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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:
Option: 1 b=a
Option: 2 \left | b-a \right |=1
Option: 3 \left | b+a \right |=1
Option: 4 b=\frac{\pi}{2}+a

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Slope of tangent to the curve y = x + sin y

\\\text { at }(a, b) \text { is } \frac{2-\frac{3}{2}}{\frac{1}{2}-0}=1 \\ \left.\Rightarrow \frac{d y}{d x}\right]_{x=a}=1

\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=1+\operatorname{cosy} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}(\text { from equation of curve })\\ &\left.\left.\frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}\right]_{\mathrm{x}=\mathrm{a}}=1+\cos \mathrm{b} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=\mathrm{a}} \end{aligned}

\begin{aligned} &\Rightarrow \cos \mathrm{b}=0\\ &\Rightarrow \sin \mathrm{b}=\pm 1\\ &\text { Now, from curve } y=x+\sin y\\ &b=a+\sin b\\ &\Rightarrow|b-a|=|\sin b|=1 \end{aligned}

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