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  • Find the points on the curve y = (cosx - 1) in [0, 2pi ], where the tangent is parallel to x-axis.

Find the points on the curve y = (cosx - 1) in [0, 2\pi], where the tangent is parallel to x-axis.

Answers (1)

Given: Equation of curve, y = cos x - 1

Firstly, we differentiate the above equation with respect to x, we get

\\ \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{y}=\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}-\frac{\mathrm{d}}{\mathrm{dx}}(1) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x}-0\left[\because \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=-\sin \mathrm{x}\right] \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x}

Given tangent to the curve is parallel to the x - axis

This means, Slope of tangent = Slope of x - axis
\begin{aligned} &\frac{d y}{d x}=0\\ &\Rightarrow-\sin x=0\\ &-\sin x=0\\ &\Rightarrow \mathrm{x}=\sin ^{-1}(0)\\ &-x=\pi\lfloor(0,2 \pi)\\ &\text { Put } x=\pi \text { in } y=\cos x-1, \text { we have }\\ &\mathrm{y}=\cos \pi-1=-1-1=-2[\because \cos \pi=-1] \end{aligned}

Hence, the tangent to the curve is parallel to the x -axis at

(π, -2)

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infoexpert22

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