The coefficient of the middle term in the binomial expansion in powers of `x` of`(1+alphax)^4` and of`(1-alphax)^6` is the same, if `alpha` equals (a) `-5/3` (b) `10/3` (c) `-3/10` (d) `3/5`

Options

A. `-5/3`
B. `10/3`
C. `-3/10`
D. `3/5`

Explanation

To solve the problem, we need to find the value of \(\alpha\) such that the coefficients of the middle terms in the binomial expansions of \((1 + \alpha x)^4\) and \((1 - \alpha x)^6\) are equal.

### Step-by-Step Solution:

1. **Identify the Middle Term for \((1 + \alpha x)^4\)**:
- The power \(n = 4\) is even.
- The middle term is given by the formula for the \(\left(\frac{n}{2} + 1\right)\)th term.
- Thus, the middle term is the \(2 + 1 = 3\)rd term.
- The general term \(T_r\) in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
- Here, \(a = 1\), \(b = \alpha x\), and \(n = 4\).
- Therefore, the 3rd term \(T_3\) is:
\[ T_3 = \binom{4}{2} (1)^{4-2} (\alpha x)^2 = \binom{4}{2} \alpha^2 x^2 \]
- Calculating \(\binom{4}{2}\):
\[ \binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \]
- Thus, the 3rd term becomes:
\[ T_3 = 6 \alpha^2 x^2 \]

2. **Identify the Middle Term for \((1 - \alpha x)^6\)**:
- The power \(n = 6\) is even.
- The middle term is given by the \(\left(\frac{n}{2} + 1\right)\)th term.
- Thus, the middle term is the \(3 + 1 = 4\)th term.
- The general term \(T_r\) in the expansion is:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
- Here, \(a = 1\), \(b = -\alpha x\), and \(n = 6\).
- Therefore, the 4th term \(T_4\) is:
\[ T_4 = \binom{6}{3} (1)^{6-3} (-\alpha x)^3 = \binom{6}{3} (-\alpha)^3 x^3 \]
- Calculating \(\binom{6}{3}\):
\[ \binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
- Thus, the 4th term becomes:
\[ T_4 = 20 (-\alpha)^3 x^3 = -20 \alpha^3 x^3 \]

3. **Set the Coefficients Equal**:
- We want the coefficients of \(x^2\) from \(T_3\) and \(x^3\) from \(T_4\) to be equal:
\[ 6 \alpha^2 = -20 \alpha^3 \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ 20 \alpha^3 + 6 \alpha^2 = 0 \]
- Factoring out \(\alpha^2\):
\[ \alpha^2 (20 \alpha + 6) = 0 \]

5. **Finding the Solutions**:
- Setting each factor to zero gives:
- \(\alpha^2 = 0 \Rightarrow \alpha = 0\)
- \(20 \alpha + 6 = 0 \Rightarrow \alpha = -\frac{6}{20} = -\frac{3}{10}\)

6. **Conclusion**:
- The values of \(\alpha\) that satisfy the condition are \(\alpha = 0\) or \(\alpha = -\frac{3}{10}\).
- Among the options provided, the correct answer is:
\[ \boxed{-\frac{3}{10}} \]