JEE Main 2020 (Online) 2nd September Morning ShiftQuestion 1
If the tangent to the curve $y=x+\sin y$ at a point $(a,b)$ is parallel to the line joining $\left(0,\frac32\right)$ and $\left(\frac12,2\right)$, then :
Differentiation and its Application
Solution
Concept. The chord has slope $1$; setting $y'=1$ gives $\cos b=0$, and the curve then gives $b-a$. Formulas used. (1) $y'=1+\cos y\,y'$ $\Rightarrow$ $y'=\dfrac{1}{1-\cos y}$Step 1 - Slope of the line. $\dfrac{2-\frac32}{\frac12-0}=1$. Step 2 - Condition. By (1), $\frac{1}{1-\cos b}=1$, so $\cos b=0$ and $\sin b=\pm1$. Step 3 - On the curve. $b=a+\sin b$, so $b-a=\pm1$ and $|b-a|=1$. Answer: (A)