If α ∈ (0, π/2), then √(x² + x) + tan²α / √(x² + x) is always greater than or equal to (A) 2 tan α (

Q: Why can't we just use calculus? Can't we differentiate with respect to $x$?

You absolutely could! But in the JEE, time is your most precious currency. Calculus is a sledgehammer; AM-GM is a scalpel. It is faster, cleaner, and less prone to algebraic errors. Using AM-GM shows you understand the underlying structure of the expression.

Q: Wait, why do we reject the negative root here?

We do not reject it; we analyze the domain. The square root function $\sqrt{f(x)}$ is defined to be non-negative in the real number system. If $x^2+x$ were negative, the expression would be undefined. We assume the domain where the expression exists, which ensures our terms are positive.

Q: Is the answer always $2 \tan \alpha$?

The minimum value is $2 \tan \alpha$ when the equality condition of AM-GM is met, which occurs when $a = b$. Here, that means $\sqrt{x^2 + x} = \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$, or $x^2 + x = \tan^2 \alpha$. Since $x$ can vary, we can always find an $x$ to satisfy this, so the minimum is indeed $2 \tan \alpha$.

Question

If $\alpha \in (0, \pi/2)$, then $\sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$ is always greater than or equal to

Options:

(A) $2 	an \alpha$

(B) $1$

(C) $2$

(D) $\sec^2 \alpha$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a problem that is not just about finding a value, but about seeing the hidden architecture of mathematics. When you look at the expression $E = \sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$, what do you see?

Do you see a terrifying mess of variables and trigonometric functions? Or do you see a beautiful, symmetric structure? The secret to mastering JEE Advanced is pattern recognition.

Notice that we have a term, let us call it $A = \sqrt{x^2 + x}$, and then we have a constant, $C = 	an^2 \alpha$, divided by that same term $A$. This is the classic $A + \frac{C}{A}$ structure. Whenever you see a variable term added to its reciprocal, your mind should immediately race to the **Arithmetic Mean-Geometric Mean (AM-GM) inequality**.

### The Power of AM-GM

Why do we choose AM-GM over calculus? Imagine you are in the exam hall. You have limited time.

You could differentiate, set the derivative to zero, solve for $x$, and then substitute back. It is a valid path, but it is a long one. AM-GM is the elegant shortcut.

It states that for any two positive numbers $a$ and $b$, the arithmetic mean is always greater than or equal to the geometric mean:

$$\frac{a + b}{2} \ge \sqrt{ab}$$

This theorem is a favorite of JEE examiners because it tests your ability to transform a complex algebraic expression into a simple, optimized form.

### The Execution

Let us define our terms. Let $a = \sqrt{x^2 + x}$ and $b = \frac{	an^2 \alpha}{\sqrt{x^2 + x}}$. Both terms are positive, provided $x^2 + x > 0$.

Now, we substitute these into our inequality:

$$\frac{\sqrt{x^2 + x} + \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}{2} \ge \sqrt{\sqrt{x^2 + x} \cdot \frac{	an^2 \alpha}{\sqrt{x^2 + x}}}$$

Look at the right-hand side. This is where the magic happens. The term $\sqrt{x^2 + x}$ in the numerator and the denominator cancels out perfectly. The variable $x$ vanishes! We are left with $\sqrt{	an^2 \alpha}$.

### The Final Step

We must be careful here. The square root of a squared quantity is the absolute value: $\sqrt{	an^2 \alpha} = |	an \alpha|$.

But look at our constraint: $\alpha \in (0, \frac{\pi}{2})$. In the first quadrant, all trigonometric ratios are positive. Thus, $|	an \alpha| = 	an \alpha$.

Our inequality simplifies to:

$$\frac{E}{2} \ge 	an \alpha \implies E \ge 2 	an \alpha$$

We have arrived at our destination. The expression is always greater than or equal to **$2 	an \alpha$**. It is not just a result; it is a testament to the power of looking for symmetry in the chaos. Keep practicing this, and soon, you will see these patterns everywhere.

Animated Solution for Mathematics - Sequence and Series: If $α \in (0, π /2)$ , then $x^{2} + x + \frac{t a n ^{2} α}{x ^{2} + x}$ is always greater than or equal to

Select Answer:

Visualized Solution

Analyze the Expression

The AM-GM Inequality

Defining the Terms

Applying AM-GM

Simplifying the Product

Evaluating the Geometric Mean

Applying the Domain Constraint

Finding the Minimum Value

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Power of AM-GM

The Execution

The Final Step

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α∈(0,π/2), then x2+x​+x2+x​tan2α​ is always greater than or equal to