Question 12 - Coefficient - Chapter 7 Class 11 Binomial Theorem
Last updated at April 16, 2024 by Teachoo
Coefficient
Ex 8.2 6 Important
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Question 9
Question 11 Important
Question 2
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Question 12 You are here
Question 7 Important
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Question 4 Important
Last updated at April 16, 2024 by Teachoo
Question 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 anβr 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 π!/2!(π β 2)! = 6 (π(π β 1)(π β 2)!)/2!(π β 2)! = 6 (π (π β 1))/2 = 6 m (m β 1) = 12 m2 β m = 12 m2 β m β 12 = 0 m2 β 4m + 3m β 12 = 0 m(m β 4) + 3 (m β 4) = 0 (m + 3) (m β 4) = 0 So, m = β3 or m = 4 But we need positive value of m Hence, m = 4