If the function f(x) = { (1 + |cos x|)^(λ/|cos x|), 0

Q: Why is the LHL an indeterminate form of $1^\infty$?

As $x \to \frac{\pi}{2}^-$, $|\cos x| \to 0$. The base $(1 + |\cos x|)$ approaches $1$, and the exponent $\frac{\lambda}{|\cos x|}$ approaches $\infty$. This is the classic $1^\infty$ form, which we resolve using the definition of the exponential constant $e$.

Q: How did $\cot(3\pi + 6h)$ simplify to $\cot(6h)$?

Remember that the cotangent function has a period of $\pi$. Therefore, $\cot(n\pi + \theta) = \cot(\theta)$ for any integer $n$. Since $3\pi$ is an integer multiple of the period, it simply vanishes, leaving us with $\cot(6h)$.

Q: Why do we equate $LHL = RHL = \mu$?

Continuity is defined by the absence of gaps. If the limit from the left didn't equal the limit from the right, or if either didn't equal the function value $\mu$, there would be a jump or a hole at $x = \frac{\pi}{2}$. Equating them ensures the graph is unbroken.

Question

If the function $f(x) = \begin{cases} (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}} & , 0 < x < \frac{\pi}{2} \ \mu & , x = \frac{\pi}{2} \ e^{\frac{\cot 6x}{\cot 4x}} & , \frac{\pi}{2} < x < \pi \end{cases}$ is continuous at $x = \frac{\pi}{2}$, then $9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}$ is equal to

Options:

(A) 11

(B) 8

(C) $2e^4 + 8$

(D) 10

Sigma Percentile · Accepted Answer

### The Elegance of Continuity: A Journey Through Limits

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are exploring the very definition of a 'smooth' existence in the world of calculus.

We are given a piecewise function, $f(x)$, which changes its identity as it crosses the threshold of $x = \frac{\pi}{2}$. Our mission is to ensure that this transition is seamless—that there are no jumps, no holes, and no sudden breaks in the road. This is the essence of continuity.

### Phase 1: The Left-Hand Limit and the $1^\infty$ Mystery

Let us begin by standing to the left of our destination, $x = \frac{\pi}{2}$. As we approach from the left, the function is defined as $f(x) = (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}}$.

As $x$ creeps closer to $\frac{\pi}{2}$, the value of $\cos x$ shrinks toward zero. Consequently, the base of our expression, $(1 + |\cos x|)$, approaches $1$, while the exponent, $\frac{\lambda}{|\cos x|}$, blows up toward infinity. We have arrived at the classic indeterminate form: $1^\infty$.

We use the standard limit definition: 
$$\lim_{t 	o 0} (1+t)^{\frac{1}{t}} = e$$

By substituting $t = |\cos x|$, our expression transforms into $\lim_{t 	o 0^+} (1+t)^{\frac{\lambda}{t}}$. This is simply $( \lim_{t 	o 0^+} (1+t)^{\frac{1}{t}} )^\lambda$, which yields the result: $e^\lambda$. We have successfully bridged the gap from the left.

### Phase 2: The Right-Hand Limit and Trigonometric Harmony

Now, let us pivot to the right side of $x = \frac{\pi}{2}$. Here, the function takes the form $f(x) = e^{\frac{\cot 6x}{\cot 4x}}$.

We evaluate the limit as $x 	o \frac{\pi}{2}^+$. Let $x = \frac{\pi}{2} + h$, where $h$ is a tiny positive value approaching zero.

Substituting this into our trigonometric terms, the exponent becomes:
$$\frac{\cot(6(\frac{\pi}{2} + h))}{\cot(4(\frac{\pi}{2} + h))} = \frac{\cot(3\pi + 6h)}{\cot(2\pi + 4h)}$$

Because the cotangent function is periodic with period $\pi$, we can discard the integer multiples of $\pi$. Thus, $\cot(3\pi + 6h)$ becomes $\cot(6h)$, and $\cot(2\pi + 4h)$ becomes $\cot(4h)$.

Our limit is now $\lim_{h 	o 0^+} \frac{\cot 6h}{\cot 4h}$. Converting to tangents, we get:
$$\lim_{h 	o 0^+} \frac{	an 4h}{	an 6h} = \lim_{h 	o 0^+} \left( \frac{	an 4h}{4h} \cdot \frac{6h}{	an 6h} \cdot \frac{4h}{6h} ight) = \frac{4}{6} = \frac{2}{3}$$

Therefore, our Right-Hand Limit is $e^{\frac{2}{3}}$.

### Phase 3: The Synthesis and the Final Calculation

For the function to be continuous, the Left-Hand Limit, the Right-Hand Limit, and the value of the function at the point must all be equal. We have established that $LHL = e^\lambda$ and $RHL = e^{\frac{2}{3}}$.

Since $f(\frac{\pi}{2}) = \mu$, we have the equality:
$$e^\lambda = \mu = e^{\frac{2}{3}}$$

By direct comparison, we find our keys to the kingdom: $\lambda = \frac{2}{3}$ and $\mu = e^{\frac{2}{3}}$.

Finally, we substitute these values into the expression $9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}$:

1. The first term: $9(\frac{2}{3}) = 6$.
2. The second term: $6 \log_e(e^{\frac{2}{3}}) = 6 \cdot \frac{2}{3} = 4$.
3. The third term: $(e^{\frac{2}{3}})^6 = e^4$.
4. The fourth term: $e^{6(\frac{2}{3})} = e^4$.

Look at the final expression: $6 + 4 + e^4 - e^4$. The exponential terms perfectly cancel each other out.

We are left with $6 + 4 = 10$. The final answer is **10**.

Select Answer:

Visualized Solution

The Concept of Continuity

Mathematical Condition for Continuity

Setting up the Left Hand Limit

Evaluating the LHL

Setting up the Right Hand Limit

Substitution for RHL

Simplifying Trigonometric Terms

Evaluating the RHL Limit

Finding $λ$ and $μ$

Setting up the Final Expression

Simplifying the Terms

Final Answer

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

The Elegance of Continuity

Phase 1

Phase 2

Phase 3

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