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

To determine the number of ordered pairs \((a, b)\) such that the function \(f(x)\) is continuous at \(x = 3\), we need to ensure that the left-hand limit, right-hand limit, and the value of the function at \(x = 3\) are equal.

Step 1: Check Continuity at \(x = 3\) 

The function is given by:

• \(f(x) = \frac{a(7x - 12 - x^2)}{b(x^2 - 7x + 12)}\) for \(x < 3\)
• \(f(x) = \frac{\sin(x - 3)}{2^{x - \lfloor x \rfloor}}\) for \(x > 3\)
• \(f(x) = b\) for \(x = 3\)

Step 2: Evaluate the Limit as \(x \to 3^-\)

For \(x < 3\), the denominator becomes:

\(x^2 - 7x + 12 = (x - 3)(x - 4)\)

So, the function simplifies to:

\(\frac{a(7x - 12 - x^2)}{b(x - 4)}\)

As \(x \to 3^-\), simplify the numerator:

\(7x - 12 - x^2 = -(x - 3)(x + 4)\)

The expression becomes:

\(\frac{a(-(x-3)(x+4))}{b(x-4)} = \frac{a(-(3-4)(3+4))}{b(3-4)}\)

Thus, the left-hand limit is:

\(L_1 = \frac{-7a}{b}\)

Step 3: Evaluate the Limit as \(x \to 3^+\)

For \(x > 3\), the function becomes:

\(f(x) = \frac{\sin(x-3)}{2^{x - \lfloor x \rfloor}}\)

Since \(\lfloor x \rfloor = 3\), it simplifies to:

\(\frac{\sin(x-3)}{1} = \sin(x-3)\)

The right-hand limit as \(x \to 3^+\) is:

\(L_2 = \sin(0) = 0\)

Step 4: Equate Limits and the Function Value

For continuity, \(L_1 = L_2 = f(3)\):

• \(\frac{-7a}{b} = 0 \Rightarrow a = 0\)
• And, \(f(3) = b = 0\)

Thus, the ordered pair is \((a, b) = (0, 0)\).

Conclusion

There is only one ordered pair \((0, 0)\) where the function \(f(x)\) is continuous at \(x = 3\). Therefore, the number of elements in set \(S\) is:

1

Answer 2

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 \( (a, b) = (-4, 2) \).