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Ex 8.2
Ex 8.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 8.2, 3 Deleted for CBSE Board 2022 Exams
Ex 8.2, 4 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,5 Deleted for CBSE Board 2022 Exams
Ex 8.2 6 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,7 Deleted for CBSE Board 2022 Exams
Ex 8.2,8 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,9 Deleted for CBSE Board 2022 Exams
Ex 8.2,10 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,11 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,12 Deleted for CBSE Board 2022 Exams You are here
Ex 8.2
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