Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
99 Percentile Qs Bank for JEE MainMathematicsBinomial Theorem

If the k^ th term in the expansion of ( 3 2 x^2- 1 3 x )^6 is independent of x , then the numerically greatest term in the expansion of ( 3 2 x^2- 1 3 x )^k when x= 2 3 , is

Options

  1. A40 81
  2. B( 7 6 )^5
  3. C20 27
  4. D( 7 6 )^4

Correct answer

C. 20 27

Step-by-step solution

We have, k^ th term is the expansion of ( 3 2 x^2- 1 3 x )^6 is independent of x Now, aligned T_ p+1 & = ^6 C_p ( 3 2 )^ 6-p (x)^ 12-2^p (- 1 3 )^p (x^ -p ) T_ p+1 & = ^6 C_p ( 3 2 )^ 6-p (x)^ 12-3 p (- 1 3 )^ aligned T_ p+i is independent of x aligned & 12-3 r & =0, r=4 & k =r+1=5 aligned aligned & Greatest term of ( 3 2 x^2- 1 3 x )^5 & r= [ n+1 1+ | 9 x^3 2 | ]= [ 6 1+ 9 2 8 27 ] [ x= 2 3 ] & r= [ 18 7 ]=2 & T₃= ^5 C₂ ( 3 2 x^2 )^3 (- 1 3 x )^2=10 27 8 x^4 9 & T₃=10 27 8 ( 2 3 )^4 1 9 = 20 27 aligned

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If 26 ( 2^3 3 12 2 + 2^5 5 12 4 + 2^7 7 12 6 + + 2¹³ 13 12 12 ) = 3¹³ - , then is equal to: 2026If (1 - x^3)¹⁰ = _ r=0 ¹⁰ a_r x^r (1-x)^ 30-2r , then 9a₉ a₁₀ is equal to __________. 2026If the coefficients of the middle terms in the binomial expansions of (1 + x)²⁶ and (1 - x)²⁸ , 0 , are equal, then the value of is: 2026The coefficient of x^2 in the expansion of (2x^2 + 1 x )¹⁰ , x 0 , is : 2026If the sum of the coefficients of x^7 and x¹⁴ in the expansion of ( 1 x^3 - x^4 )^n , x 0 , is zero, then the value of n is __________. 2026In the expansion of (9x- 1 3 x )¹⁸ , x>0 , if the term independent of x is (221)k , then k is equal to: 2026Let the smallest value of k N , for which the coefficient of x^3 in (1+x)^3 + (1+x)^4 + (1+x)^5 + + (1+x)⁹⁹ + (1+kx)¹⁰⁰ , x 0 , is (43n + 101 4 ) (¹⁰⁰C₃ ) for some n N , be p . The 2026If for 3 r 30 , 30 30-r + 3 30 31-r + 3 30 32-r + 30 33-r = m r , then m equals: 2026 Full Binomial Theorem list All 99 Percentile Qs Bank for JEE Main PYQs