JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $f$ be a differentiable function such that $x^{2} f (x) - x = 4 \int_{0}^{x} t f (t) d t, f (1) = \frac{2}{3}$ . Then $18 f (3)$ is equal to:

Choose the correct answer:

Explanation

x^2 f(x) - x = 4 \int_0^x t f(t) dt

Differentiating with respect to $x$ , we get:

x^2 f'(x) + 2x f(x) - 1 = 4 x f(x) \times 1 - 0

\Rightarrow x^2 f'(x) - 2x f(x) - 1 = 0

or

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

or

\frac{dy}{dx} - \frac{2}{x} y = \frac{1}{x^2}

(as $y = f(x)$ and $\frac{dy}{dx} = f'(x)$ )

which is a linear differential equation where

P = \frac{-2}{x}, Q = \frac{1}{x^2}

\Rightarrow I.F. = e^{\int P dx} = e^{\int \frac{-2}{x} dx} = e^{-2 \log x} = \frac{1}{x^2}

So, the required solution is

y \times I.F. = \int Q \times I.F. dx + C

or

y \times \frac{1}{x^2} = \int \frac{1}{x^2} \times \frac{1}{x^2} dx + C

\frac{y}{x^2} = \int \frac{1}{x^4} dx + C

\frac{y}{x^2} = -\frac{1}{3x^3} + C

\frac{y}{x^2} = -\frac{1}{3x^3} + C

Choose the correct answer:

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