JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Definite Integrals

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)

Answer: (D)

Solution

Rewrite the integral equation:

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

. Let

A = 1 + \int_{0}^{1} f (t) d t

and

B = \int_{0}^{1} t f (t) d t

. Then

f (x) = A x - B

, a linear function. From

A = 1 + \int_{0}^{1} (A t - B) d t = 1 + \frac{A}{2} - B

, we get

A = 2 (1 - B)

. From

B = \int_{0}^{1} t (A t - B) d t = \frac{A}{3} - \frac{B}{2}

, we get

\frac{3 B}{2} = \frac{A}{3}

, so

A = \frac{9 B}{2}

. Solving:

\frac{9 B}{2} = 2 (1 - B)

gives

B = \frac{4}{13}

and

A = \frac{18}{13}

. Thus

f (x) = \frac{18 x - 4}{13}

. Checking:

f (6) = \frac{108 - 4}{13} = \frac{104}{13} = 8

. So

(6, 8)

lies on the curve.

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option