Consider `f(x) = {{:(2 sin (a cos^(-1) x)",","if",x in (0, 1)),(sqrt(3)",","if",x = 0),(ax + b",","if",x lt 0):}`
Statement I If b = `sqrt(3) and a = (2)/(3)`, then f(x) is continuous in `(-oo, 1)`.
Statement II If a function is defined on an interval I and limit exists at every point of interval I, then function is continuou in I.

Options

A. Statement I is correct, Statement II is also correct, Statement II is the correct explanation of Statement I
B. Statement I is correct, Statement II is also correct, Statement II is not the correct explanation of Statement I
C. Statement I is correct, Statement II is incorrect
D. Statement I is incorrect, Statement II is correct.

Answer

To determine the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} 2 \sin(a \cos^{-1}(x)) & \text{if } x \in (0, 1) \\ \sqrt{3} & \text{if } x = 0 \\ ax + b & \text{if } x < 0 \end{cases} \]

we need to analyze the continuity at \( x = 0 \) since that is where the function changes its definition.

### Step 1: Calculate the left-hand limit as \( x \) approaches 0
The left-hand limit is calculated using the piece of the function defined for \( x < 0 \):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (ax + b) = a(0) + b = b \]

### Step 2: Calculate the right-hand limit as \( x \) approaches 0
The right-hand limit is calculated using the piece of the function defined for \( x \in (0, 1) \):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2 \sin(a \cos^{-1}(x))) \]

Since \( \cos^{-1}(0) = \frac{\pi}{2} \), we have:

\[ \lim_{x \to 0^+} f(x) = 2 \sin\left(a \cdot \frac{\pi}{2}\right) = 2 \sin\left(\frac{a \pi}{2}\right) \]

### Step 3: Evaluate the function at \( x = 0 \)
The value of the function at \( x = 0 \) is given directly by the function's definition:

\[ f(0) = \sqrt{3} \]

### Step 4: Set up the continuity condition
For \( f(x) \) to be continuous at \( x = 0 \), the left-hand limit, right-hand limit, and the function value must all be equal:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

This gives us the equations:

1. \( b = \sqrt{3} \)
2. \( 2 \sin\left(\frac{a \pi}{2}\right) = \sqrt{3} \)

### Step 5: Solve for \( b \) and \( a \)
From the first equation, we have:

\[ b = \sqrt{3} \]

From the second equation, we can isolate \( \sin\left(\frac{a \pi}{2}\right) \):

\[ \sin\left(\frac{a \pi}{2}\right) = \frac{\sqrt{3}}{2} \]

The angle whose sine is \( \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \). Therefore, we can set:

\[ \frac{a \pi}{2} = \frac{\pi}{3} \]

Solving for \( a \):

\[ a = \frac{2}{3} \]

### Conclusion
We have found that:

- \( b = \sqrt{3} \)
- \( a = \frac{2}{3} \)

Thus, \( f(x) \) is continuous at \( x = 0 \) when \( b = \sqrt{3} \) and \( a = \frac{2}{3} \).

### Verification of Statements
- **Statement I**: True, since we found \( b = \sqrt{3} \) and \( a = \frac{2}{3} \).
- **Statement II**: False, because the existence of limits does not guarantee continuity unless the limits equal the function value.