Misc 6 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Misc 6 (Method 1) Differentiate w.r.t. x the function, 〖𝑐𝑜𝑡〗^(−1 ) [(√(1 + sin⁡𝑥 ) + √(1 − sin⁡𝑥 ))/(√(1 + sin⁡〖𝑥 〗 ) − √(1 −sin⁡𝑥 )) ] , 0<𝑥< 𝜋/2 Let 𝑦= 〖𝑐𝑜𝑡〗^(−1 ) [(√(1 + sin⁡𝑥 ) +√(1 − sin⁡𝑥 ))/(√(1 + sin⁡〖𝑥 〗 )−√(1 − sin⁡𝑥 )) ] Rationalizing the sum 𝑦= 〖𝑐𝑜𝑡〗^(−1 ) [((√(1 + sin⁡𝑥 ) + √(1 − sin⁡𝑥 )))/((√(1 + sin⁡〖𝑥 〗 )− √(1 − sin⁡𝑥 )) ) ×((√(1 + sin⁡𝑥 ) + √(1 − sin⁡𝑥 )))/((√(1+sin⁡〖𝑥 〗 )+ √(1 − sin⁡𝑥 )) )] 𝑦= 〖𝑐𝑜𝑡〗^(−1 ) [(√(1 + sin⁡𝑥 ) + √(1 − sin⁡𝑥 ))^2/((√(1 + sin⁡〖𝑥 〗 )− √(1 −〖 sin〗⁡𝑥 )) (√(1 + sin⁡〖𝑥 〗 )+ √(1 −〖 sin〗⁡𝑥 )) ) ] = 〖𝑐𝑜𝑡〗^(−1 ) [((√(1 + sin⁡𝑥 ) )^2 + (√(1 − sin⁡𝑥 ) )^2+ 2(√(1 + sin⁡〖𝑥 〗 ))(√(1 −〖 sin〗⁡𝑥 )))/((√(1 + sin⁡〖𝑥 〗 )− √(1 − 〖 sin〗⁡𝑥 )) (√(1 + sin⁡〖𝑥 〗 )+ √(1 − sin⁡𝑥 )) ) ] = 〖𝑐𝑜𝑡〗^(−1 ) [((1 + sin⁡𝑥 ) + (1 − sin⁡𝑥 ) + 2√((1 + sin⁡𝑥 ) (1 − sin⁡𝑥 ) ))/((√(1 + sin⁡〖𝑥 〗 ))^2 − (√(1 − sin⁡𝑥 ))^2 ) ] = 〖𝑐𝑜𝑡〗^(−1 ) [(1 + sin⁡𝑥 + 1 − sin⁡𝑥 + 2√((1)^2 − sin^2⁡𝑥 ))/(1 + sin⁡𝑥 − 1 + sin⁡𝑥 ) ] = 〖𝑐𝑜𝑡〗^(−1 ) [(2 + 2√(1 − sin^2⁡𝑥 ))/(2 sin⁡𝑥 ) ] = 〖𝑐𝑜𝑡〗^(−1 ) [(2 (1 + √(𝟏 − 〖𝒔𝒊𝒏〗^𝟐⁡𝒙 ) ) )/(2 sin⁡𝑥 )] = 〖𝑐𝑜𝑡〗^(−1 ) [(1 + √(〖𝒄𝒐𝒔〗^𝟐⁡𝒙 ) )/sin⁡𝑥 ] = 〖𝑐𝑜𝑡〗^(−1 ) [(1 + 𝒄𝒐𝒔⁡𝒙 )/𝒔𝒊𝒏⁡𝒙 ] = 〖𝑐𝑜𝑡〗^(−1 ) [(1 + 𝟐 〖𝒄𝒐𝒔〗^𝟐⁡〖𝒙/𝟐〗 − 𝟏 )/(𝟐 𝒔𝒊𝒏⁡〖 𝒙/𝟐 〗 〖𝐜𝐨𝐬 〗⁡〖𝒙/𝟐〗 )] = 〖𝑐𝑜𝑡〗^(−1 ) [(2 cos^2⁡〖𝑥/2〗 )/(2 sin⁡〖 𝑥/2 〗 〖cos 〗⁡〖𝑥/2〗 )] = 〖𝑐𝑜𝑡〗^(−1 ) [(〖cos 〗⁡〖𝑥/2〗 )/sin⁡〖 𝑥/2 〗 ] = 〖𝑐𝑜𝑡〗^(−1 ) [cot⁡(𝑥/2) ] = 𝑥/2 We know that sin 2θ = 2 sin θ cos θ Replacing θ by 𝜃/2 sin θ = 2 𝒔𝒊𝒏⁡〖𝜽/𝟐〗 𝒄𝒐𝒔⁡〖𝜽/𝟐〗 and cos 2θ = 2cos2 θ – 1 Replacing θ by 𝜃/2 cos θ = 2cos2 𝜽/𝟐 – 1 Thus, 𝒚= 𝒙/𝟐 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = 𝑑/𝑑𝑥 (𝑥/2) 𝑑𝑦/𝑑𝑥 = 1/2 𝑑𝑥/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝟏/𝟐 Misc 6 (Method 2) Differentiate w.r.t. x the function, 〖𝑐𝑜𝑡〗^(−1 ) [(√(1 + sin⁡𝑥 ) + √(1 − sin⁡𝑥 ))/(√(1 + sin⁡〖𝑥 〗 ) − √(1 − sin⁡𝑥 )) ] , 0<𝑥< 𝜋/2 Let 𝑦= 〖𝑐𝑜𝑡〗^(−1 ) [(√(1 + sin⁡𝑥 ) + √(1 − sin⁡𝑥 ))/(√(1 +〖 sin〗⁡〖𝑥 〗 ) − √(1 − sin⁡𝑥 )) ] Finding √(𝟏 + 𝒔𝒊𝒏⁡𝒙 ) and √(𝟏 − 𝒔𝒊𝒏⁡𝒙 ) 〖𝒔𝒊𝒏〗^𝟐 𝜽+〖𝒄𝒐𝒔〗^𝟐⁡𝜽=1 Replacing 𝜃 by 𝑥/2 𝑠𝑖𝑛2 𝑥/2 + 〖𝑐𝑜𝑠〗^2 𝑥/2 = 1 𝒔𝒊𝒏⁡𝟐𝜽=2 𝑠𝑖𝑛⁡〖𝜃 𝑐𝑜𝑠⁡𝜃 〗 Replacing 𝜃 by 𝑥/2 𝒔𝒊𝒏⁡𝒙 = 2 𝑠𝑖𝑛⁡𝑥/2 𝑐𝑜𝑠⁡𝑥/2 √("1 + sin x" ) = √((〖𝑠𝑖𝑛〗^2 𝑥/2 + 〖𝑐𝑜𝑠〗^2 𝑥/2)" + 2 sin " 𝑥/2 " cos " 𝑥/2) = √((𝑐𝑜𝑠 𝑥/2 +sin⁡〖𝑥/2〗 )^2 ) = 𝒄𝒐𝒔 𝒙/𝟐 +𝒔𝒊𝒏⁡〖𝒙/𝟐〗 √("1 " −" sin x" ) = √((〖𝑠𝑖𝑛〗^2 𝑥/2 + 〖𝑐𝑜𝑠〗^2 𝑥/2)" − 2 sin " 𝑥/2 " cos " 𝑥/2) = √((𝑐𝑜𝑠 𝑥/2 −sin⁡〖𝑥/2〗 )^2 ) = 𝒄𝒐𝒔 𝒙/𝟐 +𝒔𝒊𝒏⁡〖𝒙/𝟐〗 √("1 " −" sin x" ) = √((〖𝑠𝑖𝑛〗^2 𝑥/2 + 〖𝑐𝑜𝑠〗^2 𝑥/2)" − 2 sin " 𝑥/2 " cos " 𝑥/2) = √((𝑐𝑜𝑠 𝑥/2 −sin⁡〖𝑥/2〗 )^2 ) = 𝒄𝒐𝒔 𝒙/𝟐 +𝒔𝒊𝒏⁡〖𝒙/𝟐〗 Thus, our equation becomes y = 〖𝑐𝑜𝑠〗^(−1) |(√(1 + sin⁡𝑥 ) + √(1 − sin⁡𝑥 ))/(√(1 + sin⁡𝑥 ) − √(1 − sin⁡𝑥 ))| Substituting value of √(1+𝑠𝑖𝑛⁡𝑥 ) & √(1−𝑠𝑖𝑛⁡𝑥 ) from (1) & (2). y = cot−1 [((〖cos 〗⁡〖𝑥/2〗 + 〖sin 〗⁡〖𝑥/2〗 ) + (〖cos 〗⁡〖𝑥/2〗 − 〖sin 〗⁡〖𝑥/2〗 ))/((〖cos 〗⁡〖𝑥/2〗 + 〖sin 〗⁡〖𝑥/2〗 ) + (〖cos 〗⁡〖𝑥/2〗 − 〖sin 〗⁡〖𝑥/2〗 ) )] 𝑦= 〖𝑐𝑜𝑡〗^(−1 ) [(𝑐𝑜𝑠⁡〖 𝑥/2〗 + 〖𝑠𝑖𝑛 〗⁡〖𝑥/2〗 + 𝑐𝑜𝑠⁡〖 𝑥/2〗 − 〖𝑠𝑖𝑛 〗⁡〖𝑥/2〗)/(𝑐𝑜𝑠⁡〖 𝑥/2〗 + 𝑠𝑖𝑛⁡〖 𝑥/2〗 − 〖𝑐𝑜𝑠 〗⁡〖𝑥/2〗 − 𝑠𝑖𝑛⁡〖 𝑥/2〗 ) ] 𝑦= 〖𝑐𝑜𝑡〗^(−1 ) [(2 〖𝑐𝑜𝑠 〗⁡〖𝑥/2〗 )/(2 𝑠𝑖𝑛⁡〖 𝑥/2〗 ) ] 𝑦= 〖𝑐𝑜𝑡〗^(−1 ) [𝑐𝑜𝑡 (𝑥 )/2 ] 𝒚= 𝒙/𝟐 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(𝑥/2)/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝟏/𝟐