If the coefficients of x⁷ in (x² + 1/bx)¹¹ and x⁻⁷ in (x - 1/bx²)¹¹, b ≠ 0, are equal, then the valu

Q: Wait, why do we use $r$ for the first expansion and $k$ for the second?

That is a fantastic question! We use different variables ($r$ and $k$) because the two expansions are completely independent. Using the same variable for both would imply that the term index must be the same, which is not necessarily true. By using $r$ and $k$, we keep our algebraic bookkeeping clean and avoid accidental errors.

Q: What happened to the negative sign in the second expansion? Did it just disappear?

It definitely did not disappear! In the second expansion, the term $B$ is $-\frac{1}{bx^2}$. When we raise this to the power of $k$, we get $(-1)^k \cdot (\frac{1}{bx^2})^k$. Since we found $k=6$, and $6$ is an even number, $(-1)^6$ becomes positive $1$. That is why it seems to vanish—it was absorbed by the even power!

Q: Can I just calculate $\binom{11}{5}$ and $\binom{11}{6}$ to solve this?

You absolutely could, but in the JEE, time is your most precious resource. Calculating $\binom{11}{5}$ and $\binom{11}{6}$ involves factorials that, while manageable, are prone to arithmetic slips. Using the symmetry property $\binom{n}{r} = \binom{n}{n-r}$ is not just faster; it is the elegant, 'JEE-style' way to solve the problem, as it allows the coefficients to cancel out entirely.

Question

If the coefficients of $x^7$ in $(x^2 + \frac{1}{bx})^{11}$ and $x^{-7}$ in $(x - \frac{1}{bx^2})^{11}$, $b 
e 0$, are equal, then the value of $b$ is equal to:

Options:

(A) 2

(B) $-1$

(C) 1

(D) $-2$

Sigma Percentile · Accepted Answer

### The Art of the Binomial Hunt: Mastering Coefficients

Imagine you are standing before two complex algebraic expressions. They look intimidating, filled with powers and constants, but they are governed by a beautiful, underlying symmetry.

This is the essence of the Binomial Theorem—a tool that allows us to dissect any power of a binomial with surgical precision. Today, we are going to solve a classic JEE problem that tests not just your algebraic skills, but your ability to see the structure hidden within the symbols.

### The Master Key: The General Term

Every binomial expansion $(A + B)^n$ follows a predictable rhythm. The general term, $T_{r+1} = \binom{n}{r} A^{n-r} B^r$, is our master key. It unlocks any specific term we desire.

In our problem, we have two distinct expansions: $(x^2 + \frac{1}{bx})^{11}$ and $(x - \frac{1}{bx^2})^{11}$. Our mission is to find the value of $b$ such that the coefficient of $x^7$ in the first matches the coefficient of $x^{-7}$ in the second.

### Phase 1: The First Expansion

Let us focus on the first expression: $(x^2 + \frac{1}{bx})^{11}$. Using our master key, the general term is:

$$T_{r+1} = \binom{11}{r} (x^2)^{11-r} \left( \frac{1}{bx} ight)^r$$

Now, we must isolate the variable $x$. By separating the constants from the variables, we get:

$$T_{r+1} = \binom{11}{r} b^{-r} \cdot x^{22-2r} \cdot x^{-r} = \binom{11}{r} b^{-r} \cdot x^{22-3r}$$

We want the coefficient of $x^7$, so we set the exponent $22-3r = 7$. A quick calculation reveals $3r = 15$, so $r = 5$. Our first coefficient, $C_1$, is:

$$C_1 = \binom{11}{5} b^{-5}$$

### Phase 2: The Second Expansion

Now, we turn our attention to the second expression: $(x - \frac{1}{bx^2})^{11}$. We use a new index, $k$, to keep our work organized. The general term is:

$$T_{k+1} = \binom{11}{k} (x)^{11-k} \left( -\frac{1}{bx^2} ight)^k$$

Notice that crucial negative sign! It is part of the term $B$. Simplifying this, we get:

$$T_{k+1} = \binom{11}{k} (-1)^k b^{-k} x^{11-k} x^{-2k} = \binom{11}{k} (-1)^k b^{-k} x^{11-3k}$$

We need the coefficient of $x^{-7}$, so we set $11-3k = -7$. This gives $3k = 18$, or $k = 6$. Our second coefficient, $C_2$, is:

$$C_2 = \binom{11}{6} (-1)^6 b^{-6} = \binom{11}{6} b^{-6}$$

### The Grand Finale: Symmetry and Cancellation

We are at the final step. The problem states that $C_1 = C_2$. Therefore:

$$\binom{11}{5} b^{-5} = \binom{11}{6} b^{-6}$$

Here is where the beauty of mathematics shines. We know the property $\binom{n}{r} = \binom{n}{n-r}$. Thus, $\binom{11}{5}$ is exactly equal to $\binom{11}{6}$.

They cancel out perfectly from both sides! We are left with the elegant equation:

$$b^{-5} = b^{-6}$$

Rewriting this as $\frac{1}{b^5} = \frac{1}{b^6}$, and knowing $b 
e 0$, we multiply both sides by $b^6$ to find the final result:

**$b = 1$**

Animated Solution for Mathematics - Binomial Theorem: If the coefficients of $x^{7}$ in $(x^{2} + \frac{1}{b x})^{11}$ and $x^{- 7}$ in $(x - \frac{1}{b x ^{2}})^{11}$ , $b \neq = 0$ , are equal, then the value of $b$ is equal to:

Select Answer:

Visualized Solution

Problem Overview

General Term Formula

General Term of First Expansion

Simplifying Powers of $x$

Equating Power to $7$

Coefficient $C_{1}$

General Term of Second Expansion

Simplifying Powers of $x$

Equating Power to $- 7$

Coefficient $C_{2}$

Equating the Coefficients

Using Binomial Properties

Final Calculation

The Sigma Insight: General Term and Middle Term

The Art of the Binomial Hunt

The Master Key

Phase 1

Phase 2

The Grand Finale

Similar Questions

If the coefficient of $x^{7}$ in $(a x^{2} - \frac{1}{b x})^{13}$ and the coefficient of $x^{- 5}$ in $(a x + \frac{1}{b x ^{2}})^{13}$ are equal, then $a^{4} b^{4}$ is equal to:

If the co-efficient of $x^{9}$ in $(α x^{3} + \frac{1}{β x})^{11}$ and the co-efficient of $x^{- 9}$ in $(α x - \frac{1}{β x ^{3}})^{11}$ are equal, then $(α β)^{2}$ is equal to ______.

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^{5}$ in the expansion of $(a x^{2} + \frac{70}{27 b x})^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of $(a x - \frac{1}{b x ^{2}})^{7}$ , then the value of $2 b$ is

The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of $(2 x^{2} + \frac{1}{2 x})^{11}$ is equal to

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If the coefficient of $x^{15}$ in the expansion of $(a x^{3} + \frac{1}{b x ^{1/3}})^{15}$ is equal to the coefficient of $x^{- 15}$ in the expansion of $(a x^{1/3} - \frac{1}{b x ^{3}})^{15}$ , where $a$ and $b$ are positive real numbers, then for each such ordered pair $(a, b)$ :

If the coefficients of $x^{3}$ and $x^{4}$ in the expansion of $(1 + a x + b x^{2}) (1 - 2 x)^{18}$ in powers of $x$ are both zero, then $(a, b)$ is equal to

Suppose A and B are the coefficients of $3 0^{t h}$ and $1 2^{t h}$ terms respectively in the binomial expansion of $(1 + x)^{2 n - 1}$ . If $2 A = 5 B$ , then n is equal to :

The coefficient of $x^{70}$ in $x^{2} (1 + x)^{98} + x^{3} (1 + x)^{97} + x^{4} (1 + x)^{96} + \dots + x^{54} (1 + x)^{46}$ is $^{99} C_{p} -^{46} C_{q}$ . Then a possible value of $p + q$ is :

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + α x)^{4}$ and of $(1 - α x)^{6}$ is the same if $α$ equals

Select Answer:

Visualized Solution

Problem Overview

General Term Formula

General Term of First Expansion

Simplifying Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Equating Power to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7

Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

General Term of Second Expansion

Simplifying Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Equating Power to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −7

Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Equating the Coefficients

Using Binomial Properties

Final Calculation

The Sigma Insight: General Term and Middle Term

The Art of the Binomial Hunt

The Master Key

Phase 1

Phase 2

The Grand Finale

Similar Questions

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Coefficient $C_{1}$

Simplifying Powers of $x$

Equating Power to $- 7$

Coefficient $C_{2}$

If the coefficient of $x^{7}$ in $(a x^{2} - \frac{1}{b x})^{13}$ and the coefficient of $x^{- 5}$ in $(a x + \frac{1}{b x ^{2}})^{13}$ are equal, then $a^{4} b^{4}$ is equal to:

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