


Ex 8.1
Ex 8.1,2 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,3 Deleted for CBSE Board 2022 Exams
Ex 8.1,4 Important Deleted for CBSE Board 2022 Exams
Ex 8.1, 5 Deleted for CBSE Board 2022 Exams
Ex 8.1 6 Deleted for CBSE Board 2022 Exams
Ex 8.1,7 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,8 Deleted for CBSE Board 2022 Exams
Ex 8.1,9 Deleted for CBSE Board 2022 Exams
Ex 8.1,10 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,11 Deleted for CBSE Board 2022 Exams You are here
Ex 8.1,12 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,13 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,14 Important Deleted for CBSE Board 2022 Exams
Ex 8.1
Ex 8.1, 11 Find (a + b)4 – (a – b)4. Hence, evaluate (√3+√2)^4– (√3−√2)^4 . We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence, (a + b)4 = 4C0 a4 + 4C1 a3 b1 + 4C2 a2 b2 + 4C3 a1b3 + 4C4 b4 = 4!/0!(4 − 0)! a4 + 4!/(1 × (4 − 1)!) a3b1 + 4!/2!(4 − 2)! a2b2 + 4!/(3!(4 − 3)!) a3 b1 + 4!/4!(4 −4)! b4 = 4!/(1 × 4!) a4 + 4!/(1 × 3!) a3 b + 4!/(2! × 2!) a2 b2 + 4!/(3! × 1!) ab3 + 4!/(4! 0!) b4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Hence, (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Replacing b with –b (a + (–b))4 = a4 + 4a3 (–b) + 6a2 (–b)2 + 4a (–b)3 + (–b)4 (a – b)4 = a4 – 4a3 b + 6a2b2 – 4ab3 + b4 We need to find (a + b)4 – (a – b)4 (a + b)4 – (a – b)4 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4 ) – (a4 – 4a3b + 6a2b2 – 4ab3 + b4 ) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 – a4 + 4a3b – 6a2b2 + 4ab3 – b4 = a4 – a4 + 6a2b2 – 6a2b2 + b4 – b4 + 4a3b + 4a3b + 4ab3 + 4ab3 = 8a3b + 8ab3 = 8ab (a2 + b2) Thus, (a + b)4 – (a – b)4 = 8ab (a2 + b2) We need to find (√𝟑+√𝟐)^𝟒– (√𝟑−√𝟐)^𝟒 Putting a = √3 and b = √2 (√3+√2)^4– (√3−√2)^4= 8(√3 )(√2 ) ((√3 )^2 "+ " (√2 )^2 ) = 8√3 √2 (3+2) = 8√3 √2 (5) = (8 × 5)√(3 ×2) = 40 √𝟔