A) \[-\frac{5}{3}\]
B) \[\frac{10}{3}\]
C) \[-\frac{3}{10}\]
D) \[\frac{3}{5}\]
Correct Answer: C
Solution :
[c] Middle term of \[{{(1+\alpha x)}^{4}}\]is \[{{T}_{3}}\]. Its coefficient is \[^{4}{{C}_{2}}{{(\alpha )}^{2}}=6{{\alpha }^{2.}}\] Middle term of \[{{(1-\alpha x)}^{6}}\]is \[{{T}_{4}}\]. Its coefficient is \[^{6}{{C}_{3}}{{(-\alpha )}^{3}}=-20{{\alpha }^{3}}\]. According to question, \[6{{\alpha }^{2}}=-20{{\alpha }^{3}}\] or\[3{{\alpha }^{2}}+10{{\alpha }^{2}}=0\] or \[{{\alpha }^{2}}(3+10\alpha )=0\] or \[\alpha =-\frac{3}{10}\]
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