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

Answer: (D)

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