If f(x) = (x²-1)/(x²+1), for every real number x, then the minimum value of f (A) does not exist bec

Q: Wait, why can't I just use the quotient rule to find the derivative? Isn't that the standard way to find minima?

You absolutely can! The quotient rule is a powerful tool, and it will give you the correct answer. However, in JEE Advanced, time is your most precious resource. Recognizing that you can manipulate the expression algebraically is a 'pro-move' that saves you precious minutes and reduces the chance of calculation errors. Always look for the algebraic shortcut first!

Q: Why do we care about the denominator being $x^2 + 1$? Can't $x^2$ be negative?

That is a classic trap! Remember, we are working with real numbers ($x \in \mathbb{R}$). For any real number, the square $x^2$ is always non-negative ($x^2 \ge 0$). Therefore, $x^2 + 1$ is always at least $1$. This constraint is the key to unlocking the minimum value.

Q: Is the minimum value always attained? What if the function approaches the value but never reaches it?

Great question! In this case, the minimum value is attained because the function reaches $-1$ exactly when $x = 0$. If the function only approached $-1$ as $x \to \infty$ (like $1/x$), we would say the infimum is $-1$, but the minimum does not exist. Here, the function actually hits the value, so it is a true minimum.

Question

If $f(x) = \frac{x^2 - 1}{x^2 + 1}$, for every real number $x$, then the minimum value of $f$

Options:

(A) does not exist because $f$ is unbounded

(B) is not attained even though $f$ is bounded

(C) is equal to 1

(D) is equal to -1

Sigma Percentile · Accepted Answer

### The Art of Algebraic Surgery

Welcome, future engineers! Today, we are not just solving a problem; we are performing surgery on a function. We are looking at $f(x) = \frac{x^2 - 1}{x^2 + 1}$.

When you see a rational function like this, your instinct might be to reach for the calculus toolkit—the quotient rule, the derivative, the critical points. And you would be right; that path leads to the answer. But in the JEE Advanced arena, speed and elegance are just as important as accuracy.

### The Algebraic Insight

Imagine you are standing before this expression. The numerator is $x^2 - 1$ and the denominator is $x^2 + 1$. They are almost identical, aren't they? The difference is just a constant.

This is the spark of genius. We want to force the numerator to look like the denominator so we can cancel terms. We can rewrite $x^2 - 1$ as $(x^2 + 1) - 2$.

This simple, elegant adjustment transforms our function into:

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

Now, watch the magic happen as we split this into two separate fractions:

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

### The Logic of Extremes

This simplification gives us $f(x) = 1 - \frac{2}{x^2 + 1}$. Now, look at this expression with fresh eyes. We have a constant, $1$, and we are subtracting a variable fraction, $\frac{2}{x^2 + 1}$.

To make the entire expression as small as possible (the minimum), we need to subtract the largest possible amount. This means we need to maximize the fraction $\frac{2}{x^2 + 1}$.

How do we maximize a fraction with a constant numerator? By making the denominator as small as possible!

### The Final Calculation

We know that for any real number $x$, the square $x^2$ is always non-negative, meaning $x^2 \ge 0$. Therefore, the smallest value $x^2$ can take is $0$, which happens when $x = 0$.

If the minimum of $x^2$ is $0$, then the minimum of our denominator, $x^2 + 1$, must be $0 + 1 = 1$. Substituting this back into our simplified function, we get:

$$f_{min} = 1 - \frac{2}{1} = 1 - 2 = -1$$

We have arrived at the answer with precision and grace. The function reaches its **absolute minimum of $-1$ at $x = 0$**. Keep this mindset—always look for the structure before you start the calculation. You are doing great!

Animated Solution for Mathematics - Differentiation: If $f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}$ , for every real number $x$ , then the minimum value of $f$

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

Choosing the Right Approach

Manipulating the Numerator

Splitting into Two Terms

The Simplified Form

Logic for Minimizing $f (x)$

Maximizing the Fractional Term

Minimizing the Denominator

Evaluating the Denominator's Minimum

Final Calculation

Visualizing the Minimum

Graph of $f (x)$ and Asymptote

The Sigma Insight: Maxima and Minima

The Art of Algebraic Surgery

The Algebraic Insight

The Logic of Extremes

The Final Calculation

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x2+1x2−1​, for every real number x, then the minimum value of f