If the tangent to the curve y = x + sin y .npoint( a, b) is para/le/to the line joining (0, 3/2) and (1/2, 2), then

A
$| b - a | = 1$
B
$b = \frac{π}{2} + a$
C
$| a + b | = 1$
D
$b = a$

Solution:

We have, $y = x + \sin y$

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

lt is given that the tangent at $(a, b)$ is parallel to the line joinin
$(0, 3 / 2)$ and $(1 / 2, 2) .$

$\begin{array}{l} ∴ {(\frac{d y}{d x})}_{(n, b)} = \frac{2 - 3 / 2}{1 / 2 - 0} \\ \Rightarrow \frac{1}{1 - \cos b} = 1 \Rightarrow \cos b = 0 \Rightarrow b = (2 n + 1) \frac{π}{2}, n \in Z \end{array}$

Point $(a, b)$ lies on $y = x + \sin y$

$\begin{array}{l} ∴ b = a + \sin b \Rightarrow b = a + \sin (2 n + 1) \frac{π}{2} \\ \Rightarrow b - a = \sin (2 n + 1) \frac{π}{2} \Rightarrow b - a = \pm 1 \Rightarrow | b - a | = 1 \end{array}$

If the tangent to the curve $y = x + \sin y . n$ point $(a, b)$ is para/le/to the line joining $(0, 3 / 2)$ and (1/2, 2), then

Solution:

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