If the function f defined as f(x) = 1/x - (k-1)/(e²ˣ-1), x ≠ 0, is continuous at x = 0, then the ord

Q: Wait, why do we reject the negative root here? Or why can't we just use L'Hôpital's rule?

L'Hôpital's rule is a valid tool, but it can become algebraically messy when you have an unknown parameter like $k$ in the numerator. Using the Taylor series expansion is often more elegant and less prone to calculation errors in JEE problems because it allows you to see the 'power balance' between the numerator and denominator directly.

Q: Why must the coefficient of $x$ in the numerator be zero?

Great question! If the coefficient of $x$ in the numerator is not zero, the limit as $x \to 0$ will behave like $C/x$ (where $C$ is a constant), which approaches $\pm \infty$. For the function to be continuous at $x=0$, the limit must be a finite, real number. Therefore, we must eliminate any terms that would cause the expression to blow up to infinity.

Q: How do we know the denominator's lowest power is $x^2$?

Look at the denominator: $x(e^{2x}-1)$. The Taylor expansion of $e^{2x}$ is $1 + 2x + 2x^2 + \dots$. So, $e^{2x}-1$ is $2x + 2x^2 + \dots$. When you multiply this by the $x$ outside, you get $2x^2 + 2x^3 + \dots$. The lowest power is indeed $x^2$.

Question

If the function f defined as $f(x) = \frac{1}{x} - \frac{k-1}{e^{2x}-1}, x 
eq 0$, is continuous at $x = 0$, then the ordered pair $(k, f(0))$ is equal to :

Options:

(A) $(\frac{1}{3}, 2)$

(B) $(3, 2)$

(C) $(2, 1)$

(D) $(3, 1)$

Sigma Percentile · Accepted Answer

### The Mystery of the Limit

Imagine you are standing on the edge of a mathematical cliff. You are looking at the function $f(x) = \frac{1}{x} - \frac{k-1}{e^{2x}-1}$.

You are told that this function is continuous at $x=0$. If you plug in $x=0$ directly, you get $\frac{1}{0} - \frac{k-1}{0}$, which is an undefined, indeterminate $\infty - \infty$ form.

The secret lies in the fact that the limit exists and is finite. Our mission is to find the value of $k$ that makes this possible and then determine the value of $f(0)$.

### The Indeterminate Trap

The first step in our journey is to tame this expression. We cannot work with two separate fractions that are both blowing up to infinity.

We must combine them by finding a common denominator, which is $x(e^{2x}-1)$. This transforms our expression into a single, unified fraction:

$$f(x) = \frac{(e^{2x}-1) - (k-1)x}{x(e^{2x}-1)}$$

As $x$ approaches zero, both the numerator and the denominator approach zero. We have a classic $0/0$ indeterminate form.

### The Taylor Series Weapon

To break through this $0/0$ barrier, we need a scalpel. The Taylor series expansion for $e^{z}$ is our perfect tool, where $e^{z} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots$.

By substituting $z = 2x$, we get:

$$e^{2x} = 1 + 2x + \frac{(2x)^2}{2} + \frac{(2x)^3}{6} + \dots = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots$$

This expansion allows us to see the "DNA" of the function near $x=0$.

### The Balancing Act

Now, let's substitute this expansion back into our numerator:

$$	ext{Numerator} = (1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots) - 1 - (k-1)x$$

The constant terms $1$ and $-1$ cancel out beautifully. We are left with:

$$	ext{Numerator} = (2 - (k-1))x + 2x^2 + \frac{4}{3}x^3 + \dots$$

Here is the crucial insight: for the limit to be finite, the lowest power of $x$ in the numerator must be at least as large as the lowest power of $x$ in the denominator.

If we look at the denominator, $x(e^{2x}-1) \approx x(2x) = 2x^2$. Since the lowest power is $x^2$, the coefficient of the $x$ term in our numerator must be zero.

This gives us the equation $2 - (k-1) = 0$, which simplifies to $3 - k = 0$, or **$k = 3$**.

### The Final Calculation

With $k=3$, the $x$ term in the numerator vanishes, and we are left with:

$$f(0) = \lim_{x 	o 0} \frac{2x^2 + \frac{4}{3}x^3 + \dots}{2x^2 + 2x^3 + \dots}$$

Dividing both the numerator and the denominator by $x^2$, we get:

$$f(0) = \lim_{x 	o 0} \frac{2 + \frac{4}{3}x + \dots}{2 + 2x + \dots} = \frac{2}{2} = 1$$

We have successfully found **$k=3$** and **$f(0)=1$**. The final result is the ordered pair **$(3, 1)$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function f defined as $f (x) = \frac{1}{x} - \frac{k - 1}{e ^{2 x} - 1}, x \neq = 0$ , is continuous at $x = 0$ , then the ordered pair $(k, f (0))$ is equal to :

Select Answer:

Visualized Solution

Continuity Condition at $x = 0$

Combining the Fractions

Taylor Expansion of $e^{2 x}$

Substituting into the Numerator

Condition for Finite Limit

Solving for $k$

Expanding the Denominator

Evaluating the Limit for $f (0)$

Final Ordered Pair $(k, f (0))$

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

The Mystery of the Limit

The Indeterminate Trap

The Taylor Series Weapon

The Balancing Act

The Final Calculation

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Select Answer:

Visualized Solution

Continuity Condition at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Combining the Fractions

Taylor Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2x

Substituting into the Numerator

Condition for Finite Limit

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Expanding the Denominator

Evaluating the Limit for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

Final Ordered Pair .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (k,f(0))