If f(x) = frac{1}{x^2} int_3^x (2t - 3f'(t)) | vedclass

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(C) Given that,$f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$. Multiplying by $x^2$,we get $x^2 f(x) = \int_3^x (2t - 3f'(t)) dt$. Differentiating b

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$\frac{1}{2}$

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(C) Given that,$f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$. Multiplying by $x^2$,we get $x^2 f(x) = \int_3^x (2t - 3f'(t)) dt$. Differentiating both sides with respect to $x$ using the product rule and the Leibniz integral rule: $2x f(x) + x^2 f'(x) = 2x - 3f'(x)$. At $x = 3$,we have $2(3) f(3) + 3^2 f'(3) = 2(3) - 3f'(3)$. From the original integral,$f(3) = \frac{1}{3^2} \int_3^3 (2t - 3f'(t)) dt = 0$. Substituting $f(3) = 0$ into the differentiated equation: $2(3)(0) + 9f'(3) = 6 - 3f'(3)$. $9f'(3) + 3f'(3) = 6$. $12f'(3) = 6$. $f'(3) = \frac{6}{12} = \frac{1}{2}$.

If $f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$,then $f'(3)$ is equal to

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