Let `f(x) = {{:(Ax - B,x le 1),(2x^(2) + 3Ax + B,x in (-1, 1]),(4,x gt 1):}`
Statement I f(x) is continuous at all x if `A = (3)/(4), B = - (1)/(4)`. Because
Statement II Polynomial function is always continuous.

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 if the function \( f(x) \) is continuous at all \( x \) given the conditions for \( A \) and \( B \), we will analyze the function piecewise and check the continuity at the critical points \( x = -1 \) and \( x = 1 \).

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} Ax - B & \text{if } x \leq -1 \\ 2x^2 + 3Ax + B & \text{if } -1 < x \leq 1 \\ 4 & \text{if } x > 1 \end{cases} \]

### Step 2: Check continuity at \( x = -1 \)
For \( f(x) \) to be continuous at \( x = -1 \), the left-hand limit (LHL) and right-hand limit (RHL) at \( x = -1 \) must be equal to \( f(-1) \).

- **Left-hand limit** as \( x \to -1^- \):
\[ \lim_{x \to -1^-} f(x) = A(-1) - B = -A - B \]

- **Right-hand limit** as \( x \to -1^+ \):
\[ \lim_{x \to -1^+} f(x) = 2(-1)^2 + 3A(-1) + B = 2 - 3A + B \]

Setting the left-hand limit equal to the right-hand limit:
\[ -A - B = 2 - 3A + B \]
Rearranging gives:
\[ 2A - 2B = 2 \quad \Rightarrow \quad A - B = 1 \quad \text{(Equation 1)} \]

### Step 3: Check continuity at \( x = 1 \)
For \( f(x) \) to be continuous at \( x = 1 \), we again set the left-hand limit equal to the right-hand limit.

- **Left-hand limit** as \( x \to 1^- \):
\[ \lim_{x \to 1^-} f(x) = 2(1)^2 + 3A(1) + B = 2 + 3A + B \]

- **Right-hand limit** as \( x \to 1^+ \):
\[ \lim_{x \to 1^+} f(x) = 4 \]

Setting these equal gives:
\[ 2 + 3A + B = 4 \]
Rearranging gives:
\[ 3A + B = 2 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( A - B = 1 \)
2. \( 3A + B = 2 \)

From Equation 1, we can express \( B \) in terms of \( A \):
\[ B = A - 1 \]

Substituting \( B \) into Equation 2:
\[ 3A + (A - 1) = 2 \]
Simplifying gives:
\[ 4A - 1 = 2 \quad \Rightarrow \quad 4A = 3 \quad \Rightarrow \quad A = \frac{3}{4} \]

Substituting \( A \) back into the expression for \( B \):
\[ B = \frac{3}{4} - 1 = -\frac{1}{4} \]

### Conclusion
Thus, the values of \( A \) and \( B \) for which \( f(x) \) is continuous at all \( x \) are:
\[ A = \frac{3}{4}, \quad B = -\frac{1}{4} \]

### Verification of Statements
- **Statement I**: True, because we found \( A = \frac{3}{4} \) and \( B = -\frac{1}{4} \) makes \( f(x) \) continuous.
- **Statement II**: True, because polynomial functions are continuous everywhere.

However, Statement II does not explain why the specific values of \( A \) and \( B \) ensure continuity at the critical points.