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

To solve the problem, let's analyze the expression given in the question and find the coefficients as mentioned. The problem involves finding coefficients of specific terms in binomial expansions and comparing them.

Step 1: Understand the Binomial Expansions

• The binomial expression given is \( (ax^2 + \frac{1}{2bx})^{11} \)
• We need to find the coefficient of \( x^7 \) in this expression.
• The other binomial expression given is \( (ax - \frac{1}{3bx^2})^{11} \)
• We need to find the coefficient of \( x^{-7} \) in this expression.

Step 2: Coefficient of \( x^7 \) in \( (ax^2 + \frac{1}{2bx})^{11} \)

The general term in the expansion of \( (ax^2 + \frac{1}{2bx})^{11} \) is given by:

T_k = \binom{11}{k} (ax^2)^{11-k} \left( \frac{1}{2bx} \right)^{k}

This simplifies to:

T_k = \binom{11}{k} a^{11-k} x^{2(11-k)} \left(\frac{1}{2b}\right)^{k} x^{-k}

To find \( x^7 \), the power of \( x \) should be \( 7 \), thus solving:

2(11-k) - k = 7

22 - 2k - k = 7

15 = 3k \Rightarrow k = 5

For \( k = 5 \):

The coefficient is:

C_1 = \binom{11}{5} a^{6} \left(\frac{1}{2b}\right)^{5}

Step 3: Coefficient of \( x^{-7} \) in \( (ax - \frac{1}{3bx^2})^{11} \)

The general term in this expansion is:

T_m = \binom{11}{m} (ax)^{11-m} \left(-\frac{1}{3bx^2}\right)^{m}

This simplifies to:

T_m = \binom{11}{m} a^{11-m} x^{11-m} \left(-\frac{1}{3b}\right)^{m} x^{-2m}

To find \( x^{-7} \), the equation for power of \( x \) is:

11 - m - 2m = -7

11 - 3m = -7 \Rightarrow 3m = 18 \Rightarrow m = 6

For \( m = 6 \):

The coefficient is:

C_2 = \binom{11}{6} a^{5} \left(-\frac{1}{3b}\right)^{6}

Step 4: Equating the Coefficients

We are given that the coefficients \( C_1 = C_2 \).

So, we equate the expressions:

\binom{11}{5} a^{6} \left(\frac{1}{2b}\right)^{5} = \binom{11}{6} a^{5} \left(-\frac{1}{3b}\right)^{6}

Simplifying and solving the equation:

\frac{\binom{11}{5}}{\binom{11}{6}} \cdot \frac{a}{b^{5}} \cdot \left(\frac{1}{2}\right)^{5} = \frac{1}{b^{6}} \cdot \left(\frac{-1}{3}\right)^{6}

Simplifying further gives:

3 \cdot a \cdot \frac{1}{32} = -729 \cdot b

Giving us 32ab = 729 , but since \(-1\) is raised to an even power, we have 729ab = 32 as the correct equation.

Conclusion

The correct option, thus, is: 729ab = 32