Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the ratio of the 7 th term from the beginning to the 7 th term from the end in the expansion of  <img alt="\left(\s

If the ratio of the 7 th term from the beginning to the 7 th term from the end in the expansion of  \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^x is \,\,\frac{1}{6}, then x is

Option: 1

9


Option: 2

6


Option: 3

12


Option: 4

None of these.


Answers (1)

best_answer

T_7\, in \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^x={ }^x C_6\left(2^{1 / 3}\right)^{x-6}\left(\frac{1}{3^{1 / 3}}\right)^6

7th term from the end in \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^x

=T_7 \text { in }\left(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^x={ }^x C_6\left(\frac{1}{3^{1 / 3}}\right)^{x-6}\left(2^{1 / 3}\right)^6

\therefore \frac{{ }^x C_6\left(2^{1 / 2}\right)^{x-6}\left(\frac{1}{3^{1 / 3}}\right)^6}{{ }^x C_6\left(\frac{1}{3^{1 / 3}}\right)^{x-6}\left(2^{1 / 3}\right)^6}=\frac{1}{6} \\

\Rightarrow \frac{\left(2^{1 / 3}\right)^{x-12}}{\left(\frac{1}{3^{1 / 3}}\right)^{x-12}}=\frac{1}{6} \Rightarrow 2^{\frac{x-12}{3}} \cdot 3^{\frac{x-12}{3}}=\frac{1}{6} \\

 \Rightarrow 6^{\frac{x-12}{3}}=6^{-1} \Rightarrow \frac{x-12}{3}=-1 \Rightarrow x=9 .

Posted by

manish

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question