x \left(1 + \int_{0}^{1} f(t) \,dt \right) - \int_{0}^{1} t \cdot f(t) \,dt\)
Let
\(1 + \int_{0}^{1} f(t) \,dt = a \quad \text{and} \quad \int_{0}^{1} t \cdot f(t) \,dt = 1\)
f(x) = ax-b
Now, 
\(a = 1 + \int_{0}^{1} (at - b) \,dt = 1 + \frac{a}{2} - b \implies \frac{a}{2} + b = 1\)
\(b = \int_{0}^{1} t(at - b) \,dt = \frac{a}{3} - \frac{b}{2} \implies \frac{3b}{2} = \frac{a}{3} \implies b = \frac{2a}{9}\)
\(\frac{a}{2} + \frac{2a}{9} = 1\)
\(⇒ a = \frac{18}{13}\)  \(b = \frac{4}{13}\)
\(ƒ(x) = \frac{18x-4}{13}\)
(6,8) lies on f(x)