The problem asks for the value of \(q\) in a given piecewise function \(f(x)\) that is stated to be continuous. A piecewise function is continuous if it is continuous at the points where its definition changes.

The given function is:

\[ f(x) = \left \{ \begin{matrix} x + 6, & x \le 1 \\ px + q, & 1 < x < 2 \\ 5x , & x \ge 2 \end{matrix} \right. \]

Since \(f(x)\) is continuous, it must be continuous at \(x = 1\) and \(x = 2\).

Ensuring Continuity at \(x = 1\)

For \(f(x)\) to be continuous at \(x = 1\), the left-hand limit (LHL), the right-hand limit (RHL), and the function value at \(x = 1\) must all be equal. That is, \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\).

• Left-Hand Limit at \(x = 1\): We use the definition of \(f(x)\) for \(x \le 1\), which is \(f(x) = x + 6\). \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 6) \] Substituting \(x = 1\), we get \(1 + 6 = 7\).
• Right-Hand Limit at \(x = 1\): We use the definition of \(f(x)\) for \(1 < x < 2\), which is \(f(x) = px + q\). \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (px + q) \] Substituting \(x = 1\), we get \(p(1) + q = p + q\).
• Function Value at \(x = 1\): We use the definition of \(f(x)\) for \(x \le 1\), which is \(f(x) = x + 6\). \[ f(1) = 1 + 6 = 7 \]

Equating the LHL, RHL, and \(f(1)\) for continuity at \(x = 1\):

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \implies 7 = p + q \quad \text{(Equation 1)} \]

Ensuring Continuity at \(x = 2\)

For \(f(x)\) to be continuous at \(x = 2\), the LHL, the RHL, and the function value at \(x = 2\) must all be equal. That is, \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)\).

• Left-Hand Limit at \(x = 2\): We use the definition of \(f(x)\) for \(1 < x < 2\), which is \(f(x) = px + q\). \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (px + q) \] Substituting \(x = 2\), we get \(p(2) + q = 2p + q\).
• Right-Hand Limit at \(x = 2\): We use the definition of \(f(x)\) for \(x \ge 2\), which is \(f(x) = 5x\). \[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5x) \] Substituting \(x = 2\), we get \(5(2) = 10\).
• Function Value at \(x = 2\): We use the definition of \(f(x)\) for \(x \ge 2\), which is \(f(x) = 5x\). \[ f(2) = 5(2) = 10 \]

Equating the LHL, RHL, and \(f(2)\) for continuity at \(x = 2\):

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \implies 2p + q = 10 \quad \text{(Equation 2)} \]

Solving for p and q

We now have a system of two linear equations with two variables \(p\) and \(q\):

1. \(p + q = 7\)
2. \(2p + q = 10\)

To find the value of \(q\), we can solve this system. One way is to subtract Equation 1 from Equation 2:

\[ (2p + q) - (p + q) = 10 - 7 \]

\[ 2p + q - p - q = 3 \]

\[ p = 3 \]

Now substitute the value of \(p = 3\) into Equation 1:

\[ p + q = 7 \]

\[ 3 + q = 7 \]

\[ q = 7 - 3 \]

\[ q = 4 \]

Thus, the value of \(q\) is 4.

Revision Table: Piecewise Function Continuity

Understanding continuity for piecewise functions is key to solving this problem. Here's a quick review:

• A function \(f(x)\) is continuous at a point \(c\) if \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)\).
• For a piecewise function, continuity needs to be checked at the points where the definition of the function changes.
• At these points, the limit from the left must equal the limit from the right, and both must equal the function's value at that specific point.

Additional Information: Solving System of Equations

We solved a system of two linear equations. Here are common methods for solving such systems:

• Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation.
• Elimination Method: Multiply one or both equations by a constant so that one variable has coefficients that are opposites. Then add the equations to eliminate that variable. This is the method we primarily used here (by subtracting one equation from the other).
• Matrix Method: Represent the system as a matrix equation and use matrix operations to solve (e.g., Cramer's rule, inverse matrix).

In this problem, using elimination (subtraction) was straightforward because the coefficient of \(q\) was the same in both equations.