If the function f defined on (π/6, π/3) by f(x) = (√2cos x - 1)/(cot x - 1) for x ≠ π/4, and k for x

Q: Wait, why do we use L'Hopital's Rule instead of just simplifying the trig expression?

You absolutely could use trig identities! You could rewrite $\cot x$ as $\cos x / \sin x$ and manipulate the fraction. However, L'Hopital's Rule is often faster and less prone to algebraic errors in high-pressure exams. Both paths lead to the same truth; choose the one where your intuition feels most confident.

Q: Does the function being defined on $(\pi/6, \pi/3)$ matter for the calculation?

Great observation! That interval defines the 'domain of interest.' Since $\pi/4$ lies comfortably inside $(\pi/6, \pi/3)$, we are guaranteed that the function behaves predictably around our point of interest. It ensures we aren't dealing with any hidden discontinuities or domain restrictions.

Q: Why did the negative signs cancel out in the derivative step?

The derivative of $\cos x$ is $-\sin x$, and the derivative of $\cot x$ is $-\csc^2 x$. When you place them in the fraction, you have a negative in the numerator and a negative in the denominator. A negative divided by a negative is always a positive. Always keep track of your signs—they are the most common place to lose marks!

Q: Is it possible to solve this without calculus?

Yes! By substituting $\cot x = \cos x / \sin x$, you can transform the expression into $\frac{(\sqrt{2}\cos x - 1)\sin x}{\cos x - \sin x}$. This is a classic 'plugging the hole' problem. However, L'Hopital's Rule is the standard 'JEE-style' approach for these indeterminate forms.

Q: What if the limit didn't exist? Would the function still be continuous?

No. Continuity is a strict requirement. If the limit from the left did not equal the limit from the right, or if the limit resulted in infinity, the function would have a jump or infinite discontinuity, and no value of $k$ could ever make it continuous.

Question

If the function $f$ defined on $(\pi/6, \pi/3)$ by $f(x) = \begin{cases} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x 
eq \pi/4 \ k, & x = \pi/4 \end{cases}$ is continuous, then $k$ is equal to

Options:

(A) $1/2$

(B) 1

(C) $1/\sqrt{2}$

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an architect tasked with building a bridge. You have two segments of a road, and you need to connect them at a specific point, $x = \pi/4$.

If the segments do not meet perfectly, you have a gap—a discontinuity. In the world of calculus, we call this a 'removable discontinuity.' Our goal is to find the exact value $k$ that acts as the keystone, perfectly sealing the gap and making the function $f(x)$ continuous.

### The Condition of Harmony

For a function to be continuous at a point $x = a$, the limit of the function as it approaches that point must be exactly equal to the value of the function at that point. Mathematically, we require that:

$$\lim_{x 	o \pi/4} f(x) = f(\pi/4)$$

Since we are given $f(\pi/4) = k$, our mission is to calculate the limit of the expression as $x$ approaches $\pi/4$. We start with the expression:

$$k = \lim_{x 	o \pi/4} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}$$

### The Indeterminate Trap

Before we dive into heavy math, let's test the waters. If we plug in $x = \pi/4$ directly, the numerator becomes $\sqrt{2}\cos(\pi/4) - 1 = \sqrt{2}(1/\sqrt{2}) - 1 = 0$.

The denominator becomes $\cot(\pi/4) - 1 = 1 - 1 = 0$. We have hit the classic $0/0$ indeterminate form. This is the universe's way of telling us that there is a 'hole' in the graph that needs to be filled.

### The Surgeon's Tool: L'Hopital's Rule

When we face a $0/0$ wall, we use L'Hopital's Rule. This rule allows us to differentiate the numerator and the denominator independently to find the limit.

Let's differentiate the numerator:

$$\frac{d}{dx}(\sqrt{2}\cos x - 1) = -\sqrt{2}\sin x$$

Now, the denominator:

$$\frac{d}{dx}(\cot x - 1) = -\csc^2 x$$

By placing these back into our limit, we get:

$$k = \lim_{x 	o \pi/4} \frac{-\sqrt{2}\sin x}{-\csc^2 x}$$

### The Beauty of Simplification

Notice how the negative signs vanish, leaving us with a much cleaner expression. We know that $\csc x = 1/\sin x$, so $\csc^2 x = 1/\sin^2 x$.

Our expression transforms beautifully:

$$\frac{\sin x}{\csc^2 x} = \sin x \cdot \sin^2 x = \sin^3 x$$

Now, the limit looks much friendlier:

$$k = \lim_{x 	o \pi/4} \sqrt{2}\sin^3 x$$

### The Final Connection

We are at the finish line. We substitute $x = \pi/4$ into our simplified expression. We know that $\sin(\pi/4) = 1/\sqrt{2}$.

Therefore:

$$k = \sqrt{2} \left( \frac{1}{\sqrt{2}} ight)^3$$

Expanding the cube, we get $1/(2\sqrt{2})$. Multiplying this by the $\sqrt{2}$ outside, the square roots cancel out perfectly, leaving us with:

**$$k = \frac{1}{2}$$**

### Reflection

We took a complex, seemingly broken function and, through the systematic application of calculus, found the exact value required to restore its continuity. This is the essence of JEE mathematics: identifying the indeterminate, applying the right tool, and simplifying until the truth reveals itself. You have successfully bridged the gap.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f$ defined on $(π /6, π /3)$ by $f (x) = {\frac{2 c o s x - 1}{c o t x - 1}, k, x \neq = π /4 x = π /4$ is continuous, then $k$ is equal to

Select Answer:

Visualized Solution

Visualizing the Function $f (x)$

Condition for Continuity

Checking the Indeterminate Form

Applying L'Hopital's Rule

Differentiating the Numerator

Differentiating the Denominator

Assembling the New Limit

Simplifying the Expression

Evaluating the Limit

Final Calculation of $k$

Conclusion

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

Analyzing the Setup

The Condition of Harmony

The Indeterminate Trap

The Surgeon's Tool

The Beauty of Simplification

The Final Connection

Reflection

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