
Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 8.2
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 You are here
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
Last updated at March 16, 2023 by Teachoo
Ex 8.2, 4 Write the general term in the expansion of (x2 – yx)12, x ≠ 0 We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br For (x2 – yx), Putting n = 12, a = x2 , b = – yx Tr + 1 = 12Cr (x2)12 – r . (– yx)r = 12Cr (x)2(12 – r) . (–1)r . yr . xr = 12Cr x24 – 2r . (–1)r . yr . xr = 12Cr x24 – 2r .xr . yr . (–1)r = (–1)r 12Cr. (x)24 – 2r + r . yr = (–1)r .12Cr . x24 – r . yr Hence, General term = (−1)r . 12Cr x24 – r . yr