To solve this problem, we are given a piecewise function $f(x)$ and need to find the values of $a$ and $b$ so that $f(x)$ is continuous everywhere.

Step-by-Step Solution

1. Continuity at $x = 1$

For continuity at $x = 1$, the left-hand limit (LHL) must equal the right-hand limit (RHL), which must also equal $f(1)$.

• From the definition of $f(x)$: $\text{LHL at } x = 1: \lim_{x \to 1^-} f(x) = -5$ $\text{RHL at } x = 1: \lim_{x \to 1^+} f(x) = a(1)^2 - b = a - b$

• Continuity condition:

  \tag{1}$$

2. Continuity at $x = 2$

For continuity at $x = 2$, the left-hand limit (LHL) must equal the right-hand limit (RHL), which must also equal $f(2)$.

• From the definition of $f(x)$: $\text{LHL at } x = 2: \lim_{x \to 2^-} f(x) = a(2)^2 - b = 4a - b$ $\text{RHL at } x = 2: \lim_{x \to 2^+} f(x) = 20(2) = 40$

• Continuity condition:

  \tag{2}$$

3. Solve the System of Equations

We now solve the system of two equations:

1. $a - b = -5$
2. $4a - b = 40$

• Subtract equation (1) from equation (2): $(4a - b) - (a - b) = 40 - (-5)$ $3a = 45$ $a = 15$

• Substitute $a = 15$ into equation (1): $15 - b = -5$ $b = 20$

4. Final Answer

The values of $a$ and $b$ are: $a = 15, \quad b = 20$

Let me know if you'd like further details or explanations. Here are five related questions to expand your understanding:

1. What does continuity mean for a piecewise function?
2. How would the solution change if the function had a different expression for $x \geq 2$?
3. Can you derive a similar condition for differentiability of $f(x)$ at $x = 1$ and $x = 2$?
4. What is the graphical interpretation of the conditions we derived for $a$ and $b$?
5. How does the solution change if $f(x)$ is defined differently for $x < 1$?

Tip: When solving for continuity in piecewise functions, always focus on ensuring the limits match at the boundary points!