If the coefficient of x⁷ in [ax² + (1/bx)]¹¹ equals the coefficient of x⁻⁷ in [ax - (1/bx²)]¹¹, then

Q: Wait, why did the negative sign in the second expression just vanish? In the second term, we have $(-\frac{1}{bx^2})^k$. If $k=6$, doesn't the negative sign matter?

That is a brilliant observation! You are absolutely right to be cautious. The term is indeed $(-\frac{1}{bx^2})^k$, which expands to $(-1)^k \cdot (\frac{1}{bx^2})^k$. Since our calculated index $k$ is $6$, and $6$ is an even number, $(-1)^6$ becomes $+1$. The negative sign effectively disappears because of the even power. Always check the parity of your index!

Q: Why can't we just calculate the values of $\binom{11}{5}$ and $\binom{11}{6}$ instead of using the symmetry property?

You certainly *could*, but in the JEE Advanced arena, time is your most precious currency. Calculating $11! / (5!6!)$ is prone to arithmetic errors. Recognizing the symmetry $\binom{n}{r} = \binom{n}{n-r}$ is not just a shortcut; it is the mark of a seasoned problem solver who looks for the underlying structure rather than brute-forcing the arithmetic.

Q: What if the power of $x$ I calculated for $r$ wasn't an integer? Does that mean the term doesn't exist?

Exactly! If you solve for $r$ and get a fraction or a negative number, it implies that the specific power of $x$ you are looking for does not exist in that expansion. In such a case, the coefficient would simply be zero. Always trust your algebra—if the math says it's impossible, the binomial expansion agrees with you.

Question

If the coefficient of $x^7$ in $[ax^2 + (\frac{1}{bx})]^{11}$ equals the coefficient of $x^{-7}$ in $[ax - (\frac{1}{bx^2})]^{11}$, then $a$ and $b$ satisfy the relation

Options:

(A) $a - b = 1$

(B) $a + b = 1$

(C) $a/b = 1$

(D) $ab = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The Binomial Theorem provides a systematic way to expand expressions of the form $(X + Y)^n$. For any expansion, the $(r+1)$-th term is defined by the general term formula:

$$T_{r+1} = \binom{n}{r} X^{n-r} Y^r$$

This formula allows us to isolate specific terms without expanding the entire polynomial. We will apply this to the two given expressions to find the required coefficients.

### First Expression: $[ax^2 + (bx)^{-1}]^{11}$

For the expression $[ax^2 + (bx)^{-1}]^{11}$, we identify $X = ax^2$ and $Y = (bx)^{-1}$. Substituting these into the general term formula, we get:

$$T_{r+1} = \binom{11}{r} (ax^2)^{11-r} (bx)^{-r}$$

To isolate the variable $x$, we group the constants and the powers of $x$:

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

We are hunting for the coefficient of $x^7$. By setting the exponent $22 - 3r = 7$, we solve for $r$:

$$3r = 15 \Rightarrow r = 5$$

Substituting $r=5$ into the expression, the coefficient is $\binom{11}{5} a^6 b^{-5}$.

### Second Expression: $[ax - (bx^2)^{-1}]^{11}$

We repeat the process for the second expression, using index $k$ to avoid confusion. The general term is:

$$T_{k+1} = \binom{11}{k} (ax)^{11-k} (-(bx^2)^{-1})^k$$

Isolating the powers of $x$, we obtain $x^{11-k} \cdot x^{-2k} = x^{11-3k}$. We require the coefficient of $x^{-7}$, so we set:

$$11 - 3k = -7 \Rightarrow 3k = 18 \Rightarrow k = 6$$

The coefficient is $\binom{11}{6} a^{11-6} b^{-6} (-1)^6$. Since $(-1)^6 = 1$, the coefficient simplifies to $\binom{11}{6} a^5 b^{-6}$.

### The Grand Unification

The problem states that these two coefficients are equal. Therefore, we equate them:

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

Utilizing the symmetry property of binomial coefficients, $\binom{n}{r} = \binom{n}{n-r}$, we recognize that $\binom{11}{5} = \binom{11}{6}$. These terms cancel out from both sides of the equation.

We are left with the simplified relation:

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

Dividing both sides by $a^5$ and multiplying by $b^5$, we arrive at the final result:

**$ab = 1$**

Animated Solution for Mathematics - Binomial Theorem: If the coefficient of $x^{7}$ in $[a x^{2} + (\frac{1}{b x})]^{11}$ equals the coefficient of $x^{- 7}$ in $[a x - (\frac{1}{b x ^{2}})]^{11}$ , then $a$ and $b$ satisfy the relation

Select Answer:

Visualized Solution

Problem Setup

The General Term $T_{r + 1}$

General Term of First Expression

Grouping the Powers of $x$

Finding $r$ for $x^{7}$

Coefficient of $x^{7}$

General Term of Second Expression

Grouping the Powers of $x$

Finding $k$ for $x^{- 7}$

Coefficient of $x^{- 7}$

Applying the Given Condition

Simplifying the Binomial Coefficients

Finding the Relation between $a$ and $b$

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

First Expression: $[a x^{2} + (b x)^{- 1}]^{11}$

Second Expression: $[a x - (b x^{2})^{- 1}]^{11}$

The Grand Unification

Similar Questions

If the coefficients of $x^{7}$ in $(a x^{2} + \frac{1}{2 b x})^{11}$ and $x^{- 7}$ in $(a x - \frac{1}{3 b x ^{2}})^{11}$ are equal, then

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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

Problem Setup

The General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr+1​

General Term of First Expression

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x7

Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x7

General Term of Second Expression

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−7

Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−7

Applying the Given Condition

Simplifying the Binomial Coefficients

Finding the Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

First Expression: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [ax2+(bx)−1]11

Second Expression: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [ax−(bx2)−1]11

The Grand Unification

Similar Questions

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The General Term $T_{r + 1}$

Grouping the Powers of $x$

Finding $r$ for $x^{7}$

Coefficient of $x^{7}$

Grouping the Powers of $x$

Finding $k$ for $x^{- 7}$

Coefficient of $x^{- 7}$

Finding the Relation between $a$ and $b$

First Expression: $[a x^{2} + (b x)^{- 1}]^{11}$

Second Expression: $[a x - (b x^{2})^{- 1}]^{11}$

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