The coordinates of a point on the curve y = x | vedclass

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(D) Given curve is $y = x \log x$. Finding the derivative with respect to $x$: $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. Th

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$(e^{-2}, -2e^{-2})$

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(D) Given curve is $y = x \log x$. Finding the derivative with respect to $x$: $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$. The slope of the tangent at any point $(x, y)$ is $m_t = 1 + \log x$. The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{1 + \log x}$. The given line is $2x - 2y = 3$,which can be written as $2y = 2x - 3$ or $y = x - \frac{3}{2}$. The slope of this line is $1$. Since the normal is parallel to the line,their slopes must be equal: $-\frac{1}{1 + \log x} = 1$. $-1 = 1 + \log x \implies \log x = -2$. $x = e^{-2}$. Now,find the $y$-coordinate: $y = x \log x = e^{-2} \cdot (-2) = -2e^{-2}$. Thus,the coordinates of the point are $(e^{-2}, -2e^{-2})$.

The coordinates of a point on the curve $y = x \log x$ at which the normal is parallel to the line $2x - 2y = 3$ are

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