JEE Main & Advanced
JEE Main Paper (Held on 08-4-2019 Morning)
question_answer
The sum of the co-efficient of all even degree terms in x in the expansion of \[{{\left( x+\sqrt{{{x}^{3}}-1} \right)}^{6}}+\]\[{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}},(x>1)\]is equal to : [JEE Main 8-4-2019 Morning]
A)32
B)26
C)29
D)24
Correct Answer:
D
Solution :
\[{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}}+{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}}\] \[{{+}^{6}}{{C}_{6}}{{({{x}^{3}}-1)}^{3}}]\] \[=2{{[}^{6}}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{7}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{4}}{{x}^{2}}\] \[-2\,\,{}^{6}{{C}_{4}}{{x}^{5}}+({{x}^{9}}-1-3{{x}^{6}}+3{{x}^{3}})]\] \[\Rightarrow \]Sum of coefficient of even powers of x \[=2[115+15+1513]=24\]