If n is the degree of the polynomial, [2/(√(5x³+1) - √(5x³-1))]⁸ + [2/(√(5x³+1) + √(5x³-1))]⁸ and m

Q: Why can't we just expand the original expression directly? It looks like a standard binomial expansion.

Expanding the original expression would lead to a nightmare of nested radicals and algebraic chaos. Rationalization is a strategic choice to transform a complex fraction into a clean, manageable polynomial. In JEE, always look for the 'simplification path' before brute-forcing.

Q: Why do the odd terms cancel out in the binomial expansion of $(a+b)^n + (a-b)^n$?

Think of the expansion of $(a+b)^n$ and $(a-b)^n$. When you add them, terms like $\binom{n}{1}a^{n-1}b^1$ and $-\binom{n}{1}a^{n-1}b^1$ appear. They are additive inverses and vanish, leaving only the even-indexed terms. It is a beautiful symmetry that saves you half the work!

Q: How did we arrive at $n=12$ so quickly?

Look at the structure of the terms after rationalization. We have expressions like $(5x^3 + 1)^4$. The highest power of $x$ is $(x^3)^4 = x^{12}$. Since every term in the expansion involves powers of $x^3$ raised to a total power of 4, the degree must be 12.

Q: Why did we use $2^{k-1}$ for the sum of coefficients?

This is a standard property of binomial coefficients. The sum of all coefficients $\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$. If you only sum the even indices, the result is exactly half of the total sum, which is $2^{n-1}$. It is a powerful shortcut for these types of problems.

Question

If n is the degree of the polynomial, $\left[ \frac{2}{\sqrt{5x^3 + 1} - \sqrt{5x^3 - 1}} \right]^8 + \left[ \frac{2}{\sqrt{5x^3 + 1} + \sqrt{5x^3 - 1}} \right]^8$ and m is the coefficient of $x^n$ in it, then the ordered pair (n,m) is equal to:

Options:

(A) $(24, (10)^8)$

(B) $(12, (20)^4)$

(C) $(12, 8(10)^4)$

(D) $(8, 5(10)^4)$

Sigma Percentile · Accepted Answer

### The Art of Algebraic Simplification

Welcome, students. Today, we are going to dismantle a problem that, at first glance, looks like a mathematical monster. You see an expression like

$$P(x) = \left[ \frac{2}{\sqrt{5x^3 + 1} - \sqrt{5x^3 - 1}} \right]^8 + \left[ \frac{2}{\sqrt{5x^3 + 1} + \sqrt{5x^3 - 1}} \right]^8$$

and your instinct might be to panic. But remember, in the world of JEE Advanced, intimidation is often a mask for a very elegant, simple truth waiting to be uncovered. Let us peel back the layers together.

### Phase 1: The Power of Rationalization

When you see square roots in the denominator, your brain should immediately scream: **Rationalize!** We are not going to touch that power of eight just yet, as that would be a trap. Instead, let us focus on the inner fractions.

Consider the first term: $\frac{2}{\sqrt{5x^3 + 1} - \sqrt{5x^3 - 1}}$. To rationalize, we multiply the numerator and the denominator by the conjugate, $\sqrt{5x^3 + 1} + \sqrt{5x^3 - 1}$.

The denominator becomes a classic difference of squares: $(5x^3 + 1) - (5x^3 - 1)$, which simplifies beautifully to $2$. The $2$ in the numerator and the $2$ in the denominator cancel out perfectly, leaving us with $\sqrt{5x^3 + 1} + \sqrt{5x^3 - 1}$.

By applying the same logic to the second term, we get $\sqrt{5x^3 + 1} - \sqrt{5x^3 - 1}$. Suddenly, the "monster" has been tamed. Our expression $P(x)$ is now simply:

$$P(x) = (\sqrt{5x^3 + 1} + \sqrt{5x^3 - 1})^8 + (\sqrt{5x^3 + 1} - \sqrt{5x^3 - 1})^8$$

### Phase 2: The Binomial Symmetry

Now, look at the structure. We have something of the form $(a+b)^n + (a-b)^n$. This is a classic binomial structure.

When we expand this, the odd-indexed terms involve terms with an odd power of $b$. Because of the minus sign in the second bracket, these terms will have opposite signs and cancel each other out completely. We are left with twice the sum of the even-indexed terms:

$$P(x) = 2 \left[ \binom{8}{0} a^8 + \binom{8}{2} a^6 b^2 + \binom{8}{4} a^4 b^4 + \binom{8}{6} a^2 b^6 + \binom{8}{8} b^8 \right]$$

Here, $a = \sqrt{5x^3 + 1}$ and $b = \sqrt{5x^3 - 1}$. Notice that $a^2 = 5x^3 + 1$ and $b^2 = 5x^3 - 1$. This is the moment where the square roots vanish for good, and we are now dealing with a pure polynomial.

### Phase 3: Hunting for the Degree and Coefficient

To find the degree $n$, we look at the highest power of $x$. In any term of our expansion, we are multiplying powers of $a^2$ and $b^2$. Since $a^2$ and $b^2$ are linear in $x^3$, the highest power of $x$ in any term will be $(x^3)^4 = x^{12}$. Thus, the degree of our polynomial is **$n = 12$**.

Now, for the coefficient $m$. In each term of the expansion, we have a factor of $5^4$. For example, in the first term, $\binom{8}{0} a^8 = \binom{8}{0} (5x^3 + 1)^4$, the coefficient of $x^{12}$ is $5^4$.

In the second term, $\binom{8}{2} a^6 b^2 = \binom{8}{2} (5x^3 + 1)^3 (5x^3 - 1)$, the coefficient of $x^{12}$ is $\binom{8}{2} (5^3) \cdot (5^1) = \binom{8}{2} 5^4$. Every term contributes a factor of $5^4$.

We can factor this out:

$$m = 2 \cdot 5^4 \left[ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} \right]$$

The sum inside the bracket is the sum of even binomial coefficients for $k=8$, which is $2^{8-1} = 2^7 = 128$.

So, $m = 2 \cdot 5^4 \cdot 2^7 = 2^8 \cdot 5^4$.

To match our options, we rewrite this as $(2^2)^4 \cdot 5^4 = 4^4 \cdot 5^4 = (4 \cdot 5)^4 = 20^4$.

### Conclusion

We have arrived at our destination: the degree $n=12$ and the coefficient $m=20^4$. The ordered pair is **$(12, 20^4)$**. This problem was not about brute force; it was about recognizing patterns—rationalization, binomial symmetry, and the properties of coefficients.

Animated Solution for Mathematics - Binomial Theorem: If n is the degree of the polynomial, $[\frac{2}{5 x ^{3} + 1 - 5 x ^{3} - 1}]^{8} + [\frac{2}{5 x ^{3} + 1 + 5 x ^{3} - 1}]^{8}$ and m is the coefficient of $x^{n}$ in it, then the ordered pair (n,m) is equal to:

Select Answer:

Visualized Solution

Analyzing the Expression

Rationalizing the First Term

Rationalizing the Second Term

Simplifying the Total Expression

Applying the Binomial Identity

Expanding the Expression

Substituting Back the Values

Determining the Degree $n$

Setting up the Coefficient $m$

Summing Binomial Coefficients

Final Calculation for $m$

The Final Ordered Pair

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Art of Algebraic Simplification

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

The expression $[x + (x^{3} - 1)^{1/2}]^{5} + [x - (x^{3} - 1)^{1/2}]^{5}$ is a polynomial of degree

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

If $(1 + a x)^{n} = 1 + 8 x + 24 x^{2} + ....$ then $a = .........$ and $n = .........$

Let $α, β, γ$ and $δ$ be the coefficients of $x^{7}$ , $x^{5}$ , $x^{3}$ and $x$ respectively in the expansion of $(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}$ , $x > 1$ . If u and v satisfy the equations $α u + β v = 18$ and $γ u + δ v = 20$ , then $u + v$ equals:

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

The sum of the co-efficients of all odd degree terms in the expansion of $(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}, (x > 1)$ is :

Let $n \geq 5$ be an integer. If $9^{n} - 8 n - 1 = 64 α$ and $6^{n} - 5 n - 1 = 25 β$ , then $α - β$ is equal to:

For the natural numbers $m, n$ , if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots + a_{m + n} y^{m + n}$ and $a_{1} = a_{2} = 10$ , then the value of $(m + n)$ is equal to:

For natural numbers $m, n$ if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + ....$ and $a_{1} = a_{2} = 10$ , then $(m, n)$ is

The sum of the co-efficients of all even degree terms in $x$ in the expansion of $(x + x^{3} - 1)^{6} + (x - x^{3} - 1)^{6}, (x > 1)$ is equal to :