Question

If the function \[ f(x) = \begin{cases} 1, & x \leq 2 \\ ax + b, & 2 < x < 4 \\ 7, & x \geq 4 \end{cases} \] is continuous at \( x = 2 \) and \( x = 4 \), then the values of \( a \) and \( b \) are:

• A. \( a = 3, b = -5 \)
• B. \( a = -5, b = 3 \)
• C. \( a = -3, b = 5 \)
• D. \( a = 5, b = -3 \)

Hint:

For continuity at \( x = a \), ensure \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \).

Answer 1

The Correct Option is A

Step 1: Checking continuity at \( x = 2 \).
For continuity, \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \] \[ 1 = a(2) + b \] \[ 2a + b = 1 \] Step 2: Checking continuity at \( x = 4 \).
\[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4) \] \[ a(4) + b = 7 \] \[ 4a + b = 7 \] Step 3: Solving for \( a \) and \( b \).
Solving the system: \[ 2a + b = 1 \] \[ 4a + b = 7 \] Subtracting, \[ 2a = 6 \Rightarrow a = 3 \] Substituting \( a = 3 \): \[ 2(3) + b = 1 \] \[ 6 + b = 1 \Rightarrow b = -5 \] Thus, the correct answer is (A) \( a = 3, b = -5 \).