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

Answer

N/a

Explanation

The equation of the curve is `y = cos x - 1`
Now, we have to find a point on the curve in `[0,2pi]`,
where the tangent is parallel to X-axis i.e., the tangent to the curve at `x = c` has a slope o, where `c in`] `0,2pi`[.
Let us apply Rolle's theorem to get the point.
(i) `y = cosx-1` is a continuous function in `[0,2pi]`.
[since it is a combination of cosine function and a constant function]
(ii) `y' = -sinx`, which exists in `(0,2pi)`.
Hence, y is differentiable in `(0,2pi)`.
(iii) `y (0) = cos 0 - 1 = 0` and `y(2pi) = cos2pi - 1 = 0`,
`:. y (0) = y(2pi)`
Since, conditions of Rolle's theorem are satisfied.
Hence, there exists a real number c such that
` f(c) = 0`
`rArr -sinc = 0`
`rArr c = pi` or `0`, where `pi in (0,2pi)`
`rArr x = pi`
`:. y =cos pi - 1 = - 2`
Hence, at required point on the curve, where the tangent drawn is parallel to the X-axis is `(pi,2)`.