To ensure that the function $f(x)$ is continuous on the real number line, we need to determine values of the constants $A$ and $B$ so that the limits match at the boundaries of each interval:

1. Continuity at $x = 0$:

   • For continuity at $x = 0$, the limit of $f(x)$ as $x$ approaches 0 from the left must equal the limit as $x$ approaches 0 from the right.
   • Left-hand limit at $x = 0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{A(1 - \cos x)}{\sin^2 x}$.
   • Right-hand limit at $x = 0$: $\lim_{x \to 0^+} f(x) = f(0) = 2(0)^2 - 0 + B = B$.
   • Solve for $A$ in terms of $B$ by setting these two limits equal.

2. Continuity at $x = 1$:

   • For continuity at $x = 1$, the limit of $f(x)$ as $x$ approaches 1 from the left must equal the limit as $x$ approaches 1 from the right.
   • Left-hand limit at $x = 1$: $\lim_{x \to 1^-} f(x) = 2(1)^2 - 1 + B = 2 + B$.
   • Right-hand limit at $x = 1$: $\lim_{x \to 1^+} f(x) = \frac{1^2 + 2 \cdot 1 - 3}{1^2 - 1}$.
   • Solve for $B$ by setting these two limits equal.

After calculating both conditions, you'll get specific values for $A$ and $B$ that make $f(x)$ continuous across all intervals.

Would you like me to carry out these detailed calculations?

1. How do we handle limits approaching a point for continuity?
2. What happens if the values of $A$ and $B$ are not adjusted for continuity?
3. Can continuity always be ensured by just adjusting constants in piecewise functions?
4. Why do we use limits from the left and right to check continuity at points?
5. How would the answer change if $f(x)$ had more complex intervals?

Tip: Always check limits carefully in piecewise functions, as small differences can disrupt continuity.