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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 (x)r Thus, Tr + 1 = mCr x We need coefficient of x2 So put 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 m + 3 =0 or m – 4 = 0 m = –3 or m = 4 But we need positive value of m Hence, m = 4

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.