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

Slope of tangent to the curve y = x + siny <br><br>at (a, b) is = ${{2 - {3 \over 2}} \over {{1 \over 2} - 0}}$ = 1 <br><br>Given curve y = x + sin y <br><br>$\Rightarrow$ ${{dy} \over {dx}} = 1 + \cos y.{{dy} \over {dx}}$ <br><br>$\Rightarrow$ (1 - cos y)${{dy} \over {dx}}$ = 1 <br><br>$\Rightarrow$ ${\left. {{{dy} \over {dx}}} \right|_{\left( {a,b} \right)}}$ = ${1 \over {1 + \cos b}}$ <br><br>Now according to question, ${1 \over {1 + \cos b}} = 1$ <br><br>$\Rightarrow$ cos b = 0 <br><br>$\Rightarrow$ sin b = $\pm$ 1 <br><br>Point (a, b) lies on curve y = x + sin y <br><br>$\therefore$ b = a + sin b <br><br>$\Rightarrow$ |b - a| = |sin b| = 1