If the tangent to the curve y = x + sin y

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. $\mathrm{b}=\mathrm{a}$

  2. $\mathrm{b}=\frac{\pi}{2}+\mathrm{a}$

  3. $\mathrm{lb}-\mathrm{al}=1$

  4. $|a+b|=1$


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 $y=x+\sin y$

$b=a+\sin b$

$\Rightarrow|b-a|=|\sin b|=1$

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