First slide

Tangents and normals

Question

The point on the curve $y = \cos (x + y),$ $0 \leq x \leq π,$ at which the tangent is parallel to $x + 2 y = 0$ is

Difficult

A

$(\frac{π}{2}, 0)$
B

$(- \frac{π}{2}, 0)$
C

$(\frac{3 π}{2}, 0)$
D

$(\frac{π}{4}, \frac{π}{4})$

Solution

Given curve $y = \cos (x + y)$
$\frac{d y}{d x} = \frac{- \sin (x + y)}{1 + \sin (x + y)}$

Slope of the line $x + 2 y = 0$

Slope of the line is $= \frac{- 1}{2}$

Slope of the tangent $M = {(\frac{d y}{d x})}_{p} = \frac{- \sin (x + y)}{1 + \sin (x + y)}$

By verification slope of tangent at $(\frac{π}{2}, 0) = \frac{- 1}{2}$

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