If the curves y = frac{ln x}{x} and y = lambd | vedclass

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(C) Let the curves touch at point $(x, y)$.For the curves to touch,they must have the same value and the same derivative at the point of contact.$1$.

Vedclass · Accepted Answer

$\frac{1}{3e}$

Vedclass · Answer

(C) Let the curves touch at point $(x, y)$.For the curves to touch,they must have the same value and the same derivative at the point of contact.$1$. Equating the functions: $\frac{\ln x}{x} = \lambda x^2 \implies \ln x = \lambda x^3$.$2$. Equating the derivatives: $\frac{d}{dx}(\frac{\ln x}{x}) = \frac{d}{dx}(\lambda x^2) \implies \frac{1 - \ln x}{x^2} = 2\lambda x \implies 1 - \ln x = 2\lambda x^3$.Substitute $\ln x = \lambda x^3$ into the derivative equation:$1 - (\lambda x^3) = 2\lambda x^3 \implies 1 = 3\lambda x^3 \implies \lambda x^3 = \frac{1}{3}$.Since $\ln x = \lambda x^3$,we have $\ln x = \frac{1}{3}$,which implies $x = e^{1/3}$.Now,substitute $x^3 = (e^{1/3})^3 = e$ into $\lambda x^3 = \frac{1}{3}$:$\lambda e = \frac{1}{3} \implies \lambda = \frac{1}{3e}$.

If the curves $y = \frac{\ln x}{x}$ and $y = \lambda x^2$ (where $\lambda$ is a constant) touch each other,then $\lambda$ is

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