Let f be a differentiable function such that . Then f(2) is equal to _________. [2025]

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Solution

Q.
Let f be a differentiable function such that $2 {(x + 2)}^{2} f (x) - 3 {(x + 2)}^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ . Then f(2) is equal to _________. [2025]

Ans.
(19)

We have,

$2 {(x + 2)}^{2} f (x) - 3 {(x + 2)}^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ ... (i)

On differentiating, we get

$4 (x + 2) f (x) + 2 {(x + 2)}^{2} f' (x) - 6 (x + 2) = 10 (x + 2) f (x)$

$\Rightarrow 2 {(x + 2)}^{2} f' (x) - 6 (x + 2) f (x) = 6 (x + 2)$

$\Rightarrow (x + 2) f' (x) - 3 f (x) = 3$

$\Rightarrow (x + 2) \frac{d y}{d x} - 3 y = 3$ $[∵ f' (x) = \frac{d y}{d x} and f (x) = y]$

$\Rightarrow \frac{d y}{d x} = \frac{3 (1 + y)}{(x + 2)} \Rightarrow \frac{d y}{3 (1 + y)} = \frac{d x}{(x + 2)}$

Integrating on both sides, we get

$\int \frac{d y}{(1 + y)} = 3 \int \frac{d x}{(x + 2)} \Rightarrow \ln (1 + y) = 3 \ln (x + 2) + \ln C$

$\Rightarrow (1 + y) = C {(x + 2)}^{3}$ ... (ii)

Put x = 0 in equation (i), we get

$2 {(2)}^{2} f (0) - 3 {(2)}^{2} = 0 \Rightarrow f (0) = \frac{3}{2}$

From (ii), $1 + \frac{3}{2} = C {(2)}^{3} \Rightarrow C = \frac{5}{16}$

$∴ y = \frac{5}{16} {(x + 2)}^{3} - 1 = f (x)$

$∴ f (2) = \frac{5}{16} \times 64 - 1 = 19$

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