If α and β are the roots of the quadratic equation x² + x sin θ - 2 sin θ = 0, θ ∈ (0, π/2), then (α

Q: Wait, why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula?

You absolutely could, but in JEE Advanced, time is your most precious resource. Solving for the roots directly would introduce complex radicals that would make the subsequent power-of-12 calculation a nightmare. The problem is specifically designed to test your ability to manipulate symmetric expressions of roots without ever needing to know the roots themselves.

Q: Why does the negative sign disappear in the final step?

That is a great observation! In the final step, we are left with a term like $\left[ \frac{-2}{\sin \theta + 8} \right]^{12}$. Because the exponent $12$ is an even number, any negative value raised to this power becomes positive. $(-1)^{12} = 1$, so the negative sign effectively vanishes.

Q: Is there a trick to remembering the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$?

Think of it as the difference between the square of the sum and the square of the difference. We know $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha \beta$ and $(\alpha - \beta)^2 = \alpha^2 + \beta^2 - 2\alpha \beta$. If you take the first and subtract $4\alpha \beta$, you get exactly the second. It is a standard tool in your algebraic toolkit that you should be able to derive in seconds.

Question

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + x \sin 	heta - 2 \sin 	heta = 0, 	heta \in (0, \pi/2)$, then $\frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$ is equal to :

Options:

(A) $\frac{2^6}{(\sin 	heta + 8)^{12}}$

(B) $\frac{2^{12}}{(\sin 	heta - 8)^6}$

(C) $\frac{2^{12}}{(\sin 	heta - 4)^{12}}$

(D) $\frac{2^{12}}{(\sin 	heta + 8)^{12}}$

Sigma Percentile · Accepted Answer

### The Symphony of Roots: A Journey into Algebraic Elegance

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a quadratic equation; we are embarking on a journey to uncover the hidden symmetries within a seemingly intimidating expression.

When you first look at the expression $E = \frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$, it is natural to feel a moment of hesitation. It looks like a mountain of algebra, but in the world of JEE Advanced, the most complex-looking problems often hide the most elegant, simple solutions.

Let us break this down together.

### Phase 1: The Foundation

We begin with our given quadratic equation: $x^2 + x \sin 	heta - 2 \sin 	heta = 0$. We are told that $\alpha$ and $\beta$ are its roots.

Before we do anything else, we must invoke the most powerful tools in our arsenal: Vieta's formulas. For any quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product is $\alpha \beta = \frac{c}{a}$.

Applying this to our equation, we immediately find:

$$ \alpha + \beta = -\sin 	heta $$

$$ \alpha \beta = -2 \sin 	heta $$

These two simple relations are the keys to the entire kingdom. Keep them close.

### Phase 2: The Algebraic Dance

Now, let us turn our attention to the target expression $E$. The presence of negative exponents, $\alpha^{-12}$ and $\beta^{-12}$, is a classic distractor.

We know that $\alpha^{-12} = \frac{1}{\alpha^{12}}$ and $\beta^{-12} = \frac{1}{\beta^{12}}$. When we add these, we get:

$$ \alpha^{-12} + \beta^{-12} = \frac{1}{\alpha^{12}} + \frac{1}{\beta^{12}} = \frac{\alpha^{12} + \beta^{12}}{(\alpha \beta)^{12}} $$

Look at that! When we substitute this back into our expression $E$, the numerator $\alpha^{12} + \beta^{12}$ appears in both the top and the bottom. They cancel out perfectly, leaving us with:

$$ E = \frac{(\alpha \beta)^{12}}{(\alpha - \beta)^{24}} = \left[ \frac{\alpha \beta}{(\alpha - \beta)^2} ight]^{12} $$

Suddenly, the mountain has become a molehill. We no longer care about the 12th powers of the roots; we only need the product of the roots and the square of their difference.

### Phase 3: The Identity

We already have the product $\alpha \beta = -2 \sin 	heta$. Now we need $(\alpha - \beta)^2$.

We use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$. Substituting our known values:

$$ (\alpha - \beta)^2 = (-\sin 	heta)^2 - 4(-2 \sin 	heta) $$

$$ (\alpha - \beta)^2 = \sin^2 	heta + 8 \sin 	heta = \sin 	heta (\sin 	heta + 8) $$

### Phase 4: The Final Reveal

We are at the finish line. Let us plug these values back into our simplified expression for $E$:

$$ E = \left[ \frac{-2 \sin 	heta}{\sin 	heta (\sin 	heta + 8)} ight]^{12} $$

The $\sin 	heta$ terms cancel out, leaving us with:

$$ E = \left[ \frac{-2}{\sin 	heta + 8} ight]^{12} $$

Since the exponent $12$ is even, the negative sign disappears, and we arrive at our beautiful final answer:

$$ **E = \frac{2^{12}}{(\sin 	heta + 8)^{12}}** $$

See how the complexity dissolved? This is the essence of JEE Advanced mathematics. It is not about brute force; it is about finding the path of least resistance through the forest of variables.

Animated Solution for Mathematics - Quadratic Equations: If $α$ and $β$ are the roots of the quadratic equation $x^{2} + x sin θ - 2 sin θ = 0, θ \in (0, π /2)$ , then $\frac{α ^{12} + β ^{12}}{( α ^{- 12} + β ^{- 12} ) ( α - β ) ^{24}}$ is equal to :

Select Answer:

Visualized Solution

The Quadratic Equation

Sum and Product $α + β$ and $α β$

Target Expression $E$

Handling Negative Exponents

Simplifying the Expression

Difference of Roots Squared $(α - β)^{2}$

Substituting Sum and Product

Final Substitution

Final Answer

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Roots

Phase 1

Phase 2

Phase 3

Phase 4

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