The coefficient of the middle term in the binomial expansion in powers of `x` of `(1 + alpha x)^(4)` and of `(1- alpha x)^(6)` is the same if `alpha` equals

Options

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

Answer

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 in the First Expansion:**
For the binomial expansion of \( (1 + \alpha x)^4 \):
- The number of terms \( n = 4 \).
- The middle term is given by \( T_{(n/2) + 1} = T_{(4/2) + 1} = T_3 \).
- The general term \( T_r \) in the expansion is given by:
\[ T_r = \binom{n}{r-1} (1)^{n-(r-1)} (\alpha x)^{r-1} \]
- For \( T_3 \):
\[ T_3 = \binom{4}{2} (1)^{4-2} (\alpha x)^{2} = \binom{4}{2} \alpha^2 x^2 \]
- Calculate \( \binom{4}{2} = 6 \):
\[ T_3 = 6 \alpha^2 x^2 \]
- The coefficient of the middle term is \( 6 \alpha^2 \).

2. **Identify the Middle Term in the Second Expansion:**
For the binomial expansion of \( (1 - \alpha x)^6 \):
- The number of terms \( n = 6 \).
- The middle term is given by \( T_{(n/2) + 1} = T_{(6/2) + 1} = T_4 \).
- For \( T_4 \):
\[ T_4 = \binom{6}{3} (1)^{6-3} (-\alpha x)^{3} = \binom{6}{3} (-\alpha)^3 x^3 \]
- Calculate \( \binom{6}{3} = 20 \):
\[ T_4 = 20 (-\alpha^3) x^3 = -20 \alpha^3 x^3 \]
- The coefficient of the middle term is \( -20 \alpha^3 \).

3. **Set the Coefficients Equal:**
Since the coefficients of the middle terms are equal:
\[ 6 \alpha^2 = -20 \alpha^3 \]

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

5. **Finding the Values of \( \alpha \):**
This gives us two cases:
- \( \alpha^2 = 0 \) which implies \( \alpha = 0 \) (not a valid option).
- \( 20 \alpha + 6 = 0 \) which leads to:
\[ 20 \alpha = -6 \implies \alpha = -\frac{6}{20} = -\frac{3}{10} \]

### Final Answer:
Thus, the value of \( \alpha \) is:
\[ \alpha = -\frac{3}{10} \]