Let be a differentiable function. If for all , then the value of f(3) is : [2025]

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Solution

Q.
Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function. If $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$ for all $x \geq 1$ , then the value of f(3) is : [2025]

1 32

2 18

3 22

4 26

Ans.
(1)

We have, $10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$

On differentiating both sides, we get

$10 f (x) = 5 f (x) + 5 x f' (x) - 5 x^{4}$

$\Rightarrow 5 f (x) = 5 x f' (x) - 5 x^{4}$

$\Rightarrow f (x) = x f' (x) - x^{4}$

$\Rightarrow f' (x) - \frac{1}{x} f (x) = x^{3}$

$\Rightarrow \frac{d y}{d x} - \frac{y}{x} = x^{3}$ [Taking y = f(x)]

This is a linear differential equation

$∴ I . F . = e \int_{}^{\frac{- 1}{x} d x} = e^{- \ln x} = \frac{1}{x}$

Solution is $\frac{y}{x} = \int \frac{x^{3}}{x} d x + c$

$\Rightarrow \frac{y}{x} = \frac{x^{3}}{3} + c$

$\Rightarrow y = \frac{x^{4}}{3} + c x$

$\Rightarrow f (x) = \frac{x^{4}}{3} + c x$ ... (i)

Now, using original equation, we have

$10 \int_{1}^{x} f (t) d t = 5 x f (x) - x^{5} - 9$

Put x = 1, we get 0 = 5f(1) – 10 $\Rightarrow$ f(1) = 2

Substituting this value in equation (i), we get

$\Rightarrow 2 = \frac{1}{3} + c \Rightarrow c = \frac{5}{3}$ $∴ f (x) = \frac{x^{4}}{3} + \frac{5}{3} x$

$\Rightarrow f (3) = 27 + 5 = 32$

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