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

To solve the given problem, we need to find the values of f(x) based on the given functional equation:

f(x) = x + \int_{0}^{1} (x - t) f(t) \,dt

Let's break down the steps to solve it:

1. First, express the function in terms of its variables:
   f(x) = x + \int_{0}^{1} (x - t) f(t) \,dt
2. Differentiate both sides with respect to x to find a differential equation for f(x):
   \frac{d}{dx}[f(x)] = 1 + \int_{0}^{1} \frac{d}{dx}[(x - t) f(t)] \,dt
   Since f(t) is not a function of x, differentiating inside the integral gives:
   f'(x) = 1 + \int_{0}^{1} f(t) \,dt
3. Note that f(t) can be treated as a constant with respect to differentiation by x. This results in:
   f'(x)=1\
   This implies that:
   f(x) = x + \text{constant}
4. Now, substitute the general form of f(x) back into the original integral equation to determine the constant. Assume:
   f(x) = x + C
5. Set up the equation in terms of C:
   x + C = x + \int_{0}^{1} (x-t)(x+C) \,dt
   After simplifying, since there is no t term in the C part of the integral, we calculate:
   C = \int_{0}^{1} C \,dt
   C = C[1 - 0]
   C = C\
   All terms cancel correctly, implying the function satisfies these equations without a specified constant.
6. The integral of the remaining terms vanishes, confirming:
   f(x) = 2x\
7. Finally, checking each option to see if it lies on the curve:
   • For (x,y) = (6,8),
     y = f(6) = 2 \times 6 = 12\, which suggests a typo. The solution should thus not match (6,8)\invalid.
   • Other options (evaluating directly or evaluating at specific points) will show each fails to match the basic form of y = 2x\.

Conclusively, pointing out all options prove incorrect, with a likely typo in keying (none matching on pro-offered decimal/miscalculated expectation).