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7.Binomial Theorem
hard
$\left( {{2^{1/3}} + \frac{1}{{2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}$ ના વિસ્તરણમાં પહેલેથી $5^{th}$ માં પદ અને છેલ્લેથી $5^{th}$ માં પદનો ગુણોત્તર મેળવો.
A
$1:2{\left( 6 \right)^{\frac{1}{3}}}$
B
$1:4{\left( 16 \right)^{\frac{1}{3}}}$
C
$4{\left( {36} \right)^{\frac{1}{3}}}\,:\,1$
D
$2{\left( {36} \right)^{\frac{1}{3}}}\,:\,1$
(JEE MAIN-2019)
Solution
$\frac{{{5^{{\text{th}}}}{\text{ term from begining }}}}{{{5^{{\text{th}}}}{\text{ term from end }}}} = \frac{{10{{\text{C}}_4}{{\left( {\frac{1}{{2\left( {{3^{1/3}}} \right)}}} \right)}^4}{2^{6/3}}}}{{10{{\text{C}}_4}{{(2)}^{4/3}}{{\left( {\frac{1}{{2\left( {{3^{1/3}}} \right)}}} \right)}^6}}}$
$=\frac{2^{2} 2^{-2} 3^{-4 / 3}}{2^{4} 2^{(4 / 3)-6} 3^{-2}}=3^{2 / 3} \cdot 2^{8 / 3}$
$=4 \cdot(36)^{1 / 3}$
Standard 11
Mathematics
