Let a differentiable function f satisfy $f(x)+\int_{3}^{x} \frac{f(t)}{t} dt=\sqrt{x+1}, x \geq 3$. Then 12f(8) is equal to:

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Differential Equations

Let a differentiable function f satisfy

f (x) + \int_{3}^{x} \frac{f ( t )}{t} d t = x + 1, x \geq 3

. Then 12f(8) is equal to:

Answer: (C)

Differentiating

f (x) + \int_{3}^{x} \frac{f ( t )}{t} d t = x + 1

gives

f^{'} (x) + \frac{f ( x )}{x} = \frac{1}{2 x + 1}

. With integrating factor

x

this is

(x f (x))^{'} = \frac{x}{2 x + 1}

, which integrates, on putting

u = x + 1

, to

x f (x) = \frac{( x + 1 ) ^{3/2}}{3} - x + 1 + C

. The original relation at

x = 3

gives

f (3) = 2

, so

6 = \frac{8}{3} - 2 + C

and

C = \frac{16}{3}

. At

x = 8

8 f (8) = \frac{27}{3} - 3 + \frac{16}{3} = \frac{34}{3}

, so

f (8) = \frac{17}{12}

and

12 f (8) = 17

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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