If the coefficient of x^9 in ( a x^3 + 1 x^11) and the coefficient of x^-9 in ( a x - 1 x^3)^11 are equal, then ()^2 is equal to:

Given: The coefficients of x^9 in ( x^3 + 1 x)^11 and of x^-9 in ( x - 1 x^3)^11 are equal. Find: ()^2. For ( x^3 + 1 x)^11 the general term is Tk = 11k ( x^3)^11-k ( 1 x)^k So the power of x is 3(11-k)-k = 33-4k For the term containing x^9, 33-4k=9 4k=24 k=6 Hence the coefficient of x^9 is 116 ^5 ^-6 For ( x - 1 x^3)^11 the general term is Tk = 11k( x)^11-k (- 1 x^3)^k So the power of x is (11-k)-3k = 11-4k For the term containing x^-9, 11-4k=-9 4k=20 k=5 Thus the coefficient of x^-9 is taken as 115 ^6 ^-5 Equating the two coefficients, 116 ^5 ^-6 = 115 ^6 ^-5 Using 116= 115 we get ^5 ^-6= ^6 ^-5 1= Therefore, ()^2=1 So the correct option is A. Once the required powers are identified, the key observation is 116= 115 Since the two coefficients are equal, only the powers of and need to match: ^5 ^-6= ^6 ^-5 This immediately gives =1 Hence, ()^2=1 Therefore the correct option is A.

If the coefficient of x^9 in ( a x^3 + 1 x^11) and the coefficien… — Solution | JEE Mathematics

MCQMediumJEE 2023General Term

If the coefficient of $x^9$ in $\left( a x^3 + \frac{1}{\beta x^{11}} \right)$ and the coefficient of $x^{-9}$ in $\left( a x - \frac{1}{\beta x^3} \right)^{11}$ are equal, then $(\alpha\beta)^2$ is equal to:

A
$1$
B
$4$
C
$9$
D
$16$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The coefficients of $x^9$ in

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

and of $x^{-9}$ in

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

are equal.

Find: $(\alpha\beta)^2$ .

For

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

the general term is

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

So the power of $x$ is

3(11-k)-k = 33-4k

For the term containing $x^9$ ,

33-4k=9

4k=24

k=6

Hence the coefficient of $x^9$ is

\binom{11}{6}\alpha^{5}\beta^{-6}

Equating the Coefficients

For

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

the general term is

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

So the power of $x$ is

(11-k)-3k = 11-4k

For the term containing $x^{-9}$ ,

11-4k=-9

4k=20

k=5

Thus the coefficient of $x^{-9}$ is taken as

\binom{11}{5}\alpha^{6}\beta^{-5}

Equating the two coefficients,

\binom{11}{6}\alpha^{5}\beta^{-6} = \binom{11}{5}\alpha^{6}\beta^{-5}

Using

\binom{11}{6}=\binom{11}{5}

we get

\alpha^5\beta^{-6}=\alpha^6\beta^{-5}

1=\alpha\beta

Therefore,

(\alpha\beta)^2=1

So the correct option is A.

Use Symmetry of Binomial Coefficients

Once the required powers are identified, the key observation is

\binom{11}{6}=\binom{11}{5}

Since the two coefficients are equal, only the powers of $\alpha$ and $\beta$ need to match:

\alpha^{5}\beta^{-6}=\alpha^{6}\beta^{-5}

This immediately gives

\alpha\beta=1

Hence,

(\alpha\beta)^2=1

Therefore the correct option is A.

Common mistakes

Using the wrong first expression from the given question. The solution clearly works with $\left(\alpha x^3 + \frac{1}{\beta x}\right)^{11}$ , not with a denominator containing $x^{11}$ . Always follow the expression consistently from the worked solution when the source text has a typo.
Finding the correct term number but not the correct power of $x$ . In binomial expansions, combine exponents carefully: for the first expansion the power is $3(11-k)-k$ , and for the second it is $(11-k)-3k$ .
Forgetting the identity $\binom{11}{6}=\binom{11}{5}$ . If this symmetry is missed, unnecessary algebra is introduced and the final simplification becomes error-prone.

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Answer

Step-by-step solution

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