# Ex 8.2,12 - Chapter 8 Class 11 Binomial Theorem

Last updated at Jan. 29, 2020 by Teachoo

Ex 8.2

Ex 8.2,1
Deleted for CBSE Board 2023 Exams

Ex 8.2, 2 Important Deleted for CBSE Board 2023 Exams

Ex 8.2, 3 Deleted for CBSE Board 2023 Exams

Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams

Ex 8.2,5 Deleted for CBSE Board 2023 Exams

Ex 8.2 6 Important Deleted for CBSE Board 2023 Exams

Ex 8.2,7 Deleted for CBSE Board 2023 Exams

Ex 8.2,8 Important Deleted for CBSE Board 2023 Exams

Ex 8.2,9 Deleted for CBSE Board 2023 Exams

Ex 8.2,10 Important Deleted for CBSE Board 2023 Exams

Ex 8.2,11 Important Deleted for CBSE Board 2023 Exams

Ex 8.2,12 Deleted for CBSE Board 2023 Exams You are here

Last updated at Jan. 29, 2020 by Teachoo

Ex 8.2, 12 Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6. We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br General term of (1 + x)m is Putting n = m, a = 1, b = x Tr + 1 = mCr (1)m-r . (x)r = mCr (1) . (x)r = mCr xr We need coefficient of x2 So putting r = 2 Tr + 2 = mC2 x2 Coefficient of x2 is mC2 It is given that coefficient of x2 is 6 i.e. mC2 = 6 𝑚!/2!(𝑚 − 2)! = 6 (𝑚(𝑚 − 1)(𝑚 − 2)!)/2!(𝑚 − 2)! = 6 (𝑚 (𝑚 − 1))/2 = 6 m (m – 1) = 12 m2 – m = 12 m2 – m – 12 = 0 m2 – 4m + 3m – 12 = 0 m(m – 4) + 3 (m – 4) = 0 (m + 3) (m – 4) = 0 So, m = –3 or m = 4 But we need positive value of m Hence, m = 4