The point on the curve y = sqrt{x - 1} where | vedclass

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(C) Given the curve $y = \sqrt{x - 1}$.Finding the derivative,we get $\frac{dy}{dx} = \frac{1}{2\sqrt{x - 1}}$. Let this be the slope of the tangent $

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$(2, 1)$

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(C) Given the curve $y = \sqrt{x - 1}$.Finding the derivative,we get $\frac{dy}{dx} = \frac{1}{2\sqrt{x - 1}}$. Let this be the slope of the tangent $m_1 = \frac{1}{2\sqrt{x - 1}}$.The equation of the line is $2x + y - 5 = 0$,which can be written as $y = -2x + 5$. The slope of this line is $m_2 = -2$.Since the tangent is perpendicular to the line,the product of their slopes must be $-1$,i.e.,$m_1 	imes m_2 = -1$.Substituting the values,we have $\left(\frac{1}{2\sqrt{x - 1}}ight) 	imes (-2) = -1$.This simplifies to $\frac{-1}{\sqrt{x - 1}} = -1$,which implies $\sqrt{x - 1} = 1$.Squaring both sides,we get $x - 1 = 1$,so $x = 2$.Substituting $x = 2$ into the curve equation,$y = \sqrt{2 - 1} = \sqrt{1} = 1$.Thus,the required point is $(2, 1)$.

The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0$ is

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