Consider the function f(x) = {a(7x-12-x²)/(b|x²-7x+12|) , x 3; b , x

Q: Wait, why does the modulus just disappear in the left-hand limit?

Great question! The modulus $|x^2-7x+12|$ is defined as $(x^2-7x+12)$ if the expression is positive, and $-(x^2-7x+12)$ if it is negative. When we approach $3$ from the left ($x < 3$), we factor the quadratic into $(x-3)(x-4)$. Since both $(x-3)$ and $(x-4)$ are negative for $x < 3$, their product is positive. Thus, the expression inside the modulus is positive, and we can simply remove the bars without changing the sign.

Q: What happens to the greatest integer function $[x]$ as $x \to 3^+$?

When we approach $3$ from the right, $x$ is slightly larger than $3$ (e.g., $3.001$). By definition, the greatest integer less than or equal to $x$ is the largest integer not exceeding $x$. For any $x$ in the interval $[3, 4)$, the greatest integer $[x]$ is exactly $3$. So, we simply replace $[x]$ with $3$ in our expression.

Q: Can we just assume $a$ and $b$ are constants?

Yes, the problem treats $a$ and $b$ as fixed parameters. Our goal is to find the specific values of $a$ and $b$ that satisfy the continuity condition. Once we equate the limits, we get a system of equations that allows us to solve for these constants uniquely.

Question

Consider the function $f(x) = \begin{cases} \frac{a(7x-12-x^2)}{b|x^2-7x+12|} & , x < 3 \ 2^{\frac{\sin(x-3)}{x-[x]}} & , x > 3 \ b & , x = 3 \end{cases}$ Where $[x]$ 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 :

Options:

(A) 2

(B) Infinitely many

(C) 4

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an architect designing a bridge that must span across the point $x=3$. For the bridge to be safe, the road coming from the left, the road coming from the right, and the central pillar at $x=3$ must all connect perfectly.

In the world of calculus, this is exactly what we mean by continuity. We are given a function $f(x)$ defined in three pieces, and our mission is to find the values of $a$ and $b$ that ensure this bridge has no gaps.

### The Condition of Connection

For $f(x)$ to be continuous at $x=3$, the mathematical requirement is elegant and absolute:

$$\lim_{x 	o 3^-} f(x) = \lim_{x 	o 3^+} f(x) = f(3)$$

This means the left-hand limit, the right-hand limit, and the function value at the point must all converge to the same height. Let us start with the easiest piece: the value at the point. The definition tells us $f(3) = b$. This is our target height.

### The Left-Hand Approach: Taming the Modulus

Now, let us look at the left side, where $x < 3$. The function is defined as:

$$f(x) = \frac{a(7x-12-x^2)}{b|x^2-7x+12|}$$

This looks intimidating, but let us simplify the numerator. Notice that $7x-12-x^2 = -(x^2-7x+12)$. Factoring the quadratic, we get $-(x-3)(x-4)$.

Now, consider the modulus in the denominator. As $x 	o 3^-$, $x$ is slightly less than $3$. In this region, both $(x-3)$ and $(x-4)$ are negative.

A negative times a negative is a positive, so the expression $(x-3)(x-4)$ is positive. Therefore, $|x^2-7x+12| = x^2-7x+12$. The limit becomes:

$$\lim_{x 	o 3^-} \frac{-a(x-3)(x-4)}{b(x-3)(x-4)}$$

The terms cancel beautifully, leaving us with the left-hand limit: $-\frac{a}{b}$.

### The Right-Hand Approach: The Power of Limits

Moving to the right side, where $x > 3$, we have $f(x) = 2^{\frac{\sin(x-3)}{x-[x]}}$. As $x 	o 3^+$, $x$ is in the interval $(3, 4)$, so $[x] = 3$.

The exponent becomes $\frac{\sin(x-3)}{x-3}$. This is a classic form! As $x 	o 3$, the angle $(x-3) 	o 0$.

Using the standard limit $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$, the exponent simplifies to $1$. Thus, our right-hand limit is $2^1 = 2$.

### The Final Synthesis

We now have our three components: the left-hand limit $-\frac{a}{b}$, the right-hand limit $2$, and the function value $b$. For continuity, they must all be equal:

$$-\frac{a}{b} = 2 = b$$

From $2 = b$, we immediately find **$b = 2$**. Substituting this into $-\frac{a}{b} = 2$, we get:

$$-\frac{a}{2} = 2 \Rightarrow a = -4$$

We have found the unique pair **$(a, b) = (-4, 2)$**. The bridge is complete, the gaps are closed, and the function is continuous.

Select Answer:

Visualized Solution

Condition for Continuity at $x = 3$

Value of $f (3)$

Left Hand Limit Setup

Factorizing the Quadratic

Resolving the Modulus

Evaluating the $L H L$

Right Hand Limit Setup

Evaluating the Greatest Integer Function

Evaluating the $R H L$

Applying Continuity Condition

Solving for $a$ and $b$

Final Answer

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Condition of Connection

The Left-Hand Approach

The Right-Hand Approach

The Final Synthesis

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