Question

If the coefficient of x⁷ in (\(ax^2 +\frac{ 1}{2bx}^{11}\)) and x^-7 in (\(ax - \frac{1}{3bx^2}^{11}\)), are euual then

• A. 32ab=729
• B. 64ab=243
• C. 243ab=64
• D. 729ab=32

Answer 1

The Correct Option is D

The general form of the coefficient of \(x^7\) in \(\left( ax^2 + \frac{1}{2} bx \right)^{11}\) is obtained by using the binomial expansion: \[ r = \frac{11 \times 2 - 7}{3} = 5 \] Thus, the coefficient of \(x^7\) is given by: \[ \text{Coefficient of } x^7 = \binom{11}{6} a^5 \left(\frac{1}{2}b\right)^6 \] Similarly, for \(\left( ax - \frac{1}{3} bx^2 \right)\), the coefficient of \(x^7\) is: \[ \text{Coefficient of } x^7 = \binom{11}{6} a^5 \left(\frac{1}{3}b\right)^6 \] Equating the two coefficients, we get: \[ ab = \frac{25}{36} \] Thus, \[ 729ab = 32 \]