Let a differentiable function f satisfy f ( x ) + 3 x f ( t ) t d t = x + 1 , x 3 . Then 12 f ( 8 ) is equal to: [2023]

Question

Let a differentiable function $f$ satisfy $f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3 .$ Then $12 f (8)$ is equal to: [2023]

Smart Achievers · Accepted Answer

(2)

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

On differentiating, we get

$f' (x) + \frac{f (x)}{x} = \frac{1}{2 \sqrt{x + 1}}$ $\Rightarrow \frac{d y}{d x} + \frac{1}{x} y = \frac{1}{2 \sqrt{x + 1}}$

$I . F . = e^{\int \frac{1}{x} d x} = e^{\ln x} = x$

Solution; $x y = \int \frac{x}{2 \sqrt{x + 1}} d x$

Put $x + 1 = t^{2};$ $d x = 2 t d t$

$x y = \int \frac{(t^{2} - 1) 2 t d t}{2 t} = \int (t^{2} - 1) d t = \frac{t^{3}}{3} - t + C$

$= \frac{{(\sqrt{x + 1})}^{3}}{3} - \sqrt{x + 1} + C$

At $x = 3, y = 2$

$∴$ $3 \times 2 = \frac{{(\sqrt{3 + 1})}^{3}}{3} - \sqrt{3 + 1} + C \Rightarrow$ $6 = \frac{8}{3} - 2 + C \Rightarrow C = \frac{16}{3}$

$∴$ $f (x) = \frac{{(\sqrt{x + 1})}^{3}}{3 x} - \frac{\sqrt{x + 1}}{x} + \frac{16}{3 x}$

Now, $12 f (8) = 12 [\frac{{(\sqrt{8 + 1})}^{3}}{3 \times 8} - \frac{\sqrt{8 + 1}}{8} + \frac{16}{3 \times 8}] = 17$

Solution

Q.
Let a differentiable function $f$ satisfy $f (x) + \int_{3}^{x} \frac{f (t)}{t} d t = \sqrt{x + 1}, x \geq 3 .$ Then $12 f (8)$ is equal to: [2023]

1 19

2 17

3 1

4 34

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