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

f ( x ) = x + x ∫ 0 1 f ( t ) d t − ∫ 0 1 t 0 f ( t ) d t Let 1 + ∫ 0 1 f ( t ) d t = α ∫ 0 1 t f ( t ) d t = β So, f(x) = ax - b Now, α = ∫ 0 1 f ( t ) d t + 1 α = ∫ 0 1 ( a t − β ) d t + 1 β = ∫ 0 1 t ⋅ f ( t ) d t β = 4 1 3 , α = 1 8 1 3 f(x) = ax – b = 1 8 x − 4 1 3 option (D) satisfies

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

Option 1 - (2, 4)

Option 2 - (1, 2)

Option 3 - (4, 17)

Option 4 - (6, 8)

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Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 4

Detailed Solution:

$f (x) = x + x \int_{0}^{1} f (t) d t - \int_{0}^{1} t_{0} f (t) d t$

Let $1 + \int_{0}^{1} f (t) d t = α$

$\int_{0}^{1} t f (t) d t = β$

So, f(x) = ax - b

Now, $α = \int_{0}^{1} f (t) d t + 1$

$α = \int_{0}^{1} (a t - β) d t + 1$

$β = \int_{0}^{1} t \cdot f (t) d t$

$β = \frac{4}{13}, α = \frac{18}{13}$

f(x) = ax – b

$= \frac{18 x - 4}{13}$

option (D) satisfies

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