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)?

पर्याय

• (2, 4)

• (1, 2)

• (4, 17)

• (6, 8)

Answer 1

(6, 8)

Explanation:

Given: f(x) = `x + int_0^1(x - t)f(t)dt`

⇒ f(x) = `(1 + int_0^1 f(t)dt)x - int_0^1tf(t)dt`

⇒ f(x) = Px – Q, where P = `1 + int_0^1 f(t)dt` and Q = `int_0^1 tf(t)dt`

Now, P = `1 + int_0^1(Pt - Q)dt`

⇒ P = `1 + [(Pt^2)/2 - Qt]_0^1`

⇒ P = `1 + P/2 - Q`

⇒ P = `2(1 - Q)`  ...(i)

∴ Q = `int_0^1tf(t)dt`

⇒ Q = `int_0^1t(Pt - Q)dt`

⇒ Q = `[(Pt^3)/3 - (Qt^2)/2]_0^1`

⇒ Q = `P/3 - Q/2`

⇒ Q = `(2P)/9`  ...(ii)

On Solving equation (i) and equation (ii), we get

P = `18/13`, Q = `4/13`

∴ f(x) = `18/13x - 4/13`

Since, (6, 8) satisfy the equation of f(x) = `18/13x - 4/13`

∴ Point (6, 8) lies on the carve y = f(x)