Let f(x) = (1 - tan x) / (4x - pi), x != pi/4, x in [0, pi/2]. If f(x) is continuous in [0, pi/2], t

Q: Wait, if the function is undefined at $x = \frac{\pi}{4}$, how can it be continuous?

That is a brilliant observation! In the strict sense, the function as originally written has a 'removable discontinuity'—a hole. When a problem asks for a function to be continuous at a point where it is initially undefined, it is implicitly asking you to define the value at that point such that the 'hole' is filled. By setting $f\left(\frac{\pi}{4}\right)$ equal to the limit, we are essentially 'patching' the graph to make it a smooth, unbroken line.

Q: Is L'Hopital's Rule the only way to solve this? Can I use substitution?

You absolutely can! L'Hopital's Rule is a powerful tool, but it is not the only one. You could also use the substitution $t = x - \frac{\pi}{4}$, which implies $x = t + \frac{\pi}{4}$. As $x \to \frac{\pi}{4}$, $t \to 0$. Substituting this into the expression and using the tangent addition formula $\tan(A+B)$ would lead you to the exact same result. Both methods are elegant; L'Hopital's is often faster, but substitution builds deeper algebraic intuition.

Q: Why did we get $\frac{0}{0}$? Does that mean the limit doesn't exist?

Not at all! In calculus, $\frac{0}{0}$ is an 'indeterminate form.' It doesn't mean the limit is undefined; it means the limit is 'hidden.' It tells us that the numerator and denominator are both shrinking to zero at rates that might cancel each other out. Our job is to peel back that layer of uncertainty—using L'Hopital's or algebraic manipulation—to reveal the finite value hiding underneath.

Question

Let $f(x) = \frac{1 - 	an x}{4x - \pi}, x 
eq \frac{\pi}{4}, x \in \left[ 0, \frac{\pi}{2} ight]$. If $f(x)$ is continuous in $\left[ 0, \frac{\pi}{2} ight]$, then $f\left( \frac{\pi}{4} ight)$ is

Options:

(A) -1

(B) 1/2

(C) -1/2

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are walking along a beautiful, smooth path represented by the function:

$$f(x) = \frac{1 - 	an x}{4x - \pi}$$

Everything seems perfect until you reach the coordinate $x = \frac{\pi}{4}$. Suddenly, you find a gap—a pothole in the road.

If you try to calculate the value of the function directly at this point, you are met with the dreaded division by zero. In the language of calculus, we have a **removable discontinuity**.

### The Condition of Continuity

To make a function continuous at a specific point, the graph must not have any jumps or gaps. Mathematically, this requires that the value of the function at that point must perfectly match the limit of the function as we approach that point from either side.

We are looking for a value $f\left(\frac{\pi}{4}ight)$ such that:

$$f\left(\frac{\pi}{4}ight) = \lim_{x 	o \frac{\pi}{4}} \frac{1 - 	an x}{4x - \pi}$$

This is our bridge. If we can calculate this limit, we can define the function at that single point to bridge the gap and create a perfectly continuous curve.

### The Indeterminate Crisis

Let us attempt to evaluate this limit directly. As $x$ approaches $\frac{\pi}{4}$, the numerator $1 - 	an x$ approaches $1 - 	an\left(\frac{\pi}{4}ight) = 1 - 1 = 0$.

Simultaneously, the denominator $4x - \pi$ approaches $4\left(\frac{\pi}{4}ight) - \pi = \pi - \pi = 0$. We have arrived at the $\frac{0}{0}$ indeterminate form.

Do not panic! This is not a dead end; it is a signal. It tells us that the function is behaving in a way that requires a more surgical approach to uncover the true value.

### The Surgical Strike: L'Hopital's Rule

When we face a $\frac{0}{0}$ form, we reach for one of the most powerful tools in our JEE toolkit: **L'Hopital's Rule**. This rule allows us to differentiate the numerator and the denominator independently to find the limit of their ratio.

Let us differentiate the numerator $g(x) = 1 - 	an x$. The derivative of the constant $1$ is $0$, and the derivative of $	an x$ is $\sec^2 x$. Thus, $g'(x) = -\sec^2 x$.

Now, let us look at the denominator $h(x) = 4x - \pi$. The derivative of $4x$ is $4$, and the derivative of the constant $\pi$ is $0$. So, $h'(x) = 4$.

Our limit expression has now transformed into the much friendlier form:

$$L = \lim_{x 	o \frac{\pi}{4}} \frac{-\sec^2 x}{4}$$

### The Final Resolution

Now that we have simplified the expression, the path forward is clear. We substitute $x = \frac{\pi}{4}$ into our new expression.

We know that $\cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$, which means $\sec\left(\frac{\pi}{4}ight) = \sqrt{2}$. Squaring this value gives us $(\sqrt{2})^2 = 2$.

Substituting this back into our limit, we get:

$$L = \frac{-2}{4} = -\frac{1}{2}$$

By calculating this limit, we have found the exact value needed to fill the hole. If we define $f\left(\frac{\pi}{4}ight) = -\frac{1}{2}$, the function becomes continuous across the entire interval.

We have taken a broken, undefined expression and, through the elegance of calculus, restored it to a smooth, continuous whole. This is the essence of mathematics—finding order within the chaos. The final result is **$-\frac{1}{2}$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = \frac{1 - t a n x}{4 x - π}, x \neq = \frac{π}{4}, x \in [0, \frac{π}{2}]$ . If $f (x)$ is continuous in $[0, \frac{π}{2}]$ , then $f (\frac{π}{4})$ is

Select Answer:

Visualized Solution

The Function and Its Domain

Condition for Continuity

Setting Up the Limit

Checking the Indeterminate Form

Applying L'Hopital's Rule

Differentiating the Numerator

Differentiating the Denominator

The New Limit Expression

Evaluating the Limit

Final Calculation

Filling the Hole

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

Analyzing the Setup

The Condition of Continuity

The Indeterminate Crisis

The Surgical Strike

The Final Resolution

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