Miscellaneous

Misc 1
Important

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 You are here

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 May 29, 2018 by Teachoo

Misc 14 Find the real numbers x and y if (π₯ β ππ¦) (3 + 5π) is the conjugate of β6 β 24π. Conjugate of β6 β24π = β 6 + 24π Now it is given that (π₯ β ππ¦) (3 + 5π) is conjugate of β6 + 24π Hence from (1) and (2) β 6 + 24π = (π₯ β ππ¦) (3 + 5π) β 6 + 24π = π₯ ( 3 + 5π ) β ππ¦ ( 3 + 5π) β 6 + 24π = 3π₯ + 5π₯π β 3π¦π β 5π2π¦ Putting π2 = β1 β 6 + 24π = 3π₯ + 5π₯π β 3π¦π β 5 Γ β 1 Γ π¦ β 6 + 24π = 3π₯ + 5π₯π β 3π¦π + 5π¦ β 6 + 24π = 3π₯ + 5π¦ β 3π¦π + 5π₯π β 6 + 24π = 3π₯ + 5π¦ + ( β 3π¦ + 5π₯)π Comparing real parts β 6 = 3π₯ + 5π¦ Comparing imaginary parts 24 = 5π₯ β 3π¦ We solve equation (3) and (4) to find the value of x and y From (3) 3π₯ + 5π¦ = β6 3π₯ = β6 β 5π¦ π₯ = (β6 β 5π¦)/3 Putting π₯ = (β6 β 5π¦)/3 in (4) 5π₯ β 3π¦ = 24 5 ((β6 β 5π¦)/3) β 3π¦ = 24 Multiplying 3 both sides 3Γ5((β6 β 5π¦)/3) β 3Γ3π¦ = 3Γ24 5(β6 β 5π¦) β 9π¦ = 72 5 (β 6) β 5 (5π¦) β 9π¦ = 72 β30 β 25π¦β 9π¦ = 72 β 25π¦ β 9π¦ = 72 + 30 β 34π¦ = 102 π¦ = 102/(β34) π¦ = β3 Putting y = β 3 in (4) 24 = 5π₯ β 3π¦ 24 = 5π₯ β 3(β3) 24 = 5π₯+9 24 β9= 5π₯ 15= 5π₯ 5π₯=15 π₯ = 15/5 π₯ = 3 Hence value of π₯ = 3 and π¦ = β3