Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 10

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 11

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 12 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 13 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 14 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 15 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 16

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Ex 6.5, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (iv) f (π‘₯)=sin⁑π‘₯ –cos⁑π‘₯, 0<π‘₯<2 πœ‹f (π‘₯)=sin⁑π‘₯ –cos⁑π‘₯, 0<π‘₯<2 πœ‹ Finding f’(𝒙) f’(π‘₯)=cos⁑π‘₯βˆ’(βˆ’sin⁑π‘₯ ) f’(π‘₯)=cos⁑π‘₯+sin⁑π‘₯ Putting f’(𝒙)=𝟎 cos⁑π‘₯+sin⁑π‘₯ = 0 cos⁑π‘₯=βˆ’sin⁑π‘₯ 1=(βˆ’sin⁑π‘₯)/cos⁑π‘₯ (βˆ’sin⁑π‘₯)/cos⁑π‘₯ =1 – tan π‘₯=1 tan π‘₯=βˆ’1 Since 0 < π‘₯ < 2Ο€ & tan π‘₯ is negative tan ΞΈ lies in either 2nd or 4th quadrant So, value of π‘₯ is π‘₯=3πœ‹/4 π‘œπ‘Ÿ 7πœ‹/4 Now finding f’’(π‘₯) f’’(π‘₯)=𝑑(cos⁑π‘₯ + sin⁑π‘₯ )/𝑑π‘₯ f’’(π‘₯)=βˆ’sin⁑π‘₯+cos⁑π‘₯ Putting 𝒙 = πŸ‘π…/πŸ’ f’’(3πœ‹/4)=βˆ’π‘ π‘–π‘›(3πœ‹/4)+π‘π‘œπ‘ (3πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹βˆ’πœ‹/4)+π‘π‘œπ‘ (πœ‹βˆ’ πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹/4)+(βˆ’π‘π‘œπ‘  πœ‹/4) =(βˆ’1)/√2βˆ’1/√2 =(βˆ’2)/√2 =βˆ’βˆš2 < 0 Hence f’’(π‘₯)<0 when π‘₯ = 3πœ‹/4 Thus π‘₯ = 3πœ‹/4 is point of local maxima ∴ f(π‘₯) is maximum value at π‘₯ = 3πœ‹/4 The local maximum value is f(π‘₯)=sin⁑π‘₯βˆ’cos⁑π‘₯ f(3πœ‹/4)=𝑠𝑖𝑛(3πœ‹/4)βˆ’π‘π‘œπ‘ (3πœ‹/4) =𝑠𝑖𝑛(πœ‹βˆ’πœ‹/4)βˆ’π‘π‘œπ‘ (πœ‹βˆ’πœ‹/4) =𝑠𝑖𝑛(πœ‹/4)βˆ’(βˆ’π‘π‘œπ‘  πœ‹/4) As sin (180 βˆ’ πœƒ) = sin πœƒ & cos (180 βˆ’ πœƒ) = βˆ’cos πœƒ =𝑠𝑖𝑛 πœ‹/4+π‘π‘œπ‘  πœ‹/4 =1/√2+1/√2 =2/√2 =√2 Now, for 𝒙 = πŸ•π…/πŸ’ f’’(π‘₯)=βˆ’sin⁑π‘₯+cos⁑π‘₯ Putting π‘₯ = 7πœ‹/4 f’’(7πœ‹/4)=βˆ’sin⁑(7πœ‹/4)+cos⁑(7πœ‹/4) f’’(7πœ‹/4)=βˆ’π‘ π‘–π‘›(2πœ‹βˆ’πœ‹/4)+π‘π‘œπ‘ (2πœ‹βˆ’πœ‹/4) =βˆ’(βˆ’π‘ π‘–π‘›(πœ‹/4))+π‘π‘œπ‘ (πœ‹/4) =𝑠𝑖𝑛 πœ‹/4+π‘π‘œπ‘  πœ‹/4 =1/√2 + 1/√2 =2/√2 =√2 > 0 f’’(π‘₯)>0 when π‘₯ = 7πœ‹/4 Thus π‘₯ = 7πœ‹/4 is point of local minima f(π‘₯) has minimum value at π‘₯ = 7πœ‹/4 As sin (2Ο€ βˆ’ πœƒ) = –sin πœƒ & cos (2Ο€ βˆ’ πœƒ) = cos πœƒ Local minimum value is f(π‘₯)=𝑠𝑖𝑛π‘₯βˆ’π‘π‘œπ‘ π‘₯ f(7πœ‹/4)=𝑠𝑖𝑛(7πœ‹/4)βˆ’π‘π‘œπ‘ (7πœ‹/4) =𝑠𝑖𝑛(2πœ‹βˆ’πœ‹/4)βˆ’π‘π‘œπ‘ (2πœ‹βˆ’πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹/4)βˆ’π‘π‘œπ‘ (πœ‹/4) =(βˆ’1)/√2 βˆ’ 1/√2 =(βˆ’2)/√2 =βˆ’βˆš2 Thus, f(π‘₯) is maximum at x = πŸ‘π…/πŸ’ and maximum value is √𝟐 & f(π‘₯) is minimum at x = πŸ•π…/πŸ’ and maximum value is β€“βˆšπŸ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.