Let f(x) = x² + 1/x² and g(x) = x - 1/x, x ∈ R - {-1, 0, 1}. If h(x) = f(x)/g(x), then the local min

Q: Why can't we just use the quotient rule? It is a standard formula for rational functions!

You absolutely can, but in JEE Advanced, time is your currency. The quotient rule here leads to a messy algebraic nightmare involving high-degree polynomials that are prone to calculation errors. The substitution method reveals the underlying structure of the function, turning a complex problem into a simple, elegant one. Always look for the 'smart' path before the 'hard' path.

Q: Why do we reject $t = -\sqrt{2}$?

We do not 'reject' it in the sense that it is mathematically invalid; we reject it because it corresponds to a local maximum, not a minimum. The second derivative test confirms that $h''(-\sqrt{2}) < 0$, which tells us the curve is concave down at that point. Since the question asks for the local minimum, we focus only on the point where the curve is concave up.

Q: Is there any restriction on the domain of $t$?

That is a brilliant observation! Since $t = x - \frac{1}{x}$, and $x$ can be any real number except $0, 1, -1$, we must ensure $t$ is valid. As $x$ varies across its domain, $t$ can take any real value except $0$. Since our critical points $\pm\sqrt{2}$ are non-zero, they are perfectly valid within the domain of our transformed function.

Question

Let $f(x) = x^2 + \frac{1}{x^2}$ and $g(x) = x - \frac{1}{x}, x \in \mathbb{R} - \{-1, 0, 1\}$. If $h(x) = \frac{f(x)}{g(x)}$, then the local minimum value of $h(x)$ is :

Options:

(A) 2\sqrt{2}

(B) 3

(C) -3

(D) -2\sqrt{2}

Sigma Percentile · Accepted Answer

Welcome, future engineers. Today, we stand before a classic problem that separates the 'calculators' from the 'thinkers'. We are given the function:

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

The instinct of a novice is to immediately apply the quotient rule. You might start writing down the derivative of the numerator, then the derivative of the denominator, and suddenly you are drowning in a sea of $x^4$ and $x^6$ terms.

But I want you to pause. Look at the expression. Do you see the symmetry? The numerator is a sum of squares, and the denominator is a difference. They are dancing together, waiting for you to notice their connection. This is the essence of the JEE Advanced mindset: observation before execution.

### The Philosophy of Substitution

The first step in our journey is to recognize that $x^2 + \frac{1}{x^2}$ and $x - \frac{1}{x}$ are not independent entities. They are intimately linked by the algebraic identity:

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

This is the 'Aha!' moment. If we let $t = x - \frac{1}{x}$, then our numerator $x^2 + \frac{1}{x^2}$ becomes $t^2 + 2$.

Suddenly, the entire function $h(x)$ transforms into a beautiful, simple function of $t$:

$$h(t) = \frac{t^2 + 2}{t}$$

By splitting the fraction, we get $h(t) = t + \frac{2}{t}$. Look at how the complexity has evaporated. We have moved from a rational function of $x$ to a simple sum of a linear term and a reciprocal term. This is the power of substitution.

### The Calculus of Optimization

Now that we have $h(t) = t + \frac{2}{t}$, finding the local minimum becomes a standard exercise in calculus. We need to find the critical points where the slope of the tangent is zero.

We differentiate with respect to $t$:

$$h'(t) = \frac{d}{dt}(t + 2t^{-1}) = 1 - \frac{2}{t^2}$$

To find the critical points, we set $h'(t) = 0$, which gives us $1 - \frac{2}{t^2} = 0$, or $t^2 = 2$. This yields two critical points: $t = \sqrt{2}$ and $t = -\sqrt{2}$.

But which one is the minimum? We cannot simply guess. We must use the second derivative test. We differentiate $h'(t)$ again:

$$h''(t) = \frac{d}{dt}(1 - 2t^{-2}) = \frac{4}{t^3}$$

### The Final Victory

Now, let us test our candidates. For $t = \sqrt{2}$, the second derivative is:

$$h''(\sqrt{2}) = \frac{4}{(\sqrt{2})^3} = \frac{4}{2\sqrt{2}} = \sqrt{2}$$

Since $\sqrt{2} > 0$, the function is concave up at this point, confirming that $t = \sqrt{2}$ is indeed a local minimum. Conversely, for $t = -\sqrt{2}$, the second derivative is negative, indicating a local maximum.

We have our winner. We substitute $t = \sqrt{2}$ back into our simplified function:

$$h(\sqrt{2}) = \sqrt{2} + \frac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$$

The local minimum value is **$2\sqrt{2}$**.

This, my friends, is the elegance of mathematics. We did not need to brute-force our way through a complex derivative. We used insight, substitution, and the fundamental tools of calculus to reveal the truth hidden within the function.

Keep this approach in your toolkit. Whenever you see symmetry, look for the substitution. It will save you time, reduce your errors, and, most importantly, help you appreciate the beauty of the problems you are solving. You are not just solving for $x$; you are uncovering the structure of the universe, one equation at a time.

Animated Solution for Mathematics - Differentiation: Let $f (x) = x^{2} + \frac{1}{x ^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - {- 1, 0, 1}$ . If $h (x) = \frac{f ( x )}{g ( x )}$ , then the local minimum value of $h (x)$ is :

Select Answer:

Visualized Solution

Defining the Objective

Constructing $h (x)$

Algebraic Manipulation of Numerator

Rewriting the Numerator

Simplifying $h (x)$

The Substitution Method

Finding Critical Points

Solving for $t$

Second Derivative Test

Testing $t = 2$

Testing $t = - 2$

Calculating the Minimum Value

Final Conclusion

The Sigma Insight: Maxima and Minima

The Philosophy of Substitution

The Calculus of Optimization

The Final Victory

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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; } Let f(x)=x2+x21​ and g(x)=x−x1​,x∈R−{−1,0,1}. If h(x)=g(x)f(x)​, then the local minimum value of h(x) is :