If alpha and beta be the coefficients of x^4 and x^2 respectively in the expansion of (x + sqrt(x^2

Q: Why do we only keep the even terms? What happened to the odd ones?

When you expand $(a+b)^n$ and $(a-b)^n$, the odd terms in the second expansion have a negative sign. When you add them, they cancel out perfectly! It's a beautiful symmetry that simplifies the entire expression.

Q: Can I just expand $(x + \sqrt{x^2-1})^6$ and $(x - \sqrt{x^2-1})^6$ separately?

You could, but it would be a nightmare of algebra. Using the symmetry property is the 'JEE way'—it saves time and prevents calculation errors.

Q: What if the exponent $n$ was odd? Would the same logic apply?

Great question! If $n$ were odd, the terms with even powers of $b$ would cancel out instead. The structure of the result depends entirely on whether $n$ is even or odd.

Question

If $\alpha$ and $\beta$ be the coefficients 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) $\alpha + \beta = -30$

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

(C) $\alpha + \beta = 60$

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

Sigma Percentile · Accepted Answer

### The Beauty of Symmetry: Unlocking the Binomial Expansion

Imagine standing before a mountain of algebra that looks like a nightmare. You see $(x + \sqrt{x^2 - 1})^6 + (x - \sqrt{x^2 - 1})^6$ and your first instinct might be to panic.

But wait! In the world of JEE Advanced, whenever you see a structure like $(a+b)^n + (a-b)^n$, you are not looking at a problem; you are looking at a hidden, elegant pattern. Let us break this down together.

### The Power of Pattern Recognition

The expression is of the form $(a + b)^n + (a - b)^n$, where $a = x$, $b = \sqrt{x^2 - 1}$, and $n = 6$.

The magic of this structure is that when we expand both parts, the terms with odd powers of $b$ have opposite signs and cancel each other out completely. We are left with exactly twice the sum of the even terms:

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

For $n=6$, this becomes:

$$2 \left[ {}^{6}C_{0} a^{6} + {}^{6}C_{2} a^{4} b^{2} + {}^{6}C_{4} a^{2} b^{4} + {}^{6}C_{6} b^{6} \right]$$

### The Vanishing Act

Now, let us substitute $a = x$ and $b = \sqrt{x^2 - 1}$. Look at what happens to those square roots!

Because every $b$ term is raised to an even power, the square roots will magically disappear. For instance, $(\sqrt{x^2 - 1})^2 = x^2 - 1$.

The term $(\sqrt{x^2 - 1})^4$ becomes $(x^2 - 1)^2$, and $(\sqrt{x^2 - 1})^6$ becomes $(x^2 - 1)^3$. This is where the magic happens, simplifying our scary expression into a neat polynomial.

### Crunching the Numbers

Before we expand, let us calculate our binomial coefficients: ${}^{6}C_{0} = 1$, ${}^{6}C_{2} = 15$, ${}^{6}C_{4} = 15$, and ${}^{6}C_{6} = 1$.

Now, we expand each term:

The second term is $15 x^4 (x^2 - 1) = 15x^6 - 15x^4$.

The third term is $15 x^2 (x^2 - 1)^2 = 15x^2 (x^4 - 2x^2 + 1) = 15x^6 - 30x^4 + 15x^2$.

The fourth term is $(x^2 - 1)^3 = x^6 - 3x^4 + 3x^2 - 1$.

### The Final Tally

Now, let us bring all these pieces together inside our main bracket:

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

Grouping the like terms:
*   $x^6$ terms: $1 + 15 + 15 + 1 = 32x^6$
*   $x^4$ terms: $-15 - 30 - 3 = -48x^4$
*   $x^2$ terms: $15 + 3 = 18x^2$
*   Constant: $-1$

So, the expression inside the bracket is $32x^6 - 48x^4 + 18x^2 - 1$. Finally, multiply by the factor of $2$ from our formula:

$$2 [ 32x^6 - 48x^4 + 18x^2 - 1 ] = 64x^6 - 96x^4 + 36x^2 - 2$$

Comparing this with the question, the coefficient of $x^4$ is $\alpha = -96$, and the coefficient of $x^2$ is $\beta = 36$.

Therefore, **$\alpha - \beta = -96 - 36 = -132$**. We have conquered the mountain!

Animated Solution for Mathematics - Binomial Theorem: If $α$ and $β$ be the coefficients 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

Identify the Structure

Recall the Expansion Formula

Substitute $a$ and $b$

Calculate Binomial Coefficients

Expand the Second Term

Expand the Third Term

Expand the Fourth Term

Combine All Terms Inside Bracket

Simplify the Polynomial

Final Expansion

Identify $α$ and $β$

Calculate $α - β$

Conclusion and Key Takeaway

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Beauty of Symmetry

The Power of Pattern Recognition

The Vanishing Act

Crunching the Numbers

The Final Tally

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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 coefficients of x4 and x2 respectively in the expansion of (x+x2−1​)6+(x−x2−1​)6, then: