JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

1. Pehli Expression: $\left( \alpha x^3 + \frac{1}{\beta x} \right)^{11}$

T_{r+1} = \binom{11}{r} (\alpha x^3)^{11-r} \left( \frac{1}{\beta x} \right)^r

T_{r+1} = \binom{11}{r} \alpha^{11-r} \beta^{-r} x^{33-4r}

$x^9$ ke liye:

33 - 4r = 9 \implies 4r = 24 \implies r = 6

Coefficient ( $C_1$ ): $\binom{11}{6} \alpha^5 \beta^{-6}$

2. Dusri Expression: $\left( \alpha x - \frac{1}{\beta x^3} \right)^{11}$

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

T_{k+1} = \binom{11}{k} \alpha^{11-k} (-1)^k \beta^{-k} x^{11-4k}

$x^{-9}$ ke liye:

11 - 4k = -9 \implies 4k = 20 \implies k = 5

Coefficient ( $C_2$ ): $-\binom{11}{5} \alpha^6 \beta^{-5}$

3. Calculation

C_1 = C_2

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

$\binom{11}{6} = \binom{11}{5}$ hone ki wajah se:

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

\frac{1}{\beta} = -\alpha \implies \alpha \beta = -1

Final Result:

(\alpha \beta)^2 = (-1)^2 = 1