Let f : R → R be defined as f(x) = { [λ|x² - 5x + 6|] / [μ(5x - x² - 6)], x

Q: Wait, why does the modulus just disappear? Doesn't it always change the sign?

Great question! The modulus $|A|$ is equal to $A$ if $A \ge 0$ and $-A$ if $A < 0$. Near $x=2$, specifically for $x < 2$, the quadratic $x^2 - 5x + 6 = (x-2)(x-3)$ is the product of two negative numbers (since $x-2 < 0$ and $x-3 < 0$). A negative times a negative is positive! So, the expression inside is already positive, and the modulus is effectively invisible.

Q: Why does $[x]$ become 2 in the RHL?

The greatest integer function $[x]$ returns the largest integer less than or equal to $x$. When we approach $2$ from the right ($x \to 2^+$), $x$ is something like $2.0001$. The largest integer less than or equal to $2.0001$ is $2$. Thus, $[x] = 2$.

Q: Why do we equate the limits to $\mu$?

Because the problem defines $f(2) = \mu$. For a function to be continuous at a point, the limit as you approach that point must equal the value of the function at that point. It's the definition of a 'connected' graph.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be defined as $f(x) = \begin{cases} \frac{\lambda |x^2 - 5x + 6|}{\mu(5x - x^2 - 6)}, & x < 2 \ \frac{e^{\frac{\sin(x-2)}{x-[x]}}}{\mu}, & x > 2 \ \mu, & x = 2 \end{cases}$ where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x = 2$, then $\lambda + \mu$ is equal to:

Options:

(A) $e(-e + 1)$

(B) $e(e - 2)$

(C) None

(D) $2e - 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

For a function $f(x)$ to be continuous at $x=2$, the path approaching from the left, the path approaching from the right, and the actual point itself must meet at the same location.

Mathematically, this requires:
$$\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2)$$

Given $f(2) = \mu$, our goal is to determine the values of $\lambda$ and $\mu$ that satisfy this equality.

### The Left Hand Limit: Unmasking the Modulus

For the left side of the bridge, where $x < 2$, the function is defined as:
$$f(x) = \frac{\lambda |x^2 - 5x + 6|}{\mu(5x - x^2 - 6)}$$

Factoring the quadratic $x^2 - 5x + 6$ yields $(x-2)(x-3)$. When $x$ is slightly less than $2$, both $(x-2)$ and $(x-3)$ are negative, making their product positive.

Since the expression inside the modulus is positive, $|x^2 - 5x + 6| = x^2 - 5x + 6$. The denominator $5x - x^2 - 6$ is equivalent to $-(x^2 - 5x + 6)$.

Canceling the quadratic terms, we find the Left Hand Limit (LHL):
$$	ext{LHL} = -\frac{\lambda}{\mu}$$

### The Right Hand Limit: The Power of Standard Limits

For the right side of the bridge, where $x > 2$, the function is:
$$f(x) = \frac{e^{\frac{\sin(x-2)}{x-[x]}}}{\mu}$$

As $x$ approaches $2$ from the right, the greatest integer function $[x]$ becomes $2$. Consequently, the denominator in the exponent, $x - [x]$, simplifies to $x - 2$.

We evaluate the limit:
$$\lim_{x 	o 2^+} \frac{e^{\frac{\sin(x-2)}{x-2}}}{\mu}$$

By substituting $h = x-2$, as $x 	o 2$, $h 	o 0$. Since $\lim_{h 	o 0} \frac{\sin h}{h} = 1$, the Right Hand Limit (RHL) becomes:
$$	ext{RHL} = \frac{e^1}{\mu} = \frac{e}{\mu}$$

### The Final Synthesis

We now equate our findings: the LHL is $-\frac{\lambda}{\mu}$, the RHL is $\frac{e}{\mu}$, and $f(2) = \mu$.

Equating the RHL to $f(2)$:
$$\frac{e}{\mu} = \mu \Rightarrow \mu^2 = e \Rightarrow \mu = \sqrt{e}$$

Equating the LHL to $f(2)$:
$$-\frac{\lambda}{\mu} = \mu \Rightarrow -\lambda = \mu^2$$

Substituting $\mu^2 = e$ into the equation $-\lambda = \mu^2$, we find:
$$\lambda = -e$$

The final value requested is $\lambda + \mu$:
$$\lambda + \mu = -e + \sqrt{e}$$

**You have successfully navigated the bridge of continuity.**

Select Answer:

Visualized Solution

Condition for Continuity

Analyzing LHL: Factorization

Handling the Modulus

Evaluating the LHL

Analyzing RHL: Greatest Integer

Evaluating the RHL Limit

Finding $μ$ and $λ$

Final Calculation: $λ + μ$

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

Analyzing the Setup

The Left Hand Limit

The Right Hand Limit

The Final Synthesis

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