Question

Consider the function.\(f(x) = \begin{cases}  \frac{a(7x - 12 - x^2)}{b(x^2 - 7x + 12)} & , \quad x < 3 \\[8pt] \frac{\sin(x - 3)}{2^{x - \lfloor x \rfloor}} & , \quad x > 3 \\[8pt] b & , \quad x = 3  \end{cases}\)Where \(\lfloor x \rfloor\)denotes the greatest integer less than or equal to \(x\). If \(S\) denotes the set of all ordered pairs \((a, b)\) such that \(f(x)\) is continuous at \(x = 3\), then the number of elements in \(S\) is:

• A. 2
• B. Infinitely many
• C. 4
• D. 1

Answer 1

The Correct Option is D

Step 1. Continuity Condition at \( x = 3 \): For \( f(x) \) to be continuous at \( x = 3 \), we must have:

 \(f(3^-) = f(3) = f(3^+)\)
 

Step 2. Calculate \( f(3^-) \): For \( x < 3 \),

  \(f(x) = \frac{a(7x - 12 - x^2)}{b|x^2 - 7x + 12|} = \frac{-a(x - 3)(x - 4)}{b(x - 3)(x - 4)} = \frac{-a}{b}\)
 
  So, \(f(3^-) = -\frac{a}{b}\).

Step 3. Calculate \( f(3^+) \): For \( x > 3 \),

  \(f(x) = \frac{\sin(x - 3)}{2x - |x|} \Rightarrow \lim_{x \to 3^+} f(x) = 2\)
  

Step 4. Set Up Continuity Condition: Since \( f(3^-) = f(3) = f(3^+) \),

  \(-\frac{a}{b} = 2 \quad \text{and} \quad b = 2 \Rightarrow a = -4\)
 

  Therefore, the only solution is \( (a, b) = (-4, 2) \).