Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Find the term independent of x in $\left(2 x-\frac{1}{x}\right)^6$
Option: 1
$-120$

Option: 2
$-140$

Option: 3
$130$

Option: 4
$-160$

Answers (1)

best_answer

Given expression is $\left(2 x-\frac{1}{x}\right)^6$

The $(r+1)^{th}$ term of $(a+b)^{n}$ is $T_{r+1}=n c_r a^{n-r} b^r$

Where $a=2x$ , $b=x^{-1}$ and $n=6$

$\begin{aligned} & T_{r+1}=6 c_r(2 x)^{6-r}\left(\frac{-1}{x}\right)^r \\\\ & T_{r+1}=(-1)^r 6 c_r(2)^{6-r} x^{6-2 r} \end{aligned}$

For the term independent of x, the power of x in the above equation should be ’0’

$6-2r=0$

$r=3$ then the $(r+1)^{th}$ term is

$\begin{aligned} T_{3+1} & =(-1)^3 6 c_3 2^{6-3} x^{6-2(3)} \\ & =(-1) 6 c_3 2^3 \\ & =-6 c_3(8) \\ & =-8\left(\frac{6 !}{3 ! 3 !}\right) \\ & =-160 \end{aligned}$

Hence, the term independent of x is $T_4 = -160$

Posted by

Rakesh

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main admit card 2025 shows NTA tightening security, Here's how

anam-khan

Engineering aspirant dies by suicide in Kota, fourth case this year

anam-khan

What is non-Aadhaar declaration in JEE Main 2025? Explain mismatch between image, face

Latest Question