By the binomial symmetry identity , here .

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Binomial Theorem: JEE Main Mathematics Question with Solution

If the co-efficient of

x^9

\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}

and the co-efficient of

x^{-9}

\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}

are equal, then

(\alpha\beta)^2

is equal to

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Step-by-step solutionView

Correct answer

The value of $(\alpha\beta)^2$ is equal to 1.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Expansions $\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$ and $\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$ . The coefficient of $x^9$ in the first equals the coefficient of $x^{-9}$ in the second.

02goal

Find the value of $(\alpha\beta)^2$ .

03approach

Write the general term $T_{r+1}$ for both binomial expansions, find the value of $r$ corresponding to the desired powers ( $x^9$ and $x^{-9}$ ), equate the resulting coefficients, and solve for $\alpha\beta$ .

04execute

For the first expansion $\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$ , the general term is: $T_{r+1} = {}^{11}C_r (\alpha x^3)^{11-r} \left(\frac{1}{\beta x}\right)^r = {}^{11}C_r \alpha^{11-r} \beta^{-r} x^{33-4r}$ To find the coefficient of $x^9$ , set the exponent of $x$ to $9$ : $33 - 4r = 9 \implies 4r = 24 \implies r = 6$ Thus, the coefficient of $x^9$ is: ${}^{11}C_6 \alpha^{11-6} \beta^{-6} = {}^{11}C_6 \alpha^5 \beta^{-6}$

05execute

For the second expansion $\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$ , the general term is: $T_{k+1} = {}^{11}C_k (\alpha x)^{11-k} \left(-\frac{1}{\beta x^3}\right)^k = (-1)^k {}^{11}C_k \alpha^{11-k} \beta^{-k} x^{11-4k}$ To find the coefficient of $x^{-9}$ , set the exponent of $x$ to $-9$ : $11 - 4k = -9 \implies 4k = 20 \implies k = 5$ Thus, the coefficient of $x^{-9}$ is: $(-1)^5 {}^{11}C_5 \alpha^{11-5} \beta^{-5} = -{}^{11}C_5 \alpha^6 \beta^{-5}$

06execute

Equating the two coefficients: ${}^{11}C_6 \alpha^5 \beta^{-6} = -{}^{11}C_5 \alpha^6 \beta^{-5}$ Since ${}^{11}C_6 = {}^{11}C_{11-6} = {}^{11}C_5$ , we can cancel ${}^{11}C_6$ and ${}^{11}C_5$ : $\alpha^5 \beta^{-6} = -\alpha^6 \beta^{-5}$ Multiply both sides by $\beta^6$ and divide by $\alpha^5$ : $1 = -\alpha\beta \implies \alpha\beta = -1$ Then compute $(\alpha\beta)^2$ : $(\alpha\beta)^2 = (-1)^2 = 1$

✓verify

Check with $\alpha = 1, \beta = -1$ : In $(x^3 - x^{-1})^{11}$ , term with $x^9$ is ${}^{11}C_6 (x^3)^5 (-x^{-1})^6 = {}^{11}C_6 x^9 = 462 x^9$ . In $(x + x^{-3})^{11}$ , term with $x^{-9}$ is ${}^{11}C_5 (x)^6 (x^{-3})^5 = {}^{11}C_5 x^{-9} = 462 x^{-9}$ . The coefficients match ( $462 = 462$ ), verifying $(\alpha\beta)^2 = 1$ .

Hints that build this answer step by step

Which index $r$ gives the term containing $x^9$ in the binomial expansion of $\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$ ?
$r = 6$ , giving the term $\binom{11}{6} (\alpha x^3)^5 \left(\frac{1}{\beta x}\right)^6$
Which index $k$ gives the term containing $x^{-9}$ in the expansion of $\left(\alpha x - \frac{1}{\beta x^3}\right)^{11} = \sum_{k=0}^{11} \binom{11}{k} (\alpha x)^{11-k} \left(-\frac{1}{\beta x^3}\right)^k$ ?
$k = 5$ , giving the term $\binom{11}{5} (\alpha x)^6 \left(-\frac{1}{\beta x^3}\right)^5$
Equating the two coefficients, $\binom{11}{6}\frac{\alpha^5}{\beta^6} = -\binom{11}{5}\frac{\alpha^6}{\beta^5}$ , what relation between $\alpha$ and $\beta$ is obtained?
$\alpha\beta = -1$
Given $\alpha\beta = -1$ , what is the final value of $(\alpha\beta)^2$ ?
$1$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is

{}^{11}C_6 = {}^{11}C_5

By the binomial symmetry identity ${}^nC_r = {}^nC_{n-r}$ , here ${}^{11}C_6 = {}^{11}C_{11-6} = {}^{11}C_5$ .

Answer

The value of $(\alpha\beta)^2$ is equal to 1.

Why each option works or fails

Step-by-step solution

given: Expansions $\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$ and $\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$ . The coefficient of $x^9$ in the first equals the coefficient of $x^{-9}$ in the second.
goal: Find the value of $(\alpha\beta)^2$ .
approach: Write the general term $T_{r+1}$ for both binomial expansions, find the value of $r$ corresponding to the desired powers ( $x^9$ and $x^{-9}$ ), equate the resulting coefficients, and solve for $\alpha\beta$ .
execute: For the first expansion $\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$ , the general term is: $T_{r+1} = {}^{11}C_r (\alpha x^3)^{11-r} \left(\frac{1}{\beta x}\right)^r = {}^{11}C_r \alpha^{11-r} \beta^{-r} x^{33-4r}$ To find the coefficient of $x^9$ , set the exponent of $x$ to $9$ : $33 - 4r = 9 \implies 4r = 24 \implies r = 6$ Thus, the coefficient of $x^9$ is: ${}^{11}C_6 \alpha^{11-6} \beta^{-6} = {}^{11}C_6 \alpha^5 \beta^{-6}$
execute: For the second expansion $\left(\alpha x - \frac{1}{\beta x^3}\right)^{11}$ , the general term is: $T_{k+1} = {}^{11}C_k (\alpha x)^{11-k} \left(-\frac{1}{\beta x^3}\right)^k = (-1)^k {}^{11}C_k \alpha^{11-k} \beta^{-k} x^{11-4k}$ To find the coefficient of $x^{-9}$ , set the exponent of $x$ to $-9$ : $11 - 4k = -9 \implies 4k = 20 \implies k = 5$ Thus, the coefficient of $x^{-9}$ is: $(-1)^5 {}^{11}C_5 \alpha^{11-5} \beta^{-5} = -{}^{11}C_5 \alpha^6 \beta^{-5}$
execute: Equating the two coefficients: ${}^{11}C_6 \alpha^5 \beta^{-6} = -{}^{11}C_5 \alpha^6 \beta^{-5}$ Since ${}^{11}C_6 = {}^{11}C_{11-6} = {}^{11}C_5$ , we can cancel ${}^{11}C_6$ and ${}^{11}C_5$ : $\alpha^5 \beta^{-6} = -\alpha^6 \beta^{-5}$ Multiply both sides by $\beta^6$ and divide by $\alpha^5$ : $1 = -\alpha\beta \implies \alpha\beta = -1$ Then compute $(\alpha\beta)^2$ : $(\alpha\beta)^2 = (-1)^2 = 1$
verify: Check with $\alpha = 1, \beta = -1$ : In $(x^3 - x^{-1})^{11}$ , term with $x^9$ is ${}^{11}C_6 (x^3)^5 (-x^{-1})^6 = {}^{11}C_6 x^9 = 462 x^9$ . In $(x + x^{-3})^{11}$ , term with $x^{-9}$ is ${}^{11}C_5 (x)^6 (x^{-3})^5 = {}^{11}C_5 x^{-9} = 462 x^{-9}$ . The coefficients match ( $462 = 462$ ), verifying $(\alpha\beta)^2 = 1$ .

Binomial Theorem: JEE Main Mathematics Question with Solution

What feels right?

The value of $(\alpha\beta)^2$ is equal to 1.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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