Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET2009MathematicsBinomial Theorem

For |x| < 1 , the constant term in the expansion of 1 (x-1)^2(x-2) is

Options

  1. A2
  2. B1
  3. C0
  4. D- 1 2

Correct answer

D. - 1 2

Step-by-step solution

aligned 1 (x-1)^2(x-2) & = 1 -2(1-x)^2 (1- x 2 ) & =- 1 2 [(1-x)⁻² (1- x 2 )⁻¹ ] & =- 1 2 [(1+2 x+ ) (1+ x 2 + ) ] aligned Coefficient of constant term is - 1 2 .

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If n 13 , n 14 and n 15 are in arithmetic progression, then the positive integer value of ' n ' can be 2026If the coefficients of x^2 and x^3 in the expansion of (3 + kx)^9 are equal, then the value of ' k ' is 2026The remainder when 7¹⁰³ is divided by 25 is 2026_ r=1 ¹⁵ r^2 ( ¹⁵ C_r 15 r-1 )= 20251 81^ n - ^ 2 n C ₁ 10 81^ n + ^ 2 n C ₂ 10^2 81^ n - + 10^ 2 n 81^ n = 2025If x is positive real number and the first negative term in the expansion of (1+ x )^ 27 / 5 is t _ k then k = 2025In the binomial expansion of (p-q)¹⁴ , if the sum of 7^ th term and 8^ th term is zero, then p+q p-q = 2025The numerically greatest term in the expansion of (x+3 y)¹³ , when x= 1 2 and y= 1 3 is 2025 Full Binomial Theorem list All AP EAMCET PYQs