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(B) Given the curve $y = \cos(x + y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x + y) \cdot (1 + \frac{dy}{dx})$.Rearranging

Vedclass · Accepted Answer

$2x + 4y - \pi = 0$

Vedclass · Answer

(B) Given the curve $y = \cos(x + y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x + y) \cdot (1 + \frac{dy}{dx})$.Rearranging,$\frac{dy}{dx}(1 + \sin(x + y)) = -\sin(x + y)$,so $\frac{dy}{dx} = \frac{-\sin(x + y)}{1 + \sin(x + y)}$.The tangent is parallel to $x + 2y = 0$,which has a slope $m = -\frac{1}{2}$.Setting $\frac{dy}{dx} = -\frac{1}{2}$,we get $\frac{-\sin(x + y)}{1 + \sin(x + y)} = -\frac{1}{2} \implies 2\sin(x + y) = 1 + \sin(x + y) \implies \sin(x + y) = 1$.Since $\sin(x + y) = 1$,then $x + y = 2n\pi + \frac{\pi}{2}$.Substituting into the original equation: $y = \cos(2n\pi + \frac{\pi}{2}) = 0$.Thus,$x + 0 = 2n\pi + \frac{\pi}{2} \implies x = 2n\pi + \frac{\pi}{2}$.For $x \in [-2\pi, 2\pi]$,possible values are $x = \frac{\pi}{2}$ and $x = -\frac{3\pi}{2}$.For $x = \frac{\pi}{2}, y = 0$,the tangent equation is $y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \implies 2y = -x + \frac{\pi}{2} \implies 2x + 4y - \pi = 0$.This matches option $B$.

The equation of the tangent to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$,which is parallel to the line $x + 2y = 0$,is:

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