If the tangent to the curve y=x+sin y at a po | vedclass

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(C) The slope of the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$ is $m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \f

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$|b-a|=1$

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(C) The slope of the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$ is $m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$. Since the tangent at $(a, b)$ is parallel to this line,the slope of the tangent $\left. \frac{dy}{dx} \right|_{(a, b)} = 1$. Differentiating the curve $y = x + \sin y$ with respect to $x$,we get $\frac{dy}{dx} = 1 + \cos y \cdot \frac{dy}{dx}$. Substituting the point $(a, b)$ and the slope $1$,we have $1 = 1 + \cos b \cdot (1)$. This implies $\cos b = 0$,which means $\sin b = \pm 1$. Since $(a, b)$ lies on the curve,$b = a + \sin b$. Therefore,$b - a = \sin b$. Taking the absolute value on both sides,$|b - a| = |\sin b| = 1$.

If the tangent to the curve $y=x+\sin y$ at a point $(a, b)$ is parallel to the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right),$ then

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