Question

{"given":"We have two binomial expansions: $(1 + \\alpha x)^4$ and $(1 - \\alpha x)^6$. We need to find the value of $\\alpha$ such that the coefficients of their middle terms are equal. ","key_observation":"For $(1 + \\alpha x)^4$ (even power), the middle term is the $\\frac{4+2}{2} = 3$rd term. For $(1 - \\alpha x)^6$ (even power), the middle term is the $\\frac{6+2}{2} = 4$th term. The general term in binomial expansion $(a+b)^n$ is $T_{r+1} = \\binom{n}{r}a^{n-r}b^r$.","option_analysis":[{"label":"(A)","text":"$\\frac{3}{10}$","verdict":"incorrect","explanation":"This is the positive value of the correct answer. When we solve $6\\alpha^2 = -20\\alpha^3$, we get $\\alpha = -\\frac{3}{10}$, not positive $\\frac{3}{10}$."},{"label":"(B)","text":"$-\\frac{10}{3}$","verdict":"incorrect","explanation":"This would give coefficient of middle term in $(1+\\alpha x)^4$ as $6 \\cdot \\frac{100}{9} = \\frac{200}{3}$ and in $(1-\\alpha x)^6$ as $-20 \\cdot (-\\frac{1000}{27}) = \\frac{20000}{27}$, which are not equal."},{"label":"(C)","text":"$-\\frac{3}{10}$","verdict":"correct","explanation":"For $(1+\\alpha x)^4$: $T_3 = \\binom{4}{2}(1)^2(\\alpha x)^2 = 6\\alpha^2$. For $(1-\\alpha x)^6$: $T_4 = \\binom{6}{3}(1)^3(-\\alpha x)^3 = 20(-\\alpha^3) = -20\\alpha^3$. Setting equal: $6\\alpha^2 = -20\\alpha^3 \\Rightarrow 6 = -20\\alpha \\Rightarrow \\alpha = -\\frac{3}{10}$."},{"label":"(D)","text":"$\\frac{10}{3}$","verdict":"incorrect","explanation":"This value is too large and would make the coefficients vastly different. Also, from our equation $6 = -20\\alpha$, we need a negative value for $\\alpha$."}],"answer":"(C)","formula_steps":[]}