If α and β be the coefficient of x⁴ and x² respectively in the expansion of (x + √(x²-1))⁶ + (x - √(

Q: Why do the odd terms cancel out? I always get confused with the signs.

Think of the expansion of $(a-b)^n$. The terms with odd powers of $b$ will have a negative sign. When you add $(a+b)^n$ and $(a-b)^n$, the terms like $\binom{n}{1}a^{n-1}b^1$ and $-\binom{n}{1}a^{n-1}b^1$ perfectly cancel each other out. This is the beauty of binomial symmetry!

Q: Is there a faster way than expanding the whole thing? Can I use substitution?

You could use a substitution like $x = \cosh \theta$, but for a power of 6, direct expansion is actually faster and less prone to trigonometric identity errors. The binomial expansion is a robust, reliable tool here.

Q: How do I avoid calculation errors when expanding the powers of $(x^2-1)$?

The trick is to group the terms by powers of $x$ before multiplying by the constant 2. Don't rush the arithmetic. Expand $(x^2-1)^2$ and $(x^2-1)^3$ separately on the side, then substitute them back into the main expression.

Question

If $\alpha$ and $\beta$ be the coefficient of $x^4$ and $x^2$ respectively in the expansion of $(x + \sqrt{x^2-1})^6 + (x - \sqrt{x^2-1})^6$, then :

Options:

(A) $\alpha + \beta = 110$

(B) $\alpha + \beta = -150$

(C) $\alpha - \beta = -132$

(D) $\alpha - \beta = -60$

Sigma Percentile · Accepted Answer

### The Symphony of Binomials: Taming the Radical

My dear student, welcome to a problem that at first glance might look like a daunting, tangled mess of radicals and powers. You see $(x + \sqrt{x^2-1})^6 + (x - \sqrt{x^2-1})^6$ and your instinct might be to panic.

But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance. This problem is not a monster; it is a symphony of symmetry waiting for you to conduct it.

### Phase 1: Recognizing the Structure

Let us look at the expression again. We have the form $(a + b)^n + (a - b)^n$, where $a = x$, $b = \sqrt{x^2-1}$, and $n = 6$.

This is a classic structure. Whenever you see this, your mind should immediately jump to the binomial expansion.

When we expand $(a+b)^n$ and $(a-b)^n$, the terms with odd powers of $b$ have opposite signs. When we add them, those odd terms vanish into thin air! We are left with exactly twice the sum of the even terms:

$$2 \left[ \binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots \right]$$

This is the secret door that opens the entire problem.

### Phase 2: The Magic of Simplification

Now, let us substitute our values. We have:

$$2 \left[ \binom{6}{0} x^6 + \binom{6}{2} x^4 (\sqrt{x^2-1})^2 + \binom{6}{4} x^2 (\sqrt{x^2-1})^4 + \binom{6}{6} (\sqrt{x^2-1})^6 \right]$$

Look at how the radicals behave. The square root of $x^2-1$ squared is simply $x^2-1$. When raised to the power of four, it becomes $(x^2-1)^2$. When raised to the power of six, it becomes $(x^2-1)^3$.

The radicals are gone! We have transformed a terrifying expression into a standard polynomial expansion. This is the power of mathematical insight over brute force.

### Phase 3: The Arithmetic Grind

Now, we calculate the binomial coefficients: $\binom{6}{0} = 1$, $\binom{6}{2} = 15$, $\binom{6}{4} = 15$, and $\binom{6}{6} = 1$. Our expression becomes:

$$2 \left[ 1 \cdot x^6 + 15x^4(x^2 - 1) + 15x^2(x^2 - 1)^2 + 1 \cdot (x^2 - 1)^3 \right]$$

Now, we expand the internal brackets. We know $(x^2-1)^2 = x^4 - 2x^2 + 1$ and $(x^2-1)^3 = x^6 - 3x^4 + 3x^2 - 1$. Substituting these back, we get:

$$2 \left[ x^6 + 15x^4(x^2 - 1) + 15x^2(x^4 - 2x^2 + 1) + (x^6 - 3x^4 + 3x^2 - 1) \right]$$

Distributing the terms carefully, we have:

$$2 \left[ x^6 + 15x^6 - 15x^4 + 15x^6 - 30x^4 + 15x^2 + x^6 - 3x^4 + 3x^2 - 1 \right]$$

### Phase 4: The Final Tally

Finally, we group the like terms. For $x^6$, we have $1 + 15 + 15 + 1 = 32$. For $x^4$, we have $-15 - 30 - 3 = -48$. For $x^2$, we have $15 + 3 = 18$. And the constant is $-1$.

So, the expression is $2 [ 32x^6 - 48x^4 + 18x^2 - 1 ]$, which simplifies to:

$$64x^6 - 96x^4 + 36x^2 - 2$$

The coefficient of $x^4$ is $\alpha = -96$, and the coefficient of $x^2$ is $\beta = 36$. The question asks for $\alpha - \beta$, which is $-96 - 36 = -132$.

We have conquered the problem! Remember, the key was not the calculation, but the recognition of the binomial identity. Keep practicing this, and you will find that even the most complex problems have a simple, elegant heart. **The final answer is -132.**

Animated Solution for Mathematics - Binomial Theorem: If $α$ and $β$ be the coefficient of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x + x^{2} - 1)^{6} + (x - x^{2} - 1)^{6}$ , then :

Select Answer:

Visualized Solution

Identifying the Expression Structure

Applying the Binomial Identity

Substituting Values into the Formula

Simplifying the Powers of $b$

Calculating Binomial Coefficients

Expanding the Internal Brackets

Distributing the Terms

Grouping Like Terms

Final Simplified Expression

Calculating the Final Result

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Symphony of Binomials

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α and β be the coefficient of x4 and x2 respectively in the expansion of (x+x2−1​)6+(x−x2−1​)6, then :