Last updated at Dec. 16, 2024 by Teachoo
Misc 1 Evaluate: (๐^18+(1/i)^25 )^3 (๐^18+(1/๐)^25 )^3 = (๐^18+ 1/(๐)^๐๐ )^3 = (๐^18+ 1/(๐ ร ๐^๐๐ ))^3 = ((๐^๐ )^๐+1/(๐ ร (๐^๐ )^๐๐ ))^3 Putting i2 = โ๐ = ((โ๐)^๐+1/ใ๐ ร (โ๐)ใ^12 )^3 = (โ๐+1/(๐ ร ๐))^3 = (โ1+1/๐)^3 Removing ๐ from the denominator = (โ1+1/๐ร๐/๐)^3 = (โ1+๐/๐^๐ )^3 = (โ1+๐/((โ๐)))^3 = (โ๐ โ ๐ )๐ = (โ1(1+ ๐ ))3 = (โ๐)๐ (๐ + ๐ )๐ = (โ1)(1 + ๐ )3 = โ(๐ + ๐ )๐ Using (a + b) 3 = a3 + b3 + 3ab(a + b) = โ(13 + ๐3 + 3 ร 1 ร ๐ (1 + ๐)) = โ(1 + ๐๐ +3๐ (1 + ๐)) = โ(1 + ๐๐ ร ๐ +3๐ (1 + ๐)) Putting i2 = โ๐ = โ(1 +(โ๐) ร ๐ +3๐ (1 + ๐)) = โ(1 โ๐ +3๐ (1 + ๐)) = โ(1 โ๐ +3๐+3๐ ร ๐) = โ(1+2๐+3๐^๐ ) Putting i2 = โ๐ = โ(1+2๐+3 ร โ๐) = โ(1+2๐โ3) = โ(2๐โ2) = โ2๐+2 = ๐โ๐๐
Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5 Important
Misc 6
Misc 7
Misc 8
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12 Important
Misc 13
Misc 14 Important
Question 1 (i)
Question 1 (ii) Important
Question 2
Question 3
Question 4 Important
Question 5
Question 6 Important
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo