Miscellaneous

Misc 1
Important
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Misc 2

Misc 3

Misc 4 Important

Misc 5 (i) Deleted for CBSE Board 2022 Exams

Misc 5 (ii) Important

Misc 6

Misc 7

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11

Misc 12

Misc 13 Important Deleted for CBSE Board 2022 Exams

Misc 14

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19

Misc 20 Important

Chapter 5 Class 11 Complex Numbers (Term 1)

Serial order wise

Last updated at Jan. 14, 2022 by Teachoo

Misc 1 Evaluate: (π^18+(1/i)^25 )^3 (π^18+(1/π)^25 )^3 = (π^18+ 1/(π)^ππ )^3 = (π^18+ 1/(π Γ π^ππ ))^3 = ((π^π )^π+1/(π Γ (π^π )^ππ ))^3 Putting i2 = βπ = ((βπ)^π+1/γπ Γ (βπ)γ^12 )^3 = (βπ+1/(π Γ π))^3 = (β1+1/π)^3 Removing π from the denominator = (β1+1/πΓπ/π)^3 = (β1+π/π^π )^3 = (β1+π/((βπ)))^3 = (βπ β π )π = (β1(1+ π ))3 = (βπ)π (π + π )π = (β1)(1 + π )3 = β(π + π )π Using (a + b) 3 = a3 + b3 + 3ab(a + b) = β(13 + π3 + 3 Γ 1 Γ π (1 + π)) = β(1 + ππ +3π (1 + π)) = β(1 + ππ Γ π +3π (1 + π)) Putting i2 = βπ = β(1 +(βπ) Γ π +3π (1 + π)) = β(1 βπ +3π (1 + π)) = β(1 βπ +3π+3π Γ π) = β(1+2π+3π^π ) Putting i2 = βπ = β(1+2π+3 Γ βπ) = β(1+2πβ3) = β(2πβ2) = β2π+2 = πβππ