Key Concepts and Formulas

• Slope of a line: The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Parallel Lines: Parallel lines have equal slopes.
• Implicit Differentiation: If $y$ is implicitly defined as a function of $x$, differentiate both sides of the equation with respect to $x$, using the chain rule where necessary.
• Point on a Curve: If a point $(a, b)$ lies on a curve, then substituting $x = a$ and $y = b$ into the equation of the curve will satisfy the equation.

Step-by-Step Solution

Step 1: Calculate the slope of the reference line.

The tangent to the curve at $(a, b)$ is parallel to the line joining $(0, \frac{3}{2})$ and $(\frac{1}{2}, 2)$. We first find the slope of this line. Let $(x_1, y_1) = (0, \frac{3}{2})$ and $(x_2, y_2) = (\frac{1}{2}, 2)$. Then the slope is: $m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$ Thus, the slope of the tangent at $(a, b)$ is 1.

Step 2: Find the derivative $\frac{dy}{dx}$ using implicit differentiation.

The equation of the curve is $y = x + \sin y$. Differentiating both sides with respect to $x$, we get: $\frac{dy}{dx} = \frac{d}{dx}(x + \sin y) = \frac{dx}{dx} + \frac{d}{dx}(\sin y) = 1 + \cos y \cdot \frac{dy}{dx}$ Now we solve for $\frac{dy}{dx}$: $\frac{dy}{dx} - \cos y \cdot \frac{dy}{dx} = 1$ $\frac{dy}{dx}(1 - \cos y) = 1$ $\frac{dy}{dx} = \frac{1}{1 - \cos y}$

Step 3: Evaluate the derivative at the point (a, b).

The slope of the tangent at $(a, b)$ is obtained by substituting $y = b$ into the expression for $\frac{dy}{dx}$: $\left. \frac{dy}{dx} \right|_{(a, b)} = \frac{1}{1 - \cos b}$

Step 4: Equate the slope of the tangent to the slope of the reference line.

We know the slope of the tangent at $(a, b)$ is 1. Therefore, $\frac{1}{1 - \cos b} = 1$ $1 = 1 - \cos b$ $\cos b = 0$

Step 5: Find the value of sin b.

Since $\cos b = 0$, we have $\sin^2 b + \cos^2 b = 1$, so $\sin^2 b = 1$, and $\sin b = \pm 1$.

Step 6: Use the fact that (a, b) lies on the curve.

Since $(a, b)$ lies on the curve $y = x + \sin y$, we have $b = a + \sin b$. Therefore, $b - a = \sin b$.

Step 7: Analyze the possibilities for $\sin b$ and relate a and b.

If $\sin b = 1$, then $b - a = 1$, so $b = a + 1$. If $\sin b = -1$, then $b - a = -1$, so $b = a - 1$.

Step 8: Reconcile with the correct answer.

If $b = a + 1$, then let's check option (A): $b=a$. This would mean $a+1=a$, which is impossible.

Let's revisit the condition $\cos b = 0$. This means $b = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

If $b - a = \sin b$, then $b - a = \pm 1$, so $|b - a| = 1$.

If $b = a$, then $\sin b = 0$. But if $\sin b = 0$, then $\cos b = \pm 1$, which contradicts $\cos b = 0$. Thus, $b = a$ is not possible.

However, if we consider the case where $a = \frac{\pi}{2}$ and $b = \frac{\pi}{2}$. Then $b = a$, and $\cos b = \cos \frac{\pi}{2} = 0$. Also, $\sin b = \sin \frac{\pi}{2} = 1$. So $b = a + \sin b$ gives $\frac{\pi}{2} = \frac{\pi}{2} + 1$, which is false.

If $a = -\frac{\pi}{2}$ and $b = -\frac{\pi}{2}$, then $b = a$, and $\cos b = \cos (-\frac{\pi}{2}) = 0$. Also, $\sin b = \sin (-\frac{\pi}{2}) = -1$. So $b = a + \sin b$ gives $-\frac{\pi}{2} = -\frac{\pi}{2} - 1$, which is false.

Let's analyze the given options with $|b-a| = 1$. (A) $b=a$. We already showed this is not possible. (B) $|b-a| = 1$. This is consistent with our derivation. (C) $b = \frac{\pi}{2} + a$. If this is true, then $b - a = \frac{\pi}{2}$. However, we know $|b-a| = 1$, so $b - a = \pm 1$. Thus $\frac{\pi}{2} = \pm 1$, which is false. (D) $|a+b| = 1$. We know $|b-a| = 1$.

Let's reconcile with the correct answer of $b=a$. Since the given answer is b=a, then sin b = 0. So $b = a$ means $b - a = 0 = \sin b$. If $\sin b = 0$, then $\cos b = \pm 1$. But we found that $\cos b = 0$. Therefore, there is no solution that satisfies b=a.

There appears to be an error in the problem statement or the provided correct answer.

Common Mistakes & Tips

• Remember the chain rule when differentiating implicitly.
• Be careful when solving trigonometric equations; consider all possible solutions.
• Always check if your solution satisfies the original equation and conditions.

Summary

We calculated the slope of the reference line and then used implicit differentiation to find the derivative of the given curve. We equated these two to solve for $\cos b$ and $\sin b$. Using the point-on-curve condition, we related $a$ and $b$. We arrived at the result $|b - a| = 1$. However, upon further inspection, this contradicts the given correct answer of $b=a$. There might be an error in the question.

The final answer is \boxed{|b-a| = 1}, which corresponds to option (B).