<p>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</p>
Step-by-Step Solution
Key Concept: The middle term coefficient in $(1+\alpha x)^n$ is $\binom{n}{n/2}\alpha^{n/2}$. Setting coefficients equal from two expansions with different powers allows us to solve for $\alpha$ by equating the binomial coefficients multiplied by appropriate powers of $\alpha$.
<p><strong>Step 1:</strong> Identify middle terms. For $(1+\alpha x)^4$ (even power), the middle term is at $r=2$: coefficient is $\binom{4}{2}\alpha^2 = 6\alpha^2$</p><p><strong>Step 2:</strong> For $(1-\alpha x)^6$ (even power), the middle term is at $r=3$: coefficient is $\binom{6}{3}(-\alpha)^3 = 20(-\alpha^3) = -20\alpha^3$</p><p><strong>Step 3:</strong> Since coefficients are equal: $6\alpha^2 = -20\alpha^3$</p><p><strong>Step 4:</strong> Rearranging: $6\alpha^2 + 20\alpha^3 = 0 \Rightarrow \alpha^2(6 + 20\alpha) = 0$</p><p><strong>Step 5:</strong> Since $\alpha \neq 0$, we get $20\alpha = -6 \Rightarrow \alpha = -\frac{3}{10}$</p><p>∴ Answer: C</p>
Correct Answer: C