JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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:

Choose the correct answer:

Explanation

Solution: Dono taraf differentiate karne par:

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

Yeh ek Linear Differential Equation hai jiska I.F. $= x$ hai. Solution hoga:

x f(x) = \int \frac{x}{2\sqrt{x+1}} dx

Substitute $x+1 = t^2 \implies dx = 2t \, dt$ :

x f(x) = \int \frac{(t^2 - 1)}{2t} 2t \, dt = \int (t^2 - 1) \, dt = \frac{t^3}{3} - t + C

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

$x=3$ par original equation se $f(3) = \sqrt{3+1} = 2$ :

3(2) = \frac{(4)^{3/2}}{3} - \sqrt{4} + C \implies 6 = \frac{8}{3} - 2 + C \implies C = 8 - \frac{8}{3} = \frac{16}{3}

Ab $x=8$ par:

8 f(8) = \frac{(9)^{3/2}}{3} - \sqrt{9} + \frac{16}{3} = \frac{27}{3} - 3 + \frac{16}{3} = 9 - 3 + \frac{16}{3} = 6 + \frac{16}{3} = \frac{34}{3}

f(8) = \frac{34}{3 \times 8} = \frac{34}{24} = \frac{17}{12}

Isliye, $12f(8)$ hoga:

12 \times \frac{17}{12} = 17

Correct Option: (3)

Choose the correct answer:

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