Question

If the co-efficient of x⁹ in \((αx^3+\frac{1}{βx})^{11}\) and the co-efficient of x^-9 in \((αx-\frac{1}{βx^3})^{11}\) are equal, then \((αβ)^2\) is equal to ________.

Answer 1

Correct Answer: 1

We are given two binomial expansions, and we need to find \( (\alpha \beta)^2 \) when the co-efficients of \( x^9 \) and \( x^{-9} \) are equal in both expansions.
Step 1: Coefficient of \( x^9 \) in \( \left( \alpha x^3 + \frac{1}{\beta x} \right)^{11} \). 
Using the binomial theorem, the general term in the expansion of \( \left( \alpha x^3 + \frac{1}{\beta x} \right)^{11} \) is: \[ T_k = \binom{11}{k} \left( \alpha x^3 \right)^{11-k} \left( \frac{1}{\beta x} \right)^k. \] The power of \( x \) in the general term is: \[ 3(11 - k) - k = 33 - 3k - k = 33 - 4k. \] We want the power of \( x \) to be 9, so: \[ 33 - 4k = 9 \quad \Rightarrow \quad 4k = 24 \quad \Rightarrow \quad k = 6. \] The coefficient of \( x^9 \) is: \[ \binom{11}{6} \alpha^{11-6} \frac{1}{\beta^6} = \binom{11}{6} \alpha^5 \frac{1}{\beta^6}. \] Thus, the coefficient of \( x^9 \) is \( \binom{11}{6} \alpha^5 \beta^{-6} \). 

Step 2: Coefficient of \( x^{-9} \) in \( \left( \alpha x - \frac{1}{\beta x^3} \right)^{11 \).} 
Using the binomial theorem, the general term in the expansion of \( \left( \alpha x - \frac{1}{\beta x^3} \right)^{11} \) is: \[ T_k = \binom{11}{k} (\alpha x)^{11-k} \left( -\frac{1}{\beta x^3} \right)^k. \] The power of \( x \) in the general term is: \[ (11 - k) - 3k = 11 - k - 3k = 11 - 4k. \] We want the power of \( x \) to be \( -9 \), so: \[ 11 - 4k = -9 \quad \Rightarrow \quad 4k = 20 \quad \Rightarrow \quad k = 5. \] The coefficient of \( x^{-9} \) is: \[ \binom{11}{5} \alpha^{11-5} \left( -\frac{1}{\beta^5} \right) = \binom{11}{5} \alpha^6 \frac{1}{\beta^5}. \] Thus, the coefficient of \( x^{-9} \) is \( \binom{11}{5} \alpha^6 \beta^{-5} \). 

Step 3: Equating the two coefficients. We are given that the coefficients of \( x^9 \) and \( x^{-9} \) are equal. Therefore, we have: \[ \binom{11}{6} \alpha^5 \beta^{-6} = \binom{11}{5} \alpha^6 \beta^{-5}. \] Simplifying this equation: \[ \frac{\binom{11}{6}}{\binom{11}{5}} = \frac{\alpha^6}{\alpha^5} \cdot \frac{\beta^5}{\beta^6} \quad \Rightarrow \quad \frac{11 - 6}{6} = \frac{\alpha}{\beta} \quad \Rightarrow \quad \frac{5}{6} = \frac{\alpha}{\beta}. \] Thus, we have: \[ \alpha = \frac{5}{6} \beta. \] 

Step 4: Finding \( (\alpha \beta)^2 \). Now, we calculate \( (\alpha \beta)^2 \): \[ (\alpha \beta)^2 = \left( \frac{5}{6} \beta \cdot \beta \right)^2 = \left( \frac{5}{6} \beta^2 \right)^2 = \frac{25}{36} \beta^4. \] Therefore, the correct answer is \( 1 \).