Let a curve y=f(x) pass through the point (2, | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{2y}{x \ln x}$.Separating the variables,we get: $\frac{dy}{y} = \frac{2 dx}{x \ln x}$.Integ

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{2y}{x \ln x}$.Separating the variables,we get: $\frac{dy}{y} = \frac{2 dx}{x \ln x}$.Integrating both sides: $\int \frac{dy}{y} = \int \frac{2}{x \ln x} dx$.Let $u = \ln x$,then $du = \frac{1}{x} dx$. The integral becomes: $\ln |y| = 2 \int \frac{du}{u} = 2 \ln |u| + C = 2 \ln |\ln x| + C$.So,$\ln |y| = \ln |(\ln x)^2| + C$,which implies $y = k(\ln x)^2$ for some constant $k$.Given that the curve passes through $(2, (\ln 2)^2)$,we substitute $x=2$ and $y=(\ln 2)^2$:$(\ln 2)^2 = k(\ln 2)^2 \Rightarrow k = 1$.Thus,the function is $f(x) = (\ln x)^2$.To find $f(e)$,we substitute $x=e$: $f(e) = (\ln e)^2 = (1)^2 = 1$.

Let a curve $y=f(x)$ pass through the point $(2, (\ln 2)^2)$ and have slope $\frac{2y}{x \ln x}$ for all positive real values of $x$. Then the value of $f(e)$ is equal to:

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