Let f : 1 , be a differentiable function. If 6 1 x f ( t ) d t = 3 x f ( x ) + x 3 - 4 for all x 1 , then the value of f ( 2 ) - f ( 3 ) is [2026]

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Solution

Q.
Let $f : [1, \infty) \to ℝ$ be a differentiable function. If $6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$ for all $x \geq 1$ , then the value of $f (2) - f (3)$ is [2026]

1 $- 4$

2 $3$

3 $- 3$

4 $4$

Ans.
(2)

$6 \int_{1}^{x} f (t) d t = 3 x f (x) + x^{3} - 4$

$Differentiate both sides$

$6 f (x) = 3 x f' (x) + 3 f (x) + 3 x^{2}$

$3 f (x) = 3 x f' (x) + 3 x^{2}$

$x \frac{d y}{d x} - y = - x^{2}$

$\frac{x \frac{d y}{d x} - 9}{x^{2}} = - 1$

$\Rightarrow \frac{d}{d x} (\frac{y}{x}) = - 1$

$\frac{y}{x} = - x + C$

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

At $x = 1, y = 1$ $\Rightarrow C = 2$

$f (x) = - x^{2} + 2 x$

$f (2) - f (3) = 3$

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