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(B) Let $t = \frac{\lambda^{2} x}{3}$. Then $dt = \frac{2\lambda x}{3} d\lambda$,which implies $d\lambda = \frac{3}{2\lambda x} dt = \frac{3}{2\sqrt{3

Vedclass · Accepted Answer

$12$

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(B) Let $t = \frac{\lambda^{2} x}{3}$. Then $dt = \frac{2\lambda x}{3} d\lambda$,which implies $d\lambda = \frac{3}{2\lambda x} dt = \frac{3}{2\sqrt{3tx/3} x} dt = \frac{3}{2x\sqrt{tx/3}} dt = \frac{\sqrt{3}}{2\sqrt{x}\sqrt{t}} dt$.Substituting this into the integral:$f(x) = \frac{2}{\sqrt{3}} \int_{0}^{x} f(t) \frac{\sqrt{3}}{2\sqrt{x}\sqrt{t}} dt = \frac{1}{\sqrt{x}} \int_{0}^{x} \frac{f(t)}{\sqrt{t}} dt$.Thus,$\sqrt{x} f(x) = \int_{0}^{x} \frac{f(t)}{\sqrt{t}} dt$.Differentiating both sides with respect to $x$ using the Leibniz rule:$\frac{1}{2\sqrt{x}} f(x) + \sqrt{x} f'(x) = \frac{f(x)}{\sqrt{x}}$.Multiplying by $2\sqrt{x}$:$f(x) + 2x f'(x) = 2f(x) \implies 2x f'(x) = f(x)$.$\frac{f'(x)}{f(x)} = \frac{1}{2x}$.Integrating both sides:$\ln|f(x)| = \frac{1}{2} \ln|x| + C \implies f(x) = k\sqrt{x}$.Given $f(1) = \sqrt{3}$,we have $k\sqrt{1} = \sqrt{3} \implies k = \sqrt{3}$.So,$f(x) = \sqrt{3x}$.Since $f(\alpha) = 6$,we have $\sqrt{3\alpha} = 6 \implies 3\alpha = 36 \implies \alpha = 12$.

Let $f$ be a differentiable function satisfying $f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left(\frac{\lambda^{2} x}{3}\right) d\lambda$ for $x > 0$ and $f(1) = \sqrt{3}$. If $y = f(x)$ passes through the point $(\alpha, 6)$,then $\alpha$ is equal to $.........$

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