Question:
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 :
Correct Option: 3,
Solution:
Slope of tangent to the curve $y=x+\sin y$
at $(a, b)$ is $\frac{2-\frac{3}{2}}{\frac{1}{2}-0}=1$
$\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=\mathrm{a}}=1$
$\frac{d y}{d x}=1+\cos y \cdot \frac{d y}{d x}$ (from equation of curve)
$\left.\left.\frac{d y}{d x}\right]_{x=a}=1+\cos b \cdot \frac{d y}{d x}\right]_{x=a}$
$\Rightarrow \quad \cos \mathrm{b}=0$
$\Rightarrow \quad \sin \mathrm{b}=\pm 1$
Now, from curve $y=x+\sin y$
$b=a+\sin b$
$\Rightarrow|b-a|=|\sin b|=1$