# If the tangent to the curve

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 :

1. (1) $b=a$

2. (2) $|b-a|=1$

3. (3) $|a+b|=1$

4. (4) $b=\frac{\pi}{2}+a$

Correct Option:

Solution:

The given curve $y=x+\sin y$

$\because$ The point $(a, b)$ lie on the curve

$\therefore b=a+\sin b$

$\Rightarrow \frac{d y}{d x}=1+\cos y \frac{d y}{d x} \Rightarrow(1-\cos y) \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{1}{1-\cos y}$

$\therefore\left(\frac{d y}{d x}\right)_{(a, b)}=\frac{1}{1-\cos b}$

Slope of the line joining the points $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$

$=\frac{2-\frac{3}{2}}{\frac{1}{2}-0}=1$

Now, according to the question,

$\left(\frac{d y}{d x}\right)_{(a, b)}=1 \Rightarrow \frac{1}{1-\cos b}=1$

$\Rightarrow 1-\cos b=1 \Rightarrow b=\frac{\pi}{2}$

Now, $b=a+\sin b$

$\Rightarrow a=b-\sin b=\frac{\pi}{2}-1$

$|b-a|=\left|\frac{\pi}{2}-\frac{\pi}{2}+1\right|=1$