Answer : (a) - \(\frac{3}{10}\)
The Middle term in the expansion of (1 + αx)4 is the \((\frac{4}{2} +1)\)th term, i.e., 3rd term.
∴ t3 = t 2 + 1 = 4C2 (αx)2
The Middle term in the expansion of (1 – αx)6 is the \((\frac{6}{2} +1)\)th term, i.e., 4th term.
∴ T4 = T3 + 1 = 6C3 (– 1)3 (αx)3
∴ Coefficient of t 3 = Coefficient of T4
⇒ 4C2 α2 = 6C3 (-1)3 α3
⇒ \(\frac{4\times 3}{2} \alpha^2 = (-1) \times \frac{6\times 5\times 4}{3\times 2} \alpha^3\)
⇒ 6α2 = -20α3
⇒ α = - \(\frac{6}{20}\)
= - \(\frac{3}{10}\)