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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Misc 6 (Method 1) Differentiate w.r.t. x the function, γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑〖π‘₯ γ€— ) βˆ’ √(1 βˆ’sin⁑π‘₯ )) ] , 0<π‘₯< πœ‹/2 Let 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) +√(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑〖π‘₯ γ€— )βˆ’βˆš(1 βˆ’ sin⁑π‘₯ )) ] Rationalizing the sum 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [((√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ )))/((√(1 + sin⁑〖π‘₯ γ€— )βˆ’ √(1 βˆ’ sin⁑π‘₯ )) ) Γ—((√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ )))/((√(1+sin⁑〖π‘₯ γ€— )+ √(1 βˆ’ sin⁑π‘₯ )) )] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))^2/((√(1 + sin⁑〖π‘₯ γ€— )βˆ’ √(1 βˆ’γ€– sin〗⁑π‘₯ )) (√(1 + sin⁑〖π‘₯ γ€— )+ √(1 βˆ’γ€– sin〗⁑π‘₯ )) ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [((√(1 + sin⁑π‘₯ ) )^2 + (√(1 βˆ’ sin⁑π‘₯ ) )^2+ 2(√(1 + sin⁑〖π‘₯ γ€— ))(√(1 βˆ’γ€– sin〗⁑π‘₯ )))/((√(1 + sin⁑〖π‘₯ γ€— )βˆ’ √(1 βˆ’ γ€– sin〗⁑π‘₯ )) (√(1 + sin⁑〖π‘₯ γ€— )+ √(1 βˆ’ sin⁑π‘₯ )) ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [((1 + sin⁑π‘₯ ) + (1 βˆ’ sin⁑π‘₯ ) + 2√((1 + sin⁑π‘₯ ) (1 βˆ’ sin⁑π‘₯ ) ))/((√(1 + sin⁑〖π‘₯ γ€— ))^2 βˆ’ (√(1 βˆ’ sin⁑π‘₯ ))^2 ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + sin⁑π‘₯ + 1 βˆ’ sin⁑π‘₯ + 2√((1)^2 βˆ’ sin^2⁑π‘₯ ))/(1 + sin⁑π‘₯ βˆ’ 1 + sin⁑π‘₯ ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 + 2√(1 βˆ’ sin^2⁑π‘₯ ))/(2 sin⁑π‘₯ ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 (1 + √(1 βˆ’ sin^2⁑π‘₯ ) ) )/(2 sin⁑π‘₯ )] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + √(γ€–cπ‘œπ‘ γ€—^2⁑π‘₯ ) )/sin⁑π‘₯ ] (𝐴𝑠 1βˆ’γ€–π‘ π‘–π‘›γ€—^2β‘πœƒ= γ€–π‘π‘œπ‘ γ€—^2β‘πœƒ) = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + cos⁑π‘₯ )/sin⁑π‘₯ ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + 2 cos^2⁑〖π‘₯/2γ€— βˆ’ 1 )/(2 sin⁑〖 π‘₯/2 γ€— γ€–cos 〗⁑〖π‘₯/2γ€— )] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 cos^2⁑〖π‘₯/2γ€— )/(2 sin⁑〖 π‘₯/2 γ€— γ€–cos 〗⁑〖π‘₯/2γ€— )] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(γ€–cos 〗⁑〖π‘₯/2γ€— )/sin⁑〖 π‘₯/2 γ€— ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [cot⁑(π‘₯/2) ] = π‘₯/2 We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by πœƒ/2 sin ΞΈ = 2 π’”π’Šπ’β‘γ€–πœ½/πŸγ€— π’„π’π’”β‘γ€–πœ½/πŸγ€— and cos 2ΞΈ = 2cos2 ΞΈ – 1 Replacing ΞΈ by πœƒ/2 cos ΞΈ = 2cos2 𝜽/𝟐 – 1 (𝐴𝑠 γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) (π‘π‘œπ‘‘β‘πœƒ )=πœƒ) Thus, π’š= 𝒙/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑/𝑑π‘₯ (π‘₯/2) 𝑑𝑦/𝑑π‘₯ = 1/2 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝟏/𝟐 Thus, π’š= 𝒙/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑/𝑑π‘₯ (π‘₯/2) 𝑑𝑦/𝑑π‘₯ = 1/2 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝟏/𝟐 Thus, π’š= 𝒙/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑/𝑑π‘₯ (π‘₯/2) 𝑑𝑦/𝑑π‘₯ = 1/2 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝟏/𝟐 Misc 6 (Method 2) Differentiate w.r.t. x the function, γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑〖π‘₯ γ€— ) βˆ’ √(1 βˆ’ sin⁑π‘₯ )) ] , 0<π‘₯< πœ‹/2 Let 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 +γ€– sin〗⁑〖π‘₯ γ€— ) βˆ’ √(1 βˆ’ sin⁑π‘₯ )) ] As 〖𝑠𝑖𝑛〗^2 πœƒ+γ€–π‘π‘œπ‘ γ€—^2β‘πœƒ=1 Replacing πœƒ by π‘₯/2 𝑠𝑖𝑛2 π‘₯/2 + γ€–π‘π‘œπ‘ γ€—^2 π‘₯/2 = 1 & 𝑠𝑖𝑛⁑2πœƒ=2 π‘ π‘–π‘›β‘γ€–πœƒ π‘π‘œπ‘ β‘πœƒ γ€— Replacing πœƒ by π‘₯/2 𝑠𝑖𝑛⁑π‘₯ = 2 𝑠𝑖𝑛⁑π‘₯/2 π‘π‘œπ‘ β‘π‘₯/2 1 + sin x = (〖𝑠𝑖𝑛〗^2 π‘₯/2 + γ€–π‘π‘œπ‘ γ€—^2 π‘₯/2) + 2 sin π‘₯/2 cos π‘₯/2 1 + sin x = (π‘π‘œπ‘  π‘₯/2 +sin⁑〖π‘₯/2γ€— )^2 √("1 + sin x" ) = (𝒄𝒐𝒔 𝒙/𝟐 +π’”π’Šπ’β‘γ€–π’™/πŸγ€— ) 1 – sin x = (〖𝑠𝑖𝑛〗^2 π‘₯/2 + γ€–π‘π‘œπ‘ γ€—^2 π‘₯/2) – 2 sin π‘₯/2 cos π‘₯/2 1 – sin x = (π‘π‘œπ‘  π‘₯/2 βˆ’sin⁑〖π‘₯/2γ€— )^2 √("1 – sin x" ) = (π‘π‘œπ‘  π‘₯/2 βˆ’sin⁑〖π‘₯/2γ€— ) Now, y = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) |(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑π‘₯ ) βˆ’ √(1 βˆ’ sin⁑π‘₯ ))| Substitute value of √(1+𝑠𝑖𝑛⁑π‘₯ ) & √(1βˆ’π‘ π‘–π‘›β‘π‘₯ ) from (1) & (2). y = cotβˆ’1 [((γ€–cos 〗⁑〖π‘₯/2γ€— + γ€–sin 〗⁑〖π‘₯/2γ€— ) + (γ€–cos 〗⁑〖π‘₯/2γ€— βˆ’ γ€–sin 〗⁑〖π‘₯/2γ€— ))/((γ€–cos 〗⁑〖π‘₯/2γ€— + γ€–sin 〗⁑〖π‘₯/2γ€— ) + (γ€–cos 〗⁑〖π‘₯/2γ€— βˆ’ γ€–sin 〗⁑〖π‘₯/2γ€— ) )] y = cotβˆ’1 [((γ€–cos 〗⁑〖π‘₯/2γ€— + γ€–sin 〗⁑〖π‘₯/2γ€— ) + (γ€–cos 〗⁑〖π‘₯/2γ€— βˆ’ γ€–sin 〗⁑〖π‘₯/2γ€— ))/((γ€–cos 〗⁑〖π‘₯/2γ€— + γ€–sin 〗⁑〖π‘₯/2γ€— ) + (γ€–cos 〗⁑〖π‘₯/2γ€— βˆ’ γ€–sin 〗⁑〖π‘₯/2γ€— ) )] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— + 〖𝑠𝑖𝑛 〗⁑〖π‘₯/2γ€— + π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— βˆ’ 〖𝑠𝑖𝑛 〗⁑〖π‘₯/2γ€—)/(π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— + 𝑠𝑖𝑛⁑〖 π‘₯/2γ€— βˆ’ γ€–π‘π‘œπ‘  〗⁑〖π‘₯/2γ€— βˆ’ 𝑠𝑖𝑛⁑〖 π‘₯/2γ€— ) ] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 γ€–π‘π‘œπ‘  〗⁑〖π‘₯/2γ€— )/(2 𝑠𝑖𝑛⁑〖 π‘₯/2γ€— ) ] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [π‘π‘œπ‘‘ (π‘₯ )/2 ] 𝑦= π‘₯/2 (𝐴𝑠 γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) (π‘π‘œπ‘‘β‘πœƒ )=πœƒ) Thus, π’š= 𝒙/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑/𝑑π‘₯ (π‘₯/2) 𝑑𝑦/𝑑π‘₯ = 1/2 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝟏/𝟐

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.