Solve all your doubts with Teachoo Black (new monthly pack available now!)

Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 (i) Deleted for CBSE Board 2023 Exams

Misc 5 (ii) Important Deleted for CBSE Board 2023 Exams

Misc 6 Deleted for CBSE Board 2023 Exams

Misc 7 Deleted for CBSE Board 2023 Exams

Misc 8 Important Deleted for CBSE Board 2023 Exams

Misc 9 Deleted for CBSE Board 2023 Exams

Misc 10 Important

Misc 11 You are here

Misc 12

Misc 13 Important

Misc 14

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19

Misc 20 Important

Last updated at May 29, 2018 by Teachoo

Misc 11 (Method 1) If a + ib = (x + π)2/(2x^2 + 1) , prove that π2 + π2 = (x^2+ 1)2/(2x^2+ 1)^2 π + ππ = (x + i)2/(2x2+ 1) Using ( π + π )^2 = π2 + π2 + 2ππ = (π₯2 + (π)^2 + 2π₯π)/(2π₯2+1) Putting π2 = β1 = (π₯2 β 1 + 2π₯π)/(2π₯2+ 1) = (x2 β 1)/(2x2 + 1) + π 2x/(2x2 + 1) Hence π + ππ = (x2 β 1)/(2x2 + 1) + π 2x/(2x2 + 1) Comparing real part π = (π₯^2 β 1)/(2π₯^2 + 1) Comparing imaginary part b = 2π₯/(2π₯2 + 1) Calculating π2 + π2 π2 + π2 = ((π₯^2 β 1)/(2π₯2 + 1))^2 + (2π₯/(2π₯2 + 1))^2 = ((π₯2β 1)2 + (2π₯)2)/((2π₯2 + 1)2) Using (π β π)^2 = π2 + π2 β 2ππ = ((π₯2 )2 + (1)2 β 2( π₯2)1 + 4π₯2)/( (2π₯2 + 1)2) = (π₯4 + 1 β2π₯2 + 4π₯2)/((2π₯2 +1)2) = (π₯4 + 1 + 2π₯2)/((2π₯2 + 1)2) = ((π₯2)2 + (1)2 + 2(π₯2) (1))/((2π₯^2 + 1)2) Using ( π + π )^2 = π2 + π2 + 2ππ = (π₯2+ 1)2/((2π₯2 + 1)2) Hence π2 + π2 = (π₯2+ 1)2/((2π₯2 + 1)2) Hence proved Misc 11 (Method 2) If a + ib = (x + π)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Introduction (π + ππ) ( π β ππ) Using ( a β b ) ( a + b ) = a2 β b2 = π2 β (ππ)2 = π2 β π2π2 Putting i2 = β1 = π2β (β1) π2 = π2 + π2 Hence, (π + ππ) (π β ππ) = π2 + π2 Misc 11 (Method 2) If a + ib = (x + π)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Given π + ππ = (π₯ + π)2/(2π₯2 + 1) For π β ππ Replace π by β π in (1) π β ππ = (π₯ β π)2/(2π₯2 + 1) Calculating (π β ππ) (π + ππ) (π β ππ) (π + ππ) = (π₯ β π)2/(2π₯2 + 1) Γ (π₯ + π)2/(2π₯2 + 1) π2 + π2 = ((π₯ β π)2 (π₯ + π)2)/(2π₯2 +1)2 = ( (π₯ β π) (π₯ + π))^2/(2π₯2 +1)2 Using ( a β b ) ( a + b ) = a2 β b2 = (( π₯^2 β (π)^2 )^2 )/(2π₯^2 + 1)2 = γ( π₯2β (β1)) γ^2/(2π₯2 + 1)2 = ( π₯2 + 1)2/(2π₯2 + 1)2 Hence a2 + b2 = (π₯2 + 1 )/(2π₯2 + 1) Hence proved